THE NATURE of HEAT

As we know, heat is a form of energy. In the form of infrared radiation, heat from the sun travels through space at the speed of 186,000 miles per second. Upon arriving on earth, much of the radiant heat is absorbed by different kinds of matter and is converted into heat that we can feel (sensible heat). When you sit in the sun for a period of time on a clear spring day, you may find that your clothing and other objects around you become warm. Similarly, when you walk barefoot across a beach on a summer day, you may find the sand so hot that it burns your feet. In both cases, radiant heat from the sun has been absorbed by matter, and has been converted into heat that you can feel. In this chapter, we will study the effect of heat upon matter.

HEAT AND THE MOTION OF MOLECULES

Have you ever tried to drill a hole through a piece of metal? Both the drill and the metal become very hot.

Around 1800, an English scientist named Count Rumford noted that, when a drill was used to bore a cannon, the bit of the drill and the cannon both became very hot. To keep the metals cool, he placed a cylinder of water around the end of the cannon. As the boring continued, the water became warmer and eventually boiled. Since the bit of the drill and the cannon were cold at the start, Rumford concluded that the heat produced probably came from the friction created by the particles of the metal of the bit rubbing against the particles of the metal of the cannon. Further, he theorized that the motions of the particles in the metals themselves (atoms or molecules) generated the heat.

Recall that various forms of energy can be converted into other forms: When electrical energy passes through a thin wire, as in a toaster, the wire becomes hot (electrical energy to heat energy). When you rub your hands together, heat is produced from the friction between the rubbed surfaces (mechanical energy to heat energy). In general, when any form of energy is absorbed by matter, the energy is changed to heat. This may be explained by the kinetic-molecular theory: The energy excites the molecules in the matter, causing them to move faster and to collide more frequently. As more collisions take place, more heat is; produced.

The effect of heat energy on the motion of molecules can demonstrated by using a sealed tube containing a little mercury with some glass beads floating on the surface of the mercury. At ordinary temperatures, the glass beads merely float on the surface of the mercury. However, when the tube of mercury is heated, the glass beads, bounce up and down in a violent but random fashion. As still more heat is supplied to the sealed tube, particles of mercury begin to move more swiftly. The glass beads are repeatedly struck by many mercury particles at the same time; consequently, the glass beads begin to move randomly themselves. Thus, Rumford's theory that heat is related to the motions of molecules appears to be correct.

According to the kinetic-molecular theory, heat energy acquired by a body is transformed into increased kinetic energy of the molecules of the body. We observe this increased kinetic energy whenever a solid, a liquid, or a gas expands on heating. A further increase in kinetic energy will eventually cause the particles of a solid or liquid to become a gas.

Recall that when an ice cube (a solid) is heated, it melts and becomes liquid water. When the water is heated, it vaporizes and becomes gaseous water. According to the kinetic-molecular theory, as increasing amounts of heat are supplied to a piece of ice, the water molecules move more rapidly until they gain sufficient energy to overcome the attractive forces holding them together. This permits the ice to liquefy and become water. Similarly, as still more energy is received, the water molecules move at even greater speeds. The attractive forces in the liquid are weakened and the water is converted into gaseous water.

EXPANSION OF SOLIDS

Your laboratory experience with the ball and ring apparatus indicated the effect of heat on volume. The increase in size is not due to an increase in the size of the particles that make up the solid ball, but rather to an increase in the average distance between the particles. When an object is heated, its particles vibrate faster, collide more violently, and consequently move farther apart, thereby increasing the volume of the object.

When the object is cooled, the opposite change occurs and the volume of the object decreases. This decrease in volume is called contraction.

The expansion of solids by heating may cause serious practical problems. For example, the expansion of railroad tracks, bridges, or the concrete in a roadbed can create dangerous situations. Thus, allowance for the expansion of solids daring hot weather must he made in the construction of rails, bridges, and roads. For example, when rails arc laid, gaps between the ends of the rails provide for expansion. If this were not done, consider what would happen to the railroad tracks on a very hot day. The metal would expand, making the tracks bend and buckle, which might cause an oncoming train to be derailed. In bridge construction, expansion joints allow for changes in the length of the bridge. Concrete roadbeds are built with spaces between the sections of concrete to allow for expansion.

The contraction of solids, by cooling, may also present problems. Thus telephone and electrical wires are strung loosely to prevent their snapping as contraction takes place during the colder times of the year.

UNEQUAL EXPANSION OF SOLIDS

Through extensive studies, scientists have found that different metals expand at different rates when they are heated. For example, when a piece of iron and a piece of aluminum of equal size are heated together, we find that the aluminum expands more than twice as much as the iron. When two strips of different metals are fastened together, they form a compound bar, or bimetallic strip, which is employed in, useful devices such as thermostats and metallic thermometers. In these devices the two different metals, usually brass and steel, are welded together. When the bar is heated, it bends because the brass expands more than the iron and becomes longer than the iron.

Thus, the brass strip will be on the outside of the bend. As it cools, the bar returns to its original shape.

The thermostat is a device containing a compound liar that regulates the heating systems of our homes. When the temperature in the house falls below the setting on the thermostat, the compound bar, which contracts as it cools, closes the circuit, turning on the heat. As the room is warmed, the compound bar in the thermostat expands, bends, and thereby breaks the circuit, shutting off the heat.

The bimetallic thermometer is often used as an oven thermometer to indicate the temperature within an oven, or within a piece of meat that is cooking.

EXPANSION OF LIQUIDS

Liquids, like solids, expand when heated. In the laboratory experience we demonstrated that when water is heated, it expands. When the same water is cooled to its original temperature, the water contracts to its original volume. Many other liquids, such as alcohol and mercury, behave in the same way. At lower temperatures, however, the behavior of water is an exception to this rule. As water is cooled from 100° C to 4° C, it contracts-like other liquids do. However, when water is cooled below 4° C, the water expands-unlike other liquids. Water continues to expand until it reaches 0° C, its freezing point. It has been found, as shown in that the spaces between the water molecules in ice are larger than the spaces between the water molecules in liquid water. Ice is therefore said to have an open structure. Thus, as ice is formed, the need for increased space between the molecules causes the volume of the ice to be greater than that of the water from which it was formed. (This expansion in volume begins as liquid water is cooled below 4° C.) Since the volume of ice is greater than the volume of water from which the ice is formed, the density of ice is less than the density of water. (Recall that density equals weight divided by volume.) This is why ice floats on water.

Like solids, different liquids expand at different rates. As we will see later, the expansion of liquids is used in alcohol and mercury thermometers.

The expansion of liquids must be considered in certain heating systems. In a hot-water heating system, allowances must be made for the expansion of heated water. As the furnace heats the water in the heating system, the water expands. If the expansion continues, pressure would build up in the pipes and could damage the entire system. To avoid this difficulty, an expansion tank is provided. Excess heated water enters the expansion tank and thereby reduces the pressure in the system.

EXPANSION OF GASES

Cases, like solids and liquids, expand when heated. Our laboratory experience indicated that, as air is warmed, it expands. Scientists have made similar observations with other gases which indicate that gases confined in an elastic container expand when they are heated and contract when they are cooled.

The expansion of gases by heat must be considered by automobile tire. manufacturers, since tires may burst if allowed to remain in the sun indefinitely. A less serious hazard caused by expanding gases is that bottles of soda may crack or even explode if they are ex posed to heat for a considerable length of time.

Products such as whipped cream, shaving cream, deodorants, and insect repellents are now supplied in aerosol cans. These cans contain the product itself and a gas that forces the product out of the can when the valve, is open. When the product is used up, the can still contains unused gas. If this can is thrown into incinerator, the gas becomes heated, expands, and may cause the can to explode. Such aerosol cans should be discarded in a manner that does not involve heating.

Different solids and liquids expand at different rates when heated. Gases, however, generally expand at the same rate when heated to the same temperature, at a given pressure.

TEMPERATURE

Heat and temperature are two terms that are often confused. We know that the temperature of a small sample of molten iron is considerably higher that the temperature of the water in the ocean. However, the total heat in a sample of molten iron is much less than the total heat of the water in the ocean.

Scientist now accept Rumford’s theory that heat is related to the motions of particles in matter. Thus, heat depends on the total kinetic energy of the particles in a body. Recall the equation relating kinetic energy with mass and velocity:
K.E. = ½ m v². Thus, the total kinetic energy of the particles in a body depends on the number of particles (mass) and the velocity of these particles.

Because the water in the ocean is colder than the sample of molten iron, the velocity of the particles in the water is less than the velocity of the particles in the molten iron. However, the much larger quantity (mass0 of water compensates for the smaller velocity of the particles and thus the particles of water in the ocean possess greater kinetic energy. This means that there is more heat in the water in the ocean than in a small sample of molten iron.

But why is the temperature of molten iron higher? Temperature, unlike heat, depends on the average kinetic energy of the particles, that is, the kinetic energy per particle. To find this average, we divide the total kinetic energy by the number of particles. Thus, the large mass of ocean water has a smaller average kinetic energy per particle and consequently has a lower temperature than a small sample of molten iron.

MEASURING TEMPERATURE

Instruments designed to measure temperature are called thermometers. Most thermometers are based on the principle that matter, on heating, expands and, on cooling, contracts. In general, matter expands and contacts regularly. This means that the amount of expansion or contraction in length are generally equal for the same increase or decrease in temperature. This regular expansion and contraction has made it possible to construct three different types of thermometers: gas (air), liquid (mercury and alcohol), and solid (bimetallic) thermometers.

The gas (air) thermometer

In this thermometer, the glass bulb contains air. When the bulb is warmed, the air in the bulb expands and forces some of the colored water out of the tube. This changes the level of the liquid in the tube. By placing a suitable scale alongside the tube, temperature changes can be measured. Air thermometers of this type, while interesting, are not very accurate because the volume of a gas is also influenced by the air pressure around it. (Note that the flask contains a tube open at both ends. Why?).

LIQUID THERMOMETERS

Thermometers containing liquids such as mercury and alcohol are useful and accurate because these liquids usually expand and contract uniformly (regularly).

Mercury thermometers are made by filling a thin glass tube with mercury at a temperature greater than the maximum to be measured. The tube is then cut and sealed at the top. When the mercury cools, it contracts, leaving a partial vacuum above the mercury. (Liquids expand and contract to a much greater extent than do solids. Thus, in the given temperature range, the glass tube is scarcely affected by the temperature change.) The partial vacuum eliminates the effect of air resistance to the expansion of the mercury. The scale of the. thermometer is generally fixed by locating the boiling and freezing points of water on the scale. The distance between the boiling and freezing points is then divided into units depending on the temperature scale used. This will be discussed in the next section.

Since mercury freezes at -39° C, it cannot be used to measure very low temperatures. However, mercury boils at 357° C, which means that a mercury thermometer can be used to measure temperatures above the boiling point of water. On the other hand, alcohol freezes at -114° C. Accordingly, alcohol thermometers are used to measure low temperatures. However, since alcohol boils at 78° C, alcohol thermometers cannot be used to measure high temperatures.

SOLID (BIMETALLIC)

Recall that a bimetallic strip behaves as it does because different metals expand at different rates. Because most metals melt only at very high temperatures, a thermometer that uses a bimetallic strip can measure temperatures as high as 1000° C. The dial thermometer, used in most homes as an oven thermometer, is an example of a bimetallic thermometer. A curved bimetallic strip, with the faster-expanding metal on the outside of the bend, is attached to a pointer. Upon heating, the bimetallic strip moves, causing the pointer to indicate the temperature on a circular scale.

Other metallic thermometers, called resistance thermometers, use the principle that the resistance of a wire changes with temperature. Such thermometers also measure high temperatures.

THE FAHRENHEIT AND CELSIUS TEMPERATURE SCALES

Temperature markings on thermometers are indicated in Fahrenheit degrees or Celsius degrees. The Fahrenheit and Celsius scales are named after their originators, Gabriel Fahrenheit and Anders Celsius. (The Celsius scale is also called the centigrade scale.) Both Fahrenheit and Celsius scales are calibrated by using the boiling and freezing points of water. The Fahrenheit scale is used in the English system of measurement and the Celsius scale in the metric system. In the Fahrenheit scale, the freezing point of water is 32.° F, and the boiling point of water is 212° F. The remainder of the scale between these two points is marked off into 180 equal divisions (212 - 32 = 180) . In the Celsius scale, the freezing point of water is 0° C and the boiling point of water is 100° C. The remainder of the scale between these points is divided into 100 equal divisions (100 ‑ 0 = 100). Note that there are 180 divisions between the freezing and boiling points of water in the Fahrenheit scale and 100 divisions between these points in the Celsius scale. Thus, each Celsius division ( degree ) is 9/5 as large as Fahrenheit division. This relationship, together with the fact that there are 32 Fahrenheit divisions between 0 °F and 32° F, makes it possible to convert one scale into the other by using the following formulas:

° C = 5/9(° F – 32) ° F = 9/5 °C + 32

THE KELVIN SCALE

Confined gases, like most solids and liquids, expand and contract uniformly. For this statement to be true, however, a gas must be heated or cooled in such a way that the pressure remains constant. ( Recall that the air thermometer is inaccurate because it is affected by surrounding air pressure.) If we start at 0° C, we find that, for every Celsius degree rise in temperature, the volume of a gas increases 273 of its original volume ( provided the pressure does not change). Similarly, if we again start from 0° C, we find that for every Celsius degree drop in temperature, the volume decreases 273 of its original volume. At -273° C, the volume of a gas would shrink to zero and all molecular motion would cease. This, in turn, means that the gas would contain no heat.

(Actually, gases generally liquefy before this temperature is reached.) Scientists refer to -273°C as absolute zero, a temperature that has never been attained, although some scientists have come very close to this point.

Absolute zero, -273°C, is also called 0 Kelvin ( 0 K ). The Kelvin scale, named after its originator, Lord Kelvin, is based on absolute temperatures. Since the Kelvin scale begins with absolute zero (-273° C), we use the following formula to convert the Celsius scale to the Kelvin scale:

degrees Kelvin = degrees Celsius + 273 degrees

This formula may be written as K = ° C + 273

Let us find the freezing point of water (0° C) in the Kelvin scale: K=0 + 273=273 K

Thus, 0° C is equivalent to 273 K.

Now, let us find the boiling point of water: K =100 + 273 = 373 K

Thus, 100° C is equivalent to 373 K.

TRANSFER of HEAT

When a metal spoon is placed in a bowl of hot soup, the entire spoon soon becomes hot because the heat travels from the soup to the bowl-shaped part of the spoon, and then to the handle. When ice is placed in warm water, the ice soon melts. Both of these examples show that heat travels from one body to another. Generally, when objects are at different temperatures, heat is transferred from the warmer object to the cooler object until both objects are at the same temperature. Heat transfer can occur through one of three methods: conduction, convection, or radiation.

CONDUCTION

When one end of a metal rod is held in a flame, the entire rod will become hot enough to burn the hand. The heat from the flame reaches the hand by traveling through the rod. Substances that allow heat to travel through them are called conductors. In general, as we learned before, metals are good conductors. However, some metals conduct heat more readily than others. This can be demonstrated by inserting rods of aluminum, copper, iron, nickel, and brass into a brass sphere or disk and then attaching a small ball of wax to the end of each rod. When the center of the brass disk is heated, the wax at the tip of each metal melts in the order in which the different metals conduct heat. The wax at the tip of the copper melts first and the wax at the tip of the iron melts last.

Conduction in most materials can be explained by the kinetic-molecular theory. When one end of a rod is heated, the molecules in that end of the rod vibrate faster and strike other nearby molecules, causing them to vibrate faster. In this manner, the increased molecular motion is transferred from one end of the rod to the other, permitting the heat to travel through the rod.

Substances that do not readily allow heat to pass through them are called insulators. Gases and liquids are generally poor conductors of heat because their molecules are farther apart than are the molecules in solids. Therefore, neighboring molecules in a gas or in a liquid are less affected by the increased motions of heated molecules, and consequently heat is not conducted rapidly.

Substances like wood or plastic are poor conductors of heat, so they are used to make handles for metallic objects that are to be heated. The clothing we wear is also a poor conductor of heat, enabling us to retain body warmth. Porous material is generally non-conducting because it contains layers of trapped air which do not permit heat transfer.

CONVECTION

Although gases and liquids are poor conductors of heat, heat is transferred through them by the process of convection. Convection is the transfer of heat due to the motion of the liquid or gas itself. For example, when a beaker of water is heated the water layer closest to the heat source is warmed slowly by conduction. As the water becomes warmer, it expands, becomes less dense, and rises. This brings heat to the upper layer. At the same time, cooler water from the upper portion of the beaker moves down, takes the place of the rising water, and becomes heated itself. When warm enough, this water rises and carries heat upward. As these processes continue, heat that enters the bottom of the beaker is distributed throughout the beaker until all the water is at the same temperature. The moving water in such a case is said to have set up a convection current.

Heat is also transferred through gases by convection. It is by this means, in part, that a stove or a radiator heats a room. Heat from the radiator warms the air above it, causing the air to expand, become less dense, and rise. The cooler air that moves in to take the place of the warmed air is also soon warmed. As this air rises, a convection current is established. The convection current continues to distribute heat throughout the room until the entire room is warmed.

The formation of a convection current in air is demonstrated with a convection box apparatus. First the candle is lighted, then smoking touch paper is placed over the chimney, opposite the candle. The smoke, coloring the air, can be seen to move down this chimney, across the box, and out through the other chimney. This occurs because the air over the candle is heated, becomes less dense, and rises, leaving a partial vacuum. Cooler, more dense air from the first chimney moves in to fill the partial vacuum. This cycle continues as long as heat is given off by the burning candle.

RADIATION

We know that light energy and heat energy travel from the sun to the earth through space, which is an almost perfect vacuum. These forms of energy, traveling without the aid of molecular collisions, are transferred from the sun to the earth by radiation, that is, by means of rays, or waves. You can understand this method of heat transfer by standing a short distance from an open fire or by placing your hand a little to one side of, but not touching, a hot radiator. Since neither source of heat is being touched, you cannot receive heat by conduction. Since warm air rises vertically from the heat source, the heat cannot reach you by convection. The heat that is transferred to you from the fire or radiator reaches you by radiation.

The heat radiated by one body ( the sun, for example) is most rapidly absorbed by other bodies that are black in color and rough in texture. In warm climates, white clothing which reflects the radiant heat of the sun is cooler than dark clothing which quickly absorbs the radiant heat. Similarly, bodies that are rough and dark tend to radiate heat better than shiny smooth bodies. This is why steam radiators are often dark and have a roughened surface. It is for the same reason that coal burning stoves are black.

Bodies that are shiny and smooth do not absorb heat readily. Instead, these bodies reflect heat. Thus, aluminum used for roofing keeps homes cool in the summer and warm in the winter. This principle is utilized in the thermos (vacuum) bottle, which is so constructed as to permit liquids to retain their temperatures for a long time. A thermos bottle is double walled, with a partial vacuum between the walls to prevent heat transfer by conduction or convection. A cork stopper also prevents heat transfer by conduction. The inner glass walls are silvered to reflect radiant heat back into the liquid, thereby minimizing heat loss by radiation. Thus, a hot liquid remains hot because heat is lost very slowly. A cold liquid remains cold in thermos bottles because outside heat enters very slowly by conduction, convection, or radiation.

MEASURING HEAT

We learned that temperature is a measure of the average kinetic energy of the molecules of a substance. This is the same as saying that temperature represents the average intensity of the motion of the molecules, or the degree of hotness of a substance. Average kinetic energy means the total kinetic energy divided by the total number of particles. Recall that the ocean contains much more heat than does a small amount of molten iron. However, since the ocean contains many more particles than the molten iron, the temperature of the ocean (that is, the total kinetic energy divided by the total number of particles) is lower than that of the molten iron.

Temperature, therefore, does not tell us the quantity of heat present. The quantity of heat represents the total kinetic energy contained by all of the particles of the substance.

In the metric system, we measure the quantity of heat by a unit called the calorie. The calorie is the amount of heat needed to raise the temperature of 1 gram of water 1 Celsius degree. In the English system, heat is measured by a unit called the British Thermal Unit (BTU). A BTU is the amount of heat needed to raise the temperature of 1 pound of water 1 Fahrenheit degree. The amount of heat energy present in a substance cannot be measured directly with simple measuring devices. Instead, it is measured by observing its effect on a given quantity of water in a device called a calorimeter. One type of calorimeter, consists of two polished metal cups surrounded by air, a poor conductor of heat. An insulating cover, holding a thermometer, makes up the top of the calorimeter. The polished cups reflect heat, thus maintaining the temperature of the liquid in the container.

To determine the amount of heat energy absorbed (or lost) by a given quantity of water, we multiply the weight of the water in grams by the change in temperature of the water in Celsius degrees. Thus: amount of heat = weight of water X change in temperature

In a calorimeter, when 20 grams of water at 20°C are heated to a temperature of 30°C, how much heat is absorbed?

The temperature change = the final temperature - the initial temperature = 30°C - 20°C = 10 Celsius degrees. Substituting, amount of heat = 20 grams X 10 C° = 200 calories

We conclude that 200 calories have been absorbed. (We assume that no heat has escaped from the calorimeter.)

HEAT EXCHANGE IN WATER

In a calorimeter, when a quantity of water at a given temperature is mixed with a quantity of water at a different temperature, the amount of heat lost by the "hot" water is equal to the amount of heat gained by the "cold" water.

Suppose we mix 100 grams of water at 90° C with 100 grams of water at 40° C and find that the final temperature of the mixture is 65° C. Let us calculate the number of calories lost. by the hot water and gained by the cold water.

The temperature of the hot water dropped from 90° C to 65° C, a decrease of 25 Celsius degrees. Since we began with 100 grams of hot water that underwent a temperature change of 25° C, we determine the amount of heat lost:

amount of heat = weight of water X change in temperature
(calories) (grams) (Celsius degrees)

amount of heat = 100 grams X 25° C amount of heat = 2500 calories

The minus sign in the answer indicates that 2500 calories of heat have been lost.

The temperature of the cold water increased from 40° C to 65° C, an increase of 25 Celsius degrees. Since we began with 100 grams of cold water that underwent a temperature change of 25° C, we determine the amount of heat gained:

amount of heat = weight of water X change in temperature
(calories) (grams) (Celsius degrees)

amount of heat = 100 grams X 25° C amount of heat = 2500 calories

Note that the amount of heat lost by the hot water (2500 calories) is the same as the amount of heat gained by the cold water (2500 calories). We assume that the heat exchange was "perfect" and that no heat escaped from the calorimeter.

The quantity of heat needed to raise the temperature of 1 gram of a substance 1 Celsius degree is called the specific heat of the substance. For water, the specific heat is 1. This means that 1 calorie of heat will raise the temperature of 1 gram of water 1° C. Water is the only substance for which this is true. Other substances vary in the quantity of heat needed to raise 1 gram of the substance 1 Celsius degree. Consequently the formula

amount of heat =weight of water X change in temperature applies only to water.

CALORIES AND FOOD

Your body requires energy in order to perform its daily tasks. Most of this energy comes from energy‑rich foods such as carbohydrates and fats. This energy is released when the body utilizes these foods. Using special calorimeters, scientists have measured the energy content, or the number of calories present, in fixed quantities of certain foods. For example, a slice of white bread contains about 60 000 calories; a typical chocolate bar may contain about 300 000 calories.

Nutritionists use a special kind of notation when discussing the calorie content of foods. They define a food Calorie ( written with a capital letter) as 1000 calories. On a calorie table, therefore, we would read that a slice of white bread contains about 60 Calories and that a chocolate bar contains about 300 Calories.

This information comes from "Physical Science Workbook", 1961

The History of Heat

Aristotle and the Greeks had their idea of fire as one of the 4 Primal Elements. Even the ancients realized that heat and light were not alike as aspects of fire, though. After the fire had gone out and the light gone, the heat of the kettle and its contents remained.

First modern chemist to study heat was Joseph Black (1728 - 1799). Black tried to explain heat in terms of a fluid. He explained how a kettle of water placed over a fire increased in temperature but a kettle filled with water and ice placed over a fire did not change in temperature till all the ice was melted. He said that until the ice was saturated with the heat-fluid and thus became melted could its temperature rise. Lavoisier accepted this theory and gave the name for this heat-fluid “caloric” from the Latin word for heat.

Another idea competed with the caloric theory. Scientist knew that kinetic energy of motion plus the stored energy called potential energy was given the name mechanical energy and that friction was a part of the conservation of these energies. They knew friction could warm up an object so maybe the invisible motion of invisible particles was what we call heat. Summed up; friction was converting mechanical energy into heat. The problem was this idea of really small particles of matter (i.e., atoms and molecules).

Count Rumford (really Benjamin Thompson - a spy for the British authorities during the Revolutionary War) was to supervise the boring of cannon for Bavarian army. Using a horse to work a treadmill he realized that the solid block of brass grew hot as the borer cut its way in. Rumford calculated that if the caloric theory were correct the heat released during the boring would have melted the entire block of metal first. He pointed out that heat was produced without fire, without light, without chemical combustion. It came just out of motion.

John Dalton comes along with his ideas of atoms and the kinetic energy idea of heat began to gain favor. An English physicist, James Prescott Joule (1818-1889) was attempting to find the mechanical equivalent of heat. In the end he found that a given amount of energy of whatever form always yielded that same amount of heat (at 4.18 joules per calorie). The relationship of the motion of atoms to temperature and heat was placed on firm theoretical basis about 1860 by the Scottish physicist James Clerk Maxwell.

Specific Heat

The ability of water to stabilize temperature depends on its relatively high specific heat. The specific heat of a substance is defined at the amount of heat that must be absorbed or lost for 1 g of that substance to change its temperature by 1º C. The specific heat of water is 1.00 cal/g ºC. Compared with most other substances, water has an unusually high specific heat. For example, ethyl alcohol, the type in alcoholic beverages, has a specific heat of 0.6 cal/g ºC.

Because of the high specific heat of water relative to other materials, water will change its temperature less when it absorbs or loses a given amount of heat. The reason you can burn your finger by touching the metal handle of a pot on the stove when the water in the pot is still lukewarm is that the specific heat of water is ten times greater than that of iron. In other words, it will take only 0.1 cal to raise the temperature of 1 g of iron 1ºC. Specific heat can be thought of as a measure of how well a substance resists changing its temperature when it absorbs or releases heat. Water resists changing its temperature; when it does change its temperature, it absorbs or loses a relatively large quantity of heat for each degree of change.

We can trace water’s high specific heat, like many of its other properties, to hydrogen bonding. Heat must be absorbed in order to break hydrogen bonds, and heat is released when hydrogen bonds form. A calorie of heat causes a relatively small change in the temperature because must of the heat energy is used to disrupt hydrogen bonds before the water molecules can begin moving faster. And when the temperature of water drops slightly, many additional hydrogen bonds form, releasing a considerable amount of energy in the form of heat.

What is the relevance of water’s high specific heat to life on Earth? By warming up only a few degrees, a large body of water can absorb and store a huge amount of heat from the sun in the daytime and during summer. At night and during winter, the gradual cooling water can warm the air. This is the reason coastal areas generally have milder climates than inland regions. The high specific heat of water also makes ocean temperatures quite stable, creating a favorable environment for marine life. Thus, because of its high specific heat, the water that covers most of planet Earth keeps temperature fluctuations within limits that permit life. Also, because organisms are made primarily of water, they are more able to resist changes in their own temperatures than if they were made of a liquid with a lower specific heat.

Water is one of the few substances that are less dense as a solid than as a liquid. While other materials contract when they solidify, water expands. The cause of this exotic behavior is, once again, hydrogen bonding. At temperatures above 4º C, water behaves like other liquids, expanding as it warms and contracting as it cools. Water begins to freeze when its molecules are no longer moving vigorously enough to break their hydrogen bonds. As the temperature reaches 0º C, the water becomes locked into a crystalline lattice, each water molecule bonded to the maximum of four partners. The hydrogen bonds keep the molecules far enough apart to make ice about 10% less dense than liquid water at 4º C. When ice absorbs enough heat for its temperature to increase to above 0º C, hydrogen bonds between molecules are disrupted. As the crystal collapses, the ice melts, and molecules are free to slip closer together. Water reaches it greatest density at 4º C and then begins to expand as the molecules move faster.

The ability of ice to float because of the expansion of water as it solidifies is an important factor in the fitness of the environment. If ice sank, then eventually all ponds, lakes, and even the oceans would freeze solid, making life as we know it impossible on Earth. During summer, only the upper few inches of the ocean would thaw. Instead, when a deep body of water cools, the floating ice insulates the liquid water below, preventing it from freezing and allowing life to exist under the frozen surface.

Metal	Specific Heat	Thermal Conductivity	Density	Electrical Conductivity
	c_p cal/g° C	k watt/cm K	g/cm³	^1E6/Ωm
Brass	0.09	1.09	8.5
Iron	0.11	0.803	7.87	11.2
Nickel	0.106	0.905	8.9	14.6
Copper	0.093	3.98	8.95	60.7
Aluminum	0.217	2.37	2.7	37.7
Lead	0.0305	0.352	11.2

Heat and Temperature Teaching Notes

1) Heat flows from hot to cold areas due to a temperature difference only.

example: A small hot block of a material is placed next to a larger, cooler block. Heat flows from the small hot block to the larger block till equilibrium is reached.

2) Note the difference between the heat content and temperature. A lake may be cooler in temperature than a liter flask of water but the lake has a much greater heat content due to the vast number of particles and their associated motion.

3) Human perception of heat and temperature is not adequate for scientific work. So we must investigate tools that provide the accuracy and repeatability we need.

4) For temperature we will be using thermometers. There are other devices that allow you to measure temperature.

5) For heat we will be working with simple calorimeters. We will look at these devices when we look at the transfer of heat.

Thermometers and Temperature Scales see comparisons charts

thermometers = instruments to measure temperature

see drawing: gas (air) thermometers

see drawing: liquid (Hg and alcohol)

see drawing: solid (bimetallic)

Mercury thermometer -- fill thin glass tube with Hg at a temp. greater than the maximum to be measured; tube is cut and sealed; the Hg cools and contracts leaving a partial vacuum above the Hg (eliminates effects of air resistance on expansion of Hg); then calibrate thermometer

Hg freezes at -39° C so it cannot be used for low temperatures (use alcohol which freezes at -114° C)

can use Hg for high temperature boils at 357° C (alcohol cannot be used for high temperature work due to its low boiling point - 78° C)

use alcohol thermometers in schools unless extreme accuracy is needed due to safety factor. The alcohol may be inaccurate by 1 - 2 degrees but this may not be a problem if you are looking at changes in temp.

Calibration

Both Celsius (Anders Celsius) and Fahrenheit (Gabriel Fahrenheit) scales are established by using the boiling and freezing point of water at 1 atmosphere of pressure.

step 1 -- establish a mixture of ice and water in equilibrium (0° C) mark point of liquid in thermometer at 0° C

establish a mixture of steam and water at equilibrium (both at a pressure of 1 atm.) steam condensing and water vaporizing)

label this point of liquid as 100 °C

step 2 -- divide the interval between 0° C and 100° C into 100 equal parts, each representing a change in temperature of 1° C.

using this scale you can extend your marks below 0 °C and above 100 °C as far as ,you wish.

The Fahrenheit scale labels the freezing point at 32°

(his label for the temperature he could achieve with an ice and water mixture and labeled the boiling point temperature of water at 212 ° which was a number chosen for convenience apparently creating 180 divisions.

You might ask your self about the amount of heat energy need to cause a 1 degree change in temperature on a Celsius scale compared to that needed on a Fahrenheit scale. (more heat needed to cause change of 1 degree on Celsius scale.)

KELVIN scale

We know that gases decrease in volume 1/273 of its original volume for each degree drop in temperature. Thus at -273° C the volume of gas would shrink to zero and the gas would have no molecular motion. We know this is impossible (particles have zero-point energy). To label these very low temperatures a scale called the absolute or Kelvin scale is often used. It designates -273° C at the zero point, and is called called absolute zero.

Extrapolation to absolute zero: A good research project for students or a quick review of graphing can be done by using the method to calculate absolute zero. Use a capillary tube with a trapped bubble of air between light machine oil. Measure the length of the bubble after placing it in different temperature mixtures. Plot the length versus temperature.

Since it is not easy to obtain very cold temperatures, the linear series of points that you did obtain should allow you to extrapolate, (extend a curve beyond the known data points following the apparent pattern of the curve) until it intersects the temperature axis. See actual plot!

Transfer of Heat by Conduction, Convection, Radiation

Conduction is a consequence of the kinetic behavior of matter. Faster vibrating particles collide with less energetic neighbors and transfer some of their kinetic energy to the slower moving particle.

Through successive molecular collisions energy travels through a material without the average position of the particles being changed. There must be a temperature differential (one end of some object at a higher temperature than the other) for heat to be conducted.

Gases are poor conductors of heat (compared to liquids and solids) because the molecules are relatively far apart and collisions are infrequent.

Metals have the greatest ability to conduct heat (for the same reason as their high electrical conductivity). This is due to a significant number of electrons being able to move about freely instead of being bound permanently to particular atoms.

Thermal conductivity of a material is a measure of its ability to conduct heat.

Example: wood and metal (see class discussion)

Convection involves the actual motion of a hot fluid from one place to another, displacing a colder fluid in its path and setting up a convection current. Convection is the chief mechanism of heat transfer in fluids.

Natural convection occurs when the buoyancy of heated fluids leads to motion. Heated fluids (gas or liquid) expand and becomes less dense than surrounding cooler fluids. It then rises.

Radiation is defined as the energy that is transmitted by electromagnetic waves and requires no material medium for passage.

All objects radiate electromagnetic waves with the higher the temperature of an object the shorter the predominating wavelength of its radiation.

Example: see glass lined thermos bottle in heat packet in class.

Transfer of Heat

	Conduction -- place iron rod in fire -- the end you are holding becomes warm due to conduction
	Convection -- stove heats room by convection
	Radiation -- heat the earth receives from the sun is radiation

Natural direction of heat flow is from hot bodies to cold ones.

Conduction -- conduction is a consequence of kinetic behavior of matter

faster vibrating particle collide with less energetic neighbor and transfer some of their kinetic energy the slower moving particle

example: place hand on wood and metal sample at same temperature. The metal will seem cooler because it conducts the heat away much faster than the wood

Convection -- actual motion of hot fluid from one place to another, displacing cold fluid in its path setting up a convection current = chief mechanism of heat transfer in fluids in most instances

natural convection - the buoyancy of heated fluids leads to motion ‑ heated fluid (gas or liquid) expands and becomes less dense than surrounding cooler fluids and rises

Radiation -- energy that is transmitted by electromagnetic waves and requires no material medium for passage

all objects radiate electromagnetic waves but the higher the temperature of an object the shorter the predominate wavelength of its radiation

There are many examples you can use to demonstrate these three ideas but discussing a glass lined thermos bottle will allow you discuss them as well as

High specific heat capacity material demonstrates relatively small change in temperature for a given change in internal energy content

add 1 calorie of heat to 1 gram of water, helium, ice, gold

temperature rises: water 1 °C

helium 1.3 °C

ice 2 °C

gold 33 °C

Calorimetry:

1) Use 2 polystyrene cups (one within the other) -- the polystyrene will not absorb very much heat.

2) You may find if difficult for the students to be patient when working with the calorimeters. Your step-by-step procedure must be very simple and clear. This type lab is also a very dangerous time for your thermometers.

3) Several labs have been included in the packet involving the use of the calorimeters and thermometers: a) the heat of reactions and heat of solutions labs deal with endothermic and exothermic reactions (and you can incorporate the use of the mole as review) b) the specific heat capacity lab is an excellent demonstration of the principle of heat exchange as well as specific heat capacity of different metals and liquids. 1) for better results with this lab try to use as much metal as possible and as little fluid as possible and still cover the metal in the cups. Drain the metal samples quickly after removing them from the boiling water. This is a good time to have the students watch the boiling process which will be one of the final topics in this unit. 2) the math may be a bit difficult for you at first.

Thermal Expansion of water

1) From 0° C to 4° C the volume of water in a sample decreases (the greatest density is at 4° C)

2) We know that ice floats (less dense) so that a body of water in winter freezes from the top down. The ice is a poor conductor of heat so that the initial layer of ice that freezes impedes further freezing allowing fish and plant life to live through the winter.

3) The spaces between molecules in ice are greater than the same spaces in liquids.

4) Ice has what is called an Open Structure --> each water molecule can participate in 4 bonds with other water molecules, while other solid molecules can have as many as a dozen bonds with surrounding molecules resulting in a more compact substance.

5) As stated the density of the water increases from 0 °C to 4 °C. Large clusters of water molecules break into smaller clusters that occupy less space in the aggregate as the temperature rises to 4 °C. Only above 4 °C does the normal thermal expansion show a decreasing density with increasing temperature. Above 4° C , the normal thermal expansion of materials is seen. Here as the temperature rises the density decreases.

Heat and Temperature

HEAT Heat is a form of internal energy which is transferred from one object to another due to a difference in temperature between the objects. Heat is the total energy of motion of all particles (the total kinetic energies of all the particles.)

TEMPERATURE The temperature of a body of matter is a measure of the average kinetic energy of the random motion of its particles. Temperature is the kinetic energy divided by the number of particles. Temperature is that property of a substance which determines whether it is in thermal equilibrium with another object.

Thermal Conductivity - a measure of the ability of a substance to conduct heat

THERMAL EQUILIBRIUM This is the situation in which no heat moves from one object to another.

CALORIE A 15° Calorie is the amount of heat energy needed to change the temperature of of 1 gram of water by 1° C (from 14.5° C to 15.5° C at 1 atmosphere of pressure). 1 calorie = 4.185 Joules and 1 kilocalorie = 1000 calories.

SPECIFIC HEAT CAPACITY This is the amount of heat (in calories or Joules) that must be added or removed from a unit mass of that substance to change its temperature by one degree. Different substances have different capacities because they absorb and release heat at different rates.

WATER Water has a specific heat capacity of 1.00 cal/g °C or 4.185 Joules/g °C. The SI unit would be 4185 J/kg °C.

PRINCIPLE OF HEAT EXCHANGE The heat lost by an object must equal the heat gained by the object to which the heat is transferred. There must be a temperature difference for heat to be transferred.

Q (heat energy) = m (mass) x At (temp.) x c_p (specific heat capacity)
(cal/Joules) (g) (°C) (cal/g °C) or (Joule/g °C)

Problems:
1) How much heat energy is needed to raise the temperature of 100 grams of water from 0 degrees to 30° C?

2) A calorimeter contains 300 grams of water at 10° C. After a food sample is burned in the calorimeter the water temperature changes to 15° C. How much heat was given off by the food sample?

LATENT HEAT Latent heat is the heat required to bring about a change in state.

HEAT OF FUSION The heat of fusion is the amount of heat that must be supplied to change a unit mass of the substance at its melting point from solid to liquid. The heat of fusion of water is 80 calories per gram (80 kcal/kg).

HEAT OF VAPORIZATION The heat of vaporization is the amount of heat that must be supplied to change a unit mass of the substance at its boiling point from liquid to gas or vapor state. For water this is 540 cal/g or 540 kcal/kg.

HEAT OF SUBLIMATION The heat of sublimation is the heat needed to change a solid to a gas.

HEAT OF CONDENSATION The heat of condensation is the reverse of the heat of vaporization, it is the heat given off when a gas condenses to a liquid.

Four States of Matter

Matter is defined as any material that has mass, occupies volume, and exhibits inertia (resistance to movement).

Solids definite shape and volume, resist deformation

	very close spacing of particles that make up the solid
	these particles appear to vibrate about fixed points
	particles vibrate faster at higher temperatures, slower at lower temperatures

Crystalline solids particles are arranged in regular, repeated patterns - said to have "long-range order" to their structure -example would be NaCl (table salt)

Amorphous solids solids that lack the definite arrangement in crystals are `amorphous' which means `without form'. Can be thought of as liquids whose stiffness is due to exaggerated viscosity. These solids are said to have "short range order". Examples are pitch, glass, plastics. Polymers are flexible and some will change their structure when undergoing a physical change (examples are rubber bands, Saran Warp, Lucite, DNA, fats, cellulose, glycogen. Jello is a natural glucose polymer.

Liquids definite volume, resist compression, will flow, takes the shape of its container

greater spacing between molecules, liquid particles appear to travel in straight line paths between collisions but appear to rotate or vibrate about moving points

Gases Have no definite shape or volume, takes the shape and volume of its container

can be compressed or dispersed, the particles vibrate very rapidly, are relatively far far apart, and there are no forces holding them together

Plasma very high temperature ionized gas (as high as 100 million degrees in some fusion reactors). These plasmas have no fixed volume or shape, most are mixtures that are not easily containable. They all have particles that are electrically charged and of low density. The Milky Way is a huge plasma.

Energy Definitions

Energy: having the ability to do work (move matter)

Work: a push or pull over some distance (force x distance)

Force: a push or a pull

Potential Energy:   the energy a body possesses by virtue of its position, composition, and or condition
                                stored energy or energy of position
                                P.E. = mass x gravity x height
                                examples: water behind a dam, stretched/compressed spring, explosives

Kinetic Energy:   the energy of motion (conserved in every elastic collision)
                                K.E. = 1/2 mass x velocity²
                                heat energy flows from hot objects to cooler ones through transfer of K.E. when particles collide

Momentum: mass time velocity (momentum is conserved in every collision where there is no friction)

Linear momentum of a moving body is a measure of its tendency to continue in motion at a constant velocity. The conservation of linear momentum states that in the absence of forces from outside the system the total momentum of colliding particles cannot change but the distribution of the total momentum may change. Momentum is redistributed in a collision.

Intermolecular Forces:

	potential energy forces that hold molecules together and in correct position in solids
	potential energy forces that hold molecules together in liquids
	the kinetic energy of the molecules in solids and liquids cannot overcome intermolecular forces holding the molecules together (so they do not fly apart)
	gas molecules have enough kinetic energy to break free from intermolecular forces or to keep such forces from forming

Kinetic – Molecular Theory of Gases

	gases are made up of molecules that are in continuous motion
	an increase in the temperature increases the speed of the molecules, thus increasing the kinetic energy of the substance
	All gases are compressible
	Gases display diffusion (random movement of molecules from one area to another with a net change in concentration – rate varies with temperature and molecular mass)
	Gases can be liquefied (called liquefaction)

Closed System Criteria

In using the above information we look at pressure, temperature, and volume in a closed system.

In a closed system nothing escapes or is allowed in (unless we choose to allow it)
all molecules are in motion (have K.E.)
molecules exert a uniform pressure on all surface areas of the walls of the container
Pressure = force/area (see examples given in class)
Atmospheric pressure is the cumulative effect of the force generated by the weight of the atmosphere. Given values that must be used in problems include: 14.7 lb/in², 101.3 kPa, 1 atmosphere, 760 mm of Hg, 1 033.6 g/cm²
Molecules exert pressure on other molecules inside container as they collide, push, and bounce off other molecules
The pressure a gas exerts on the walls of its container is the sum of the forces acting on the walls (equals the frequency of collisions with the walls of the container plus the force of each molecule as it pushes against the wall) due to the random collision of limitless numbers of these moving molecules.

Collisions that occur between molecules are perfectly elastic, the particles bounce off each other and exchange energy, but there is no loss of energy
* elastic atomic collisions: atoms (molecules) bounce back as far/fast as it would have had it not collided (no change in the total kinetic energy of the two particles before and after the collision)
* inelastic collisions: the normal order in which the objects lose energy and slow down

Momentum is conserved in every collision where there is no friction, energy is conserved only in elastic collisions.

Gas Laws

J.L. Gay-Lussac’s Law If the volume remains constant, the pressure is directly proportional to the absolute temperature:

P ~ T P₁/ T₁ = P₂ / T₂

Boyle’s Law If the temperature remains constant, the volume of a gas varies inversely with the pressure:

V ~ 1/P P₁ V₁ = P₂ V₂

Charles’ Law If the pressure is kept constant, the volume of a gas is directly proportional to its absolute temperature:

V ~ T V₁/ T₁ = V₂ / T₂

For each degree increase in temperature, the volume increases 1/273 of its original volume

Combined gas law:

P₁V₁ / T₁ = P₂ V₂ / T₂

5. Ideal Gas Law:

PV = nRT

Overall conclusions:

	The temperature of a gas increases when it is compressed because the average energy of its molecules increases. The molecules rebound from the inward moving piston, traveling faster than before hitting the piston.
	Molecules rebounding from fixed walls have unchanged speeds.
	The temperature of a gas decreases when its volume is expanded because the average energy of its molecules decreases. The molecules rebounding from outward moving piston move slower than before.

Gas Law Problems:

1) An insulated system is known to have a temperature of 100.0° C at a pressure of 4.00 atm. If the absolute temperature is cut in half, what will be the new:

___________ atm, __________kPa, ______________° C, _______________ K

2) The volume is given as 27.0 L. If the pressure goes from 3.00 atm. to 9.00 atm., what is the new:

___________L, _____________ kPa

3) The volume is given as 5.00 L. If the absolute temperature goes from 273 to 819 K, what is the the new __________L (if new temperature was 800. K, what is the new volume in liters?)

4) The temperature is given as 25.0° C. If the volume is decreased from 100. mL to 10.0 mL, what is the new: ____________K, ____________° C

Thought Work -- Heat & Gas Laws

1. Devise a way to remove carbon dioxide (carbonation) from soda pop. This must be done quantitatively. How does temperature affect the solubility of gases being dissolved in liquids under pressure?

2. Explain what happens to a marshmallow when it is toasted. Why does this happen? What would happen to a frozen marshmallow? Why does a marshmallow float?

3. Find the pressure you exert when standing on both feet, on one foot, and lying flat on your back.

4. Explain why a toy balloon filled with hydrogen partially deflates overnight.

5. Using a steel ball and pieces of old pottery or modeling clay, devise an experiment that would demonstrate potential energy, kinetic energy, and momentum (all of which are involved with mass and velocity).

6. Suppose you had two identical sections of glass plate before you, one heated above body temperature and the other cooled by ice. What happens when you breathe on the two of them and why? Maybe try this at home first.

7. Devise an experiment to show the concept of diffusion, another to show cohesion, adhesion, and surface tension, and one to show buoyancy and Pascal's Law.

8. Changing ice to water requires 80 calories per gram of ice, but changing water to steam requires about 540 calories per gram of water. What does this tell you about the intermolecular forces in ice and water, both qualitatively and quantitatively? Also explain why the chemical change of splitting or forming water requires about 5 times as many calories as the physical change of state.

Chemical Properties of Matter

Chemical properties are those properties of a substance that can be determined by a chemical test. Chemical properties are seen by the material's tendency to change, either alone or by interaction with other substances, and in doing so form different materials.

1.does the substance support combustion: examples are O₂ and Cl₂

2. does the substance burn (combustibility)

3. how does the substance react with acids (does it dissolve, evolve gases, explode, do nothing)

4. how does the substance react with oxygen (burn, form new compounds)

5. what is its reaction with electricity (usually it will be separated into simpler components)

examples: alcohol burns, iron rust, wood decays, sodium explodes in water

Physical Properties of matter

Physical properties are those properties used in identifying substances when we use our senses. These do not require chemical analysis.

1. color - reaction of eye and brain in recognizing combinations of certain wavelengths of visible light.

2. hardness - a measure of the ability of a substance to resist abrasion (see Mows Scale of Hardness)

3. density - the mass divided by its volume (often reported as specific gravity which is a unitless relationship between the density of the substance and the density of water)

4. texture - how object feels to touch; usually rough or smooth

5. magnetic attraction - is the material attracted to a magnet or can it be magnetized (must contain Fe, Co. Ni, or steel)

6. solubility - the amount of a substance which will dissolve in a known amount of solvent at a given temperature

7. taste - reaction of taste buds to stimuli along with the brain's recognition of the pattern

8. light transmission - is the substance transparent, translucent, or opaque

9. viscosity - a measure of the internal resistance (friction) to flow in a liquid (molasses and tar would have high viscosities)

10. refractive index amount a ray of light is bent as it passes through a substance (technically the ratio of the speed of light in that substance to the speed of light in a vacuum)

11. specific heat capacity - the amount of heat energy (calories or Joules) required to change the temperature of 1 gram of a substance by 1 °C

12. atomic radius - the distance from the center of an atom's nucleus to the outermost orbital electron

13. boiling point - the temperature at which the liquid's vapor pressure equals atmospheric pressure during the boiling of a pure substance the temperature remains constant as long as both liquid and vapor are present

14. melting-freezing point - temperature at which solid-liquid phase is in equilibrium - during melting of pure solid the temperature remains constant; when all solid is melted and only liquid is present, further heating results in a steady increase in temperature to the boiling point

15. odor - olfactory nerves are stimulated by certain molecule and send messages to the brain which remembers the pattern

16. expansion - contraction coefficients - materials expand or contract a known amount when heated or cooled

Collapsing Can Demo

area of surface of Coke can = 0.031 m²

pressure on this area = 3.1 E 3 N (about 680 lb)

1 atm = 1.0 E 5 N/m²

if we can reduce pressure by as little as 75% there would be a 500 lb difference between pressure inside and outside the can

1) when can is inverted in water bath - the water seals the opening and cools the can

2) as can cools vapor condenses - reduction of pressure inside can

3) can is sufficiently weak and water sufficiently viscous that can collapses before it fills with water

4) must use all aluminum can

Physical Changes in State

	a change in the physical properties of a substance without a change in the chemical composition
	the arrangement of molecules may be changed but the molecular make-up remains the same
	these changes deal with intermolecular forces which increase or decrease during the change.

Problems:

1. 72 grams of ice + 51 840 calories yield 72 grams of water vapor. How many calories must be removed from water vapor to condense it back to ice?

2. If 1 gram of water at room temperature evaporates, about 600 calories are taken from the surroundings to convert the liquid to a gas. How many calories are `needed' to change 1001 grams of gas to a liquid?

3. If 50 grams of water vapor loses 36 000 calories in turning to ice, how many calories would 1 gram of water vapor have to lose to be turned back to ice?

ice (0° C) + heat à water vapor (100° C)

36 g 25 920 cal 36 g

water vapor (100° C) à ice (0° C) + heat

36 g 36 g 25 920 cal

2 H₂ + O₂ à 2H₂O + heat energy released

4 g 32 g 36 g 136 600 cal

2H₂O + energy à 2H₂ + O₂

36 g 136 600 cal 4 g 32 g

Chemical Changes in State (Phase)

The molecular make-up (the specific arrangement of atoms) is changed, resulting in new substances being formed and energy changes occurring.

EXOTHERMIC - any chemical change that releases energy is exothermic

	the amount of heat released is greater than the amount of heat used to start the reaction
	bond making is exothermic (energy is released into surroundings)
	example: oxidation à wooden splint burning ( heat, light, gases like CO₂ and H₂O being given off with carbon and ashes left over)
	other examples: burning H₂ in O₂, body reactions, dissolving metals in strong acids, mixing acid and water, homogenization, plaster of Paris in water, sugar dehydration

ENDOTHERMIC - any chemical change that absorbs energy is endothermic

	energy continues to be absorbed as long as the reaction continues
	bond breaking is endothermic (energy is absorbed from surroundings)
	example: electrolysis à splitting some compound (usually water) by running an electric current through it
	other examples: photosynthesis, pasteurization, canning vegetables

Sugar dehydration demo here

The chemical change involving splitting or forming water takes about 5 times as many calories as the physical change of state. The reason is that atoms (or molecules) are bonded together in a compound; the stronger the bond the more energy holding the parts together, thus more energy required to break these bonds. A physical change needs far less energy to overcome intermolecular forces holding groups of molecules together. Much more energy is needed to break bonds within molecules than to overcome the forces between molecules.

physical change -- strength of intermolecular forces increased or decreased

chemical change -- bonds formed or broken

energy absorbed -- bonds broken or intermolecular forces overcome

energy released -- bonds formed or intermolecular forces strengthened

Problems:

Tell whether each of the following is a chemical or physical change and further describe each chemical change as endothermic or exothermic and the physical changes as absorbing or releasing energy.

	dry ice sublimates
	CO₂ + H₂O + sunlight à glucose
	air in heated tire expands
	burning coal
	water frozen into ice
	acid dissolves metal

Endothermic vs Exothermic Reactions

All chemical reactions involve bond breaking and bond making.

Bond breaking is endothermic (energy is absorbed from surroundings)

Bond making is exothermic (energy is released into surroundings)

Imagine stretching a rubber band until it breaks. You must do work to stretch the band because the tension in the band opposes your efforts. You lose energy; the band gains it. Something similar happens when bonds break in a chemical reaction. The energy required to break the bonds is absorbed from the surroundings.

Energy is absorbed or released when the heat capacities of the products and reactants differ. Usually this is small. Remember that heat capacity is best thought of with a penny and specific heat best thought of as copper metal.

Neutralization reactions are usually exothermic but when you add baking soda to vinegar it is slightly endothermic. The neutralization reaction actually does release heat:

HC₂H₃O₂ + NaHCO₃ à CO₂ + NaC₂H₃O₂ (aq) + H₂O

This is because there is net bond formation. The products collectively have lower energy than the reactants. But evaporation of the liquid occurs as the carbon dioxide escapes from solution. Evaporation absorbs heat, cooling the liquid. (The expansion of the carbon dioxide gas bubbles as they are released also helps to cool the surroundings by Joule-Thomson cooling). The net result is an endothermic reaction.

Mixing a strong acid with water is exothermic. Breaking a chemical bond requires energy (remember that stretching a spring until it breaks requires energy). Forming a chemical bond will release energy. So in a reaction that releases heat (exothermic) there must be net bond formation. Lets looks at HCl dissolved in water:

HCl à H⁺ (aq) + Cl^1- (aq)

You would think at first this would be a heat absorbing (endothermic) process, because it looks like the bond between H and Cl is broken. But there is another reaction hiding here. The hydrogen ion reacts with water to form a complex of the form: H₃O·(H₂O)⁺_n where n is a number between 1 and 9. It is much easier just to write H⁺(aq). Because the hydrogen ion is so tiny, a large amount of charge is concentrated in a very small area, and the polar water molecules are strongly attracted to it. This "hydration" of the hydrogen ion involves the formation of a covalent bond to one of the waters and a large number of strong hydrogen bonds, so it’s a strongly exothermic process. This causes the mixing of a strong acid with water to be strongly exothermic overall.

Exothermic processes	Endothermic processes
making ice cubes	melting ice cubes
formation of snow in clouds	conversion of frost to water vapor
condensation of rain from water vapor	evaporation of water
a candle flame	forming a cation from an atom in the gas phase
mixing sodium sulfite and bleach	baking bread
rusting iron	cooking an egg
burning sugar	producing sugar by photosynthesis
forming ion pairs	separating ion pairs
combining atoms to make a gas molecule	splitting a gas molecule apart
mixing strong acids and water	mixing water and ammonium nitrate
nuclear fission	melting solid salts

Heating Curve Information: This graph will aid in understanding the following information.

Phase Change Diagram

Solid - Gas Phase Change

This change involves sublimation which is the direct change of a solid to a gas (deposition is the opposite).

Examples include: moth balls (naphthalene), paradichlorobenzene, camphor, iodine crystals, and CO₂ fire extinguishers (advantages: does not conduct electricity, colder than water, replaces O₂ since CO₂ is heavier and settles on ground area and the CO₂ does not combust), will sublime away reducing cleanup - disadvantages include difficulty in keeping container pressurized over time, fact that you cannot use on living things due to extreme cold, and cost).

Liquid - Solid Phase Change

This discussion deals with melting-freezing point. A complete discussion of this concept using ice and heat units will be completed in class.

See class discussion of ice cube. The addition of 1 calorie of heat to the ice cube at 0° C does not cause a change in the temperature of the ice cube though 1 calorie would change the temperature of 1 gram of water at 0° C.

It will take 80 calories just to melt the ice cube. That heat that is consumed in melting the solid is converted into potential energy. Freely moving molecules in liquids, with respect to intermolecular attraction, possess more energy than similar molecules bound rigidly in solids at the same temperature.

Remember that temperature is a measure of the average kinetic energy only while heat content is a measure of the total kinetic energy plus potential energy possessed by that body.

See class examples of the heat energy needed to change ice at any temperature to steam at any temperature.

The melting-freezing point is defined as the temperature at which the solid and liquid phases are in equilibrium. This is the temperature at which a change of state between the solid and liquid phase can occur. Some of the solid will be melting and some of the liquid will be freezing.

When a solid is heated to its melting point, its atoms or molecules acquire enough energy to shift the bonds holding them together so they form separate clusters. This clustering in liquids is confirmed with X-ray studies but the clusters are constantly shifting their arrangement unlike the permanent arrangement in solids.

When heat is added to a solid the temperature of the solid will increase till it reaches the melting-freezing point. It will remain there until all the solid has melted and only then can the temperature of the liquid rise according to its specific heat.

Water molecules at 0° C contain more energy than the ice molecules at 0° C , not in the form of a faster more rapid motion but in the form of an ability to resist the attractive forces tending to pull them together.

Melting points also depend on pressure (though not as much as boiling points.) Ice is strange in that its melting point decreases with increasing pressure. Almost all other materials show increasing melting points with greater pressures. The pressure an ice skater exerts on the ice due to the small area of the skate blade is usually enough to melt the ice creating a thin film of water that acts as a lubricant. On unusually cold days the pressure may not be enough to melt the ice and thus skating would be impossible.

Liquid - Gas Phase Change

The change from a gas to la liquid is condensation. This is due to cooling and/or a pressure change.

In liquids, the energy of the particles is raised by adding heat. When some molecules have enough K.E. they break away from the liquid surface and become vapor.

If the temperature falls, there is a decrease in the energy of the moving molecules and the liquid may eventually freeze to the solid phase.

Process of EVAPORATION: Molecules that have enough energy of motion (K.E.) break free from intermolecular forces and escape into the air as vapor. Some may return to the liquid is their energy is lost to other atoms.

The liquid surface left behind is cooled. In evaporation the molecules that escape are the ones with the greatest velocity (heat) thus the average velocity and K.E. of the remaining particles is reduced. This results in cooling effects. Heat must be absorbed from the surroundings to continue the evaporation process.

Adding heat increases evaporation because the VAPOR PRESSURE is increased. This is the pressure exerted by the vapor (gas) of a substance when it is in equilibrium with liquid or solid phase. The system is in equilibrium when the rate of evaporation equals the rate of condensation.

The temperature at which the liquid's vapor pressure is equal to outside (atmospheric) pressure is that liquid's BOILING POINT. At this temperature the pressure of the vapor escaping from liquid equals the outside pressure.

When the vapor pressure equals outside pressure bubbles of vapor form and push through to the surface. As they move into the gas phase we say this is boiling. Conduction of heat creates the gas, which rises because it is less dense than the liquid, as it strives for equilibrium.

The boiling point varies with atmospheric pressure. In mountains, the boiling point is below 100°C because the pressure of the atmosphere is less.

Cooking requires longer times at high altitudes because of low boiling point

Pressure cookers make food cook more rapidly because the foods can be heated above the normal boiling point without actually boiling.

Intermolecular Forces and Latent Heat

if we heat a mixture of ice and water, we find that no matter how much heat is transferred to the mixture, the temperature remains at 0° C until the last of the ice is melted. Only after all the ice is melted is heat converted into kinetic energy, and only then can the temperature of the water begin to rise. Experiment shows that 80 calories of heat must be absorbed from the outside world in order that 1 gram of ice might be melted, and that no temperature rise takes place in the process. The ice at 0° C is changed to water at 0° C.

But if the heat gained by the ice is not converted into molecular kinetic energy, what does happen to it? If the Law of Conservation of Energy is valid, we know it cannot simply disappear.

The water molecules in ice are bound together by strong attractive forces that keep the substance a rigid solid. In order to convert the ice to liquid water (in which the molecules, as in all liquids, are free of mutual bonds to the extent of being able to slip and slide over, under, and beside each other) those forces must be countered. As the ice melts, the energy of heat is consumed in countering those intermolecular forces. The water molecules contain more energy than the ice molecules at the same temperature, not in the form of a more rapid motion or vibration but in the form of an ability to resist the attractive forces tending to pull them rigidly together.

The Law of Conservation of Energy requires that the energy change in freezing be the reverse of the energy change in melting. If liquid water at 0° C is allowed to lose heat to the outside world, the capacity to resist the attractive forces is lost, little by little. More and more of the molecules lock rigidly into place, and the water freezes. The amount of heat lost to the outside world in this process of freezing is 80 calories for each gram of ice formed.

In short, 1 gram of ice at 0° C, absorbing 80 calories, melts to 1 gram of water at 0° C; and 1 gram of water at 0° C giving off 80 calories, freezes to 1 gram of ice at 0° C.

The heat consumed in melting ice or any solid, is converted into a sort of potential energy of molecules. Just as a rock at the top of a cliff has, by virtue of its position with respect to gravitational attraction more energy than a similar rock at the bottom of the cliff, so do freely moving molecules in liquids, by virtue of their position with respect to intermolecular attraction, possess more energy than similar molecules bound rigidly in solids.

It is the kinetic and potential energies of the molecules that together make up the internal energy that represents the heat content. It is kinetic energy only that is measured by the temperature. By changing the potential energy only, as in melting or freezing, the total heat content is changed without changing the temperature.

In converting a gram of liquid water at 100° C to a gram of steam at 100° C what remains of the intermolecular attractions must be completely neutralized. Only then are molecules capable of displaying the typical properties of gases ‑‑ that is, virtually independent motion. In the earlier process of melting, only a minor portion of the intermolecular attractive force was countered, and the major portion remains to be dealt with. The latent heat of vaporization of water (the amount of heat required to convert 1 gram of water at 100° C to 1 gram of steam at 100° C) is 540 calories, almost seven times the earlier 80 calories needed in changing ice to water.

The energy content of steam is thus surprisingly high. 100 grams of water at 100° C can be made to yield 10 000 calories as it cools to the freezing point. 100 grams of steam at 100° C can be made to give up 54 000 calories merely by condensing it to water. The water produced can then give up another 10 000 calories if it is cooled to the freezing point. It is for this reason that steam engines are so useful and hot water engines would never do as a substitute.

If we boil water in a kettle its temperature remains at 100° C, no matter how fast we boil it, but we have to keep adding heat to keep it boiling. Heat is absorbed by the molecules as they escape their liquid state and become a gas. The amount of heat needed to pull apart liquid molecules is called heat of vaporization (calories/gram). The heat of vaporization which a molecule must absorb before it can become a gas molecule is released by it when it cools again to liquid, as heat of condensation. Liquids with low boiling points, such as alcohol or ether, chill the hand as the molecules pick up their heat of vaporization and become a gas. The same is true for a glass of water, it will be cooler than room temperature.

The kinetic molecular theory states that the kinetic energy depends on heat energy, which can be measured as temperature. A thermometer in boiling water and a thermometer in the vapor just above the boiling surface will read the same; 100° C at sea level. Therefore the average kinetic energy of the liquid molecules must be the same as the average kinetic energy of the gas molecules above it. An average molecule in the liquid state will be moving as fast as an average molecule in the gaseous state.

Gas particles move in a straight line until they collide with another bit of matter, then they bound away in some other direction but always in a straight line, and without losing any of their energy to friction in the collision. The particles have perfect resilience. However, they will change their kinetic energies in the familiar way of all normal matter, as, for instance, do billiard balls. A slow‑moving particle hit from behind by a fast one is speeded up, while the fast one is slowed down, but the sum total of their kinetic energies remains the same. In the world of normal matter, perfect elasticity is unknown, as there is friction between surfaces. Two billiard balls when they collide will change each other's speed and direction of motion, and they will also roll to a stop in a short time. The ultimate particles of matter lose not a bit of their energies in collisions. They simply exchange speeds. If two particles collide, their total heat before and after is the same, but the originally slower particle after the collision is traveling faster and is therefore hotter, while the formerly speedier particle is now cooler and moving more slowly that it was. Heat and molecular motion, according to the theory, are two ways of speaking about the same thing.

Heat/Temperature Problems:

1. Suppose a piece of iron (mass = 21.50 g at a temperature of 100.° C) is dropped into an insulated container of water (mass of water = 132 g and the temperature before adding iron was 20.0° C). What will be the final temperature of the system (at thermal equilibrium)? The specific heat of iron is 0.113 cal/g° C.

2. If 200. grams of water is to be heated from 24.0° C to 100.0° C to make a cup of tea, how much heat must be added?

3. Which is more effective in cooling a drink, 10 grams of water at 0° C or 10 grams of ice at 0° C? Explain your answer quantitatively.

4. A 3.00 kg lead bar at 100.0° C is placed in 4.00 kg of water at 20.0° C. The final temperature of the lead bar would be ___________. (c_p of lead is 0.0305 cal/g° C)

5. A 0.60 kg copper kettle holds 1.70 kg of water at 30.0° C. A 0.10 kg iron ball at 120.0° C is dropped into the water. What is the final temperature of the water? (c_p of copper = 0.377 J/g° C and iron is 0.448 J/g° C)

6. A piece of iron with a mass of 20.50 grams at a temperature of 100.0° C is dropped into 140.00 grams of water at 40.0° C. What will be the final temperature of the system. The c_p of iron is 0.45 J/g° C)

7. A sample of mercury metal is heated from 25.5° C to 52.5° C. In the process, 187 cal of heat are absorbed. What mass of mercury was in the sample? The specific heat of mercury is 0.033 cal/ g° C

8. A block of aluminum weighing 140. g is cooled from 98.4° C to 62.2° C with the release of 1080 cal of heat. From these data, calculate the specific heat of aluminum.

9. A cube of gold weighing 192.4 g is heated from 30.0° C to some higher temperature, with the absorption of 226 cal of heat. The specific heat of gold is 0.030 cal/ g° C. What was the final temperature of the gold?

10. A total of 54.0 cal of heat are absorbed as 58.3 g of lead is heated from 12.0° C to 42.0° C. From these data, what is the specific heat of lead?

11. A piece of erbium metal weighing 100.0 g and heated to 95.0° C is dropped into 200.0 g of water initially at 20.0° C. The final temperature of the mixture is 21.5° C. What is the specific heat of erbium metal?

12. A block of rhenium metal (specific heat = 0.0329 cal/ g° C) is heated to 88.2° C and then dropped into 100.0 g of water initially at 26.4° C. The final temperature of the mixture is 32.4° C. What was the mass of the block of rhenium?

13. When 258.6 g of benzene vapor is condensed to a liquid at its boiling point, 33 875 cal of heat are released. What is the heat of vaporization for benzene?

14. A sample of ethyl alcohol is converted from a liquid to a vapor with no temperature change. In the process 30 640 cal of heat are absorbed. What mass of ethyl alcohol was in the sample? The heat of vaporization of ethyl alcohol is 210. cal/g.

15. The heat of combustion of methane is 212.8 kcal per mole. How much heat will be produced in the combustion of 100.0 g of methane?

16. The heat of combustion of toluene is 934.2 kcal per mole. How much heat will be released during the combustion of 250.0 g of toluene? The formula for toluene is C₆H₅CH₃

17. Copper has a density of 8.94 g/cm³ and a specific heat of 0.090 cal/g° C. A cube of copper is heated from 10.5° C to 214° C. The cube of copper has dimensions of 5.00 cm. How much heat would the copper cube absorb?

18. The specific heat of water is 4.185 J/g° C (1.00 cal/g° C). A piece of a pure metal with a mass of 24.0 g at a temperature of 45.0° C is added to 55 mL of water at 60.0° C. The final equilibrium temperature of the mixture is 95.4° C. Find the specific heat of the pure metal in both cal/g° C and J/g° C.

19. The specific heat of ice is 2.03 J/g° C. How much heat is needed to convert 550. g of ice at –15.0° C to 10.0° C?

20. What is the total amount of heat needed (in calories and joules) to convert 2.25 kg of ice at 0.0° C to steam at 200.0° C.

21. The specific heat of silicon is 0.057 cal/g° C and the density of silicon is 4.4 g/cm³. The volume of a cylinder formula is given as p r² L. The addition of 6000. calories raises the temperature of the silicon cylinder 55.5° C. Find the radius of the cylinder. The length of the cylinder is given as 6.00 cm.

22. A piece of metal with a mass of 75.5 g is heated to 84.5° C and added to 100.0 mL of water at 5.0° C. The final temperature of the mixture is 75.0° C. Find the specific heat of the metal.

23. Granite has a specific heat of 800. J/g° C. What mass of granite is needed to store 1.50 E 6 J of heat if the temperature of the granite is to be increased by 15.5° C?

24. A 55 kg block of granite has an original temperature of 15.0° C. What will be the final temperature of this granite if 4.5 E 4 kJ of heat energy are added to the granite?

The above curve attempts to demonstrate the addition of 100 calories of heat energy per minute to 1 gram of water. A thorough class discussion will attempt to identify the amount of time needed for each change to occur. Knowledge of specific heat capacity, latent heat of fusion and vaporization is needed to determine these values.

The above chart shows a graph of the contraction of an air bubble in a capillary tube filled with oil. As the tube was cooled the length of the air bubble was measured and plotted as dots. When it could no longer be cooled to a lower temperature the graph line was extrapolated to find Absolute Zero (a temperature unobtainable in the actual world).

Short Answer Questions

Directions: Answer each question fully. Use complete sentences. Skip two lines of notebook paper (double double space if typed) between each question. Use only the front of paper.

Explain latent heat (use water as an example). Give quantitative examples and clearly explain how intermolecular forces are involved during these changes.

Discuss how specific heat capacity might be used to identify an unknown metal sample.

Compare and contrast endothermic and exothermic chemical reactions.

Compare and contrast kinetic and potential energy, using as many practical examples as possible.

Using examples, explain fully elastic and inelastic collisions. Try to include some of the problems these concepts cause in science classes. Include momentum in your discussion.

Describe how a liquid thermometer is made and how it could be calibrated without another thermometer

Explain how chemical and physical properties might be used to identify an unknown substance. Pick one common substance and give physical/chemical details for it.

Discuss the way in which gas exerts pressure on the walls of its container.

Explain the Kinetic-Molecular Theory of Gases as it relates to ordinary life. Include examples of Boyle’s Law and Charles’ Law in your explanation.

Compare intermolecular forces with chemical bonds within molecules.

Discuss atmospheric pressure and how it relates to gas laws, boiling points, and vapor pressures.

Explain how conduction, convection, and radiation might occur when using our calorimeters. Compare our calorimeters with good Thermos bottles.

Explain fully melting-freezing point theory.

Explain fully boiling point theory and related phenomena such as elevation changes, pressure cookers, super heating, and vapor pressure.

Molecular Motion Demonstrator

This demonstration tool allows us to model a variety of atomic behavior. The following are notes (these will save the student having to make them during the demonstration and will allow them to study the modeling better.

General observations

    1)     some molecules move faster than others, with constant random motion
2)     molecules collide with each other and the walls
3)     near elastic collisions
4)     rarity of 3 body collisions
5)     generally there is a large amount of space between the molecules
6)     random motion in straight lines between collisions
7)     pressure exerted on whatever they hit

Diffusion - as the particles move across the barrier they demonstrate diffusion: (movement of molecules from one area to another with a net change in concentration)

Temperature

      1)     related to average speed of molecules
2)     would 2 different gases at same temperature have same average speeds?
3)     observe different speeds of different gas particles -> temperature relates to average K.E. of molecules
4)     increase vibration rate: average speed increases, frequency of collisions increases, mean free path decreases

Similarities with real world

      1)     model and real gases involve particles in continual, random motion, with straight line paths between collisions
2)     particles in model occupy only a small fraction of total volume
3)     changes in amount of movement is associated with change in temperature
4)     rebounding forces of particles produce pressure in both model and real gases

Dissimilarities with real world

      1)     model uses particles approximately ten million times the diameter of the particles in real gases
2)     distance between collisions in real gases much greater than the distance traveled by plastic balls in relation to size
3)     the speeds of the balls in the model are at most a few miles per hour while gas molecules travel at speeds of hundreds of miles per hour
4)     real gas molecules collide elastically while our model will run down due to friction is energy is not applied
5)     real gases are three dimensional, model is two dimensional
6)     most gas molecules are not spherical
7)     spaces between particles in model taken up by air, in gases empty spaces occupies the space between molecules

Liquids and Solids

attractive forces between gas molecules are called van der Waal forces - positive charge of nucleus is not completely shielded by electron cloud so at short distances the nucleus may be attracted to electron cloud of another atom - these gas molecules may come together, slow down, and allow attractive forces to form

Boyle’s Law

Charles’ Law

Relative Humidity

When molecules from the vapor above a liquid surface impinge on the surface, they may be trapped there, so that a constant two-way traffic of molecules to and from the liquid occurs. If the density of the vapor above the liquid is sufficiently great, as many molecules return as leave it at any time, a situation that is described by saying that the region is saturated with the substance. The higher the temperature, the greater the maximum vapor density: at 0° C the density of water vapor at saturation is 5 g/m³, at 20° C it is 17 g/m³, at 100° C it is 598 g/m³, and at 300° C it is all the way up to 45.6 kg/m³.

If for any reason (such as a sudden drop in temperature) the vapor density exceeds the saturation value, condensation will be more rapid than evaporation until equilibrium is reestablished. It is for this reason that on a hot day moisture condenses on the outside of a glass that contains a cold drink. The relative humidity of a volume of air describes its degree of saturation with water vapor. Relative humidities of 0, 50%, and 100% mean respectively that no water vapor is present, that the air contains half as much moisture as the maximum possible, and that the air is saturated. On a hot day the evaporation of sweat from the skin is the chief means by which the human body dissipates heat, and a high relative humidity is uncomfortable because it impedes the process. A low relative humidity is also undesirable because it leads to the drying of the skin and mucous membranes. The regulation of relative humidity is as important a function of a heating or of an air-conditioning system as the regulation of temperature.

We know that heating air decreases its relative humidity and cooling air increases its relative humidity. For instance, between 10° C and 20° C the saturated vapor density (which corresponds to 100% relative humidity) just about doubles. This means that if outside air at 10° C whose relative humidity is say, 70% is taken inside a house and heated to 20° C, the relative humidity indoors will only be 35% since the actual vapor density stays the same. A way to humidify heated air in winter is clearly desirable. If the outside air is at 30° C with 70% relative humidity, then cooling it down to 24° C is enough to bring it to saturation, which is 100% relative humidity; further cooling will cause water to condense out. An air-conditioning system therefore should incorporate means to remove water vapor from the air being cooled.

Relative humidity is water vapor density relative to saturation density. Changing the temperature of a body of air also changes its relative humiditiy.

Principle of Heat Exchange - The Coffee-Cream Problem

Newton's law of cooling: The rate of heat conduction is proportional to the temperature difference between an object and its surroundings.

The Stefan-Boltzmann law of radiation: The rate of heat lost by radiation is proportional to the fourth power of the absolute temperature.

The historic problem:

Ah, you see, there is this business man who likes a large amount of cream in his coffee, and he wants the resultant mixture as hot as possible. (Alas, there is no microwave oven available).

He has just prepared his boiling coffee when he is called by the boss for a quick conference of ten minutes duration. The boss tolerates no coffee in his presence.

What to do? To keep the coffee as hot as possible should he add the cream now or wait until after the conference?

Which do you think he should do? _____________________________________

Using beakers for the coffee cups and water for the coffee and cream, design an experiment to test the problem. Use the computer and temperature probe to measure the temperature over time. How would you set up a graph(s) to help prove the best solution.

Formulas

K.E. = ½ m v²

P.E. = m g h

Q = mass x D t x c_p

Q = mass x H_f

Q = mass x H_v

K = ° C + 273

° C = K - 273

1 calorie = 4.185 joules

Pressure = Force
area

Temperature Scale Comparisons

	Fahrenheit	Celsius	Kelvin
Boiling p. of water	212°	100°	373
Body Temperature	98.6°	37°	310
Freezing p. water	32°	0°	273
Coincidence p.	- 40°	- 40°	233
Absolute Zero	- 460°	-273.16°	0

Heat of Solution Reactions Lab

Chemical and physical changes are usually accompanied by the liberation or absorption of energy. If energy is evolved, the reaction is said to be EXOTHERMIC. If the energy is absorbed, the reaction is ENDOTHERMIC.

Heat is a form of energy. The calorie or joule is the unit used to express heats of reaction. The calorie is defined as the amount of heat required to raise the temperature of 1 gram of water one Celsius degree. For conversions, 1 calorie = 4.185 joules.

Energy may be transformed from one kind to another within an isolated (or closed) system but the total energy does not change. If the change in energy of a system can be measured and if this change is due solely to a chemical reaction, then the energy change must be equal to that of the chemical reaction itself.

In this experiment a simple calorimeter will be used and the change in energy of the system will be measured by observing the temperature of a given weight of water before and after the reaction occurs. The specific heat capacity of water (i.e., the energy required to raise the temperature 1^oC of 1 gram of the material) is very nearly 1.00 cal/g^oC for temperatures between 0^oC and 100^oC. Thus, if a calorimeter contained 100 grams of water at 23.0^o C initially and after the reaction took place there were still 100. grams of water and the temperature was now 30.0^oC, the energy liberated in the reaction would be

Q (cal) = mass (g) x Dt (^oC) x c_p (cal/g^oC)

700. cal = 100. g x 7.00^oC x 1.00 cal/g^oC

This calculation assumes that no energy was required to raise the temperature of the calorimeter itself and that no energy was lost, or gained, by the calorimeter during the experiment.

Accurate calorimeters are very expensive and very tedious to operate. In this experiment a simple calorimeter will be made by placing one styrofoam (polystyrene) cup inside another and covering the inner cup with a piece of cardboard to minimize heat loss or gain from the surface of the liquid. Expanded polystyrene is a good insulator and has a very high specific heat capacity.

Procedure:

1. Measure as accurately as possible with a graduated cylinder 75.0 mL of water and transfer the water to the inner styrofoam cup. Record the temperature of the water.

2. Mass accurately, a quantity of NaOH between two and three grams.

3. Add the NaOH to the water, continually swishing the cup gently, and with constant observation of the temperature until it remains constant for about 15-20 seconds. Record this temperature.

4. Repeat using NH₄NO₃.

5. After completing all data collection for this experiment, move on to the second experiment.

Data Table:

mass of 75.0 mL of water                     ______
mass of NaOH                                     ______
initial temperature of water                    ______
final temperature of solution                   ______
mass of 75.0 mL of water                   ______
mass of NH₄NO₃                                 ______
initial temperature of water                    ______
final temperature of solution              ______

Questions:

1. Is the dissolution of the NaOH in water an exothermic or endothermic process? What about the ammonium nitrate?

2. Assuming 1 mL of water has a mass of 1 gram and the specific heat of the dilute solution is the same as water, calculate the number of calories involved in the dissolution of the different materials. (use actual mass of H₂O)

# of cal = 75 g x Dt x 1.00 cal/g^oC # of joules = 75 g x Dt x 4.185 J/gºC

NaOH ______________ cal NaOH _______________ J
NH₄NO₃ ______________ cal NH₄NO₃ ______________ J

3. How many moles of each substance were dissolved?

# of moles = grams dissolved | 1 mole of substance
| formula mass

NaOH ____________ mole
NH₄NO₃ _____________ mole

4. Calculate the number of calories (and joules) that would be involved if one mole of the substance were dissolved in water.

# of cal or J = # of cal or J (see question 2 above)
mole # of moles (see question 3 above)

NaOH ______________ cal/mole NaOH ____________ J/mol

NH₄NO₃ ______________ cal/mole NH₄NO₃ ____________ J/mol

5. Calculate the joules/gram absorbed or released when the sodium hydroxide and ammonium nitrate was dissolved in water.

Endothermic and Exothermic Reactions

Some chemical reactions absorb energy and are called endothermic reactions. Many chemical reactions give off energy. Chemical reactions that release energy are called exothermic reactions. You will study one endothermic reaction and one exothermic reaction in this experiment.

You will study the reaction between citric acid solution and baking soda. An equation for the reaction is

H₃C₆H₅O₇ (aq) + 3 NaHCO₃ (s) 3 CO₂(g)+ 3 H₂O (l)+ Na₃C₆H₅O₇(aq)

You will study the reaction between magnesium metal and hydrochloric acid. An equation for this reaction

Mg(s) + 2 HCl (aq) H₂ (g)+ MgCl₂(aq)

OBJECTIVES

In this experiment, you will

• observe two chemical reactions

• determine the change in temperature, Dt, for each of the reactions

• identify endothermic and exothermic reaction

PROCEDURE

1. Obtain and wear goggles.

Citric Acid Plus Baking Soda

2. Mass your calorimeter. Measure out 30.0 mL of citric acid solution in the graduated cylinder and place it in the calorimeter. Mass the calorimeter with acid solution and record so that the mass of the acid can be calculated. Place the thermometer into the citric acid solution and record the temperature after 15 seconds with no change in the reading.

3. Weigh out 10.0 g of solid baking soda on a piece of weighing paper.

4.   Add the baking soda to the citric acid solution. Stir the solution to ensure a good mixing.
   Continue to watch the temperature carefully. Record the lowest or highest temperature that is
    reached.

5. Dispose of the reaction products into the plastic waste cups.

Hydrochloric Acid Plus Magnesium

6. Measure out 30.0 mL of HCl solution into the clean graduated cylinder and transfer into a cleaned calorimeter. CAUTION: Handle this acid with care. It can cause painful burns if it comes in contact with your skin or gets into your eyes. Mass and record to find the mass of the acid solution. Place the thermometer into the HCl solution and record the temperature after 15 seconds with no change in the reading.

7. Get approximately 0.1 grams of magnesium. Record actual mass of magnesium used.

8. Add the Mg to the HCl solution.

• Stir the solution to ensure good mixing. CAUTION: Do not breathe the vapors!

• Collect data until a minimum or maximum temperature has been reached.

9. Dispose of the reaction products into the plastic waste cups.

DATA

1. Mass of calorimeter: ______ g

2. Mass of citric acid and calorimeter: ______ g

3. original temperature of citric acid solution: ______ °C

4. final temperature of citric acid/baking soda: ______ °C

5. mass of hydrochloric acid and calorimeter: ______ g

6. original temperature of hydrochloric acid: ______ °C

7. final temperature of HCl/Mg mixture: ______ °C

5. actual mass of Mg used: ______ g

OBSERVATIONS

PROCESSING THE DATA

1. Calculate the temperature change, Dt, for each reaction by subtracting the minimum temperature from the maximum temperature (Dt = t_max – t _min).

2. Tell which reaction is endothermic. Tell which reaction is exothermic. Explain.

3. For each reaction, describe three ways you could tell a chemical reaction was happening.

EXTENSIONS

1. Determine the energy effect, in joules per gram of NaHCO₃ and in joules per mole

2. Determine the energy effect, in joules per gram of Mg and in joules per mole

Quantity of Heat Lab

The principle of heat exchange is often demonstrated by heating a metal sample in boiling water, then adding the metal to a known quantity of water. The temperature of the water is measured before and after adding the metal sample. The heat lost by the metal is gained by the water. The following equation can then be used:

m_metal x Dt_metal x c_{p metal} = m_water x Dt_water x c_{p water}

Procedure: Immediately begin boiling water in the beaker. Place precisely 100.0 mL of water in the styrofoam cup (a cup within a cup) and the determine the mass and temperature of that water. Record these measurements. Mass the dry metal samples. Record. Heat the metal samples in the beaker. Heat the samples in boiling water until the temperature remains constant for 3-4 minutes to insure that the samples are heated evenly throughout. DO NOT RUSH THIS IMPORTANT STEP. Record the temperature of the boiling water (we are assuming the metal samples are at the same temperature). Using the string pick up the mesh containing the metal samples. Holding it immediately above the boiling liquid, allow it to drain FOR A FRACTION OF A SECOND. Immediately place it in the water the styrofoam cups as quickly and safely as possible. This step should not allow the samples to cool. As you read the thermometer swish the water gently for a few moments. Record the highest temperature reached by the water.

Data:
sample material .. ......................
mass of metal sample .... . ............
specific heat capacity (metal) ...........
mass of 100.0 mL of water................
initial temp. of water (in cup) ..........
initial temp. of metal (boiling water)...
final temp. of water/samples (in cup) ....

2. Repeat above experiment with 100.0 mL of permanent antifreeze. Remember that the samples must be thoroughly reheated in boiling water before being added to the antifreeze.

Data:
mass of 100.0 mL of antifreeze ............
specific heat capacity (antifreeze) ......
initial temp. of antifreeze (in cup) .....
initial temp. of metal (boiling water)...
final temp. of antifreeze/samples (cup)..

3. Repeat original procedure using 100.0 mL of the 50/50 mixture of water and antifreeze.

Data:
mass of 100.0 mL of 50/50 mixture .........
specific heat capacity (50/50 mixture)...
initial temp. of mixture (in cup) ... ....
initial temp. of metal (boiling water)...
final temp. of 50/50 mix/samples (cup)...

Safety Awareness: Be careful with the boiling water and the hot metal samples. Thermometers must be allowed to slowly adjust to changes in temperature. This is especially true in going from boiling temperatures to room temperatures. Be certain to return the pure antifreeze to the proper container as well as returning the 50/50 mixture to its original container.

Analysis:
1. Why use boiling water as the method of heating the metal samples each time. Why not just hold the samples over a burner?
2. List the materials used in this lab in order, from lowest to highest, in specific heat capacities?
3. How does the quantity of heat transferred depend on specific heat, mass, and initial temperature?
4. List the liquids, from lowest to highest, based on the amount of change in temperature. Can this be explained?
5. How did the heat lost by the metal samples in each run compare? List this and explain.
6. Which of the liquids would be the most effective coolant in a car's radiator. Why?
7. Should one run 100% coolant (antifreeze) in a car's radiator in the summer? Why?
8. How would your results differ if you ran the experiment again with a different metal?
9. Explain briefly how specific heat capacities could be used to identify an unknown metal or liquid?
10. Given a true 50/50 mixture of ethylene glycol and water and without using the lab scales, how could you calculate the mass of the mixture. (The given density of pure ethylene glycol is 1.1157 g/cm³)

Calculations:

1. Calculate the specific heat capacity of water. Use the specific heat capacity for your metal. See the chart on the board for densities and specific heat capacities.

m_metal x Dt_metal x c_{p metal} = m_water x Dt_water x [c_{p water}]

Calculate the percentage error between that of the actual value for water and what you found.

% error = theoretical value - experimental value; x 100 = %
theoretical value

2. Calculate the specific heat capacity of the metal you used. This time you must use the actual vale for water.

m_metal x Dt_metal x [ c_{p metal} ] = m_water x Dt_water x c_{p water}

Calculate the percentage error for the metal.

3. Calculate the specific heat capacity of the pure antifreeze. Use the actual value for the metal. The anti here indicates antifreeze.

m_metal x Dt_metal x c_{p metal} = m_antifreeze x Dt_antifreeze x [c_{p antifreeze}]

Calculate the percentage error for the pure ethylene glycol. Additives may have changed the value.

4. Calculate the specific heat capacity of the 50/50 mixture. Use the actual value for the metal. The 50/50 here indicates the mixture. Attempt to calculate the actual ratio of antifreeze to water.

m_metal xDt_metal x c_{p metal} = m_50/50 x Dt_50/50 x [c_{p 50/50}]

Lab Write-up: As usual you may substitute your notecard for the procedure and data table. You may leave out the equipment listing and safety rule section. Be prepared for an extensive calculation and result section. Remember to answer all the questions in the discussion section.

Heat of Combustion of Candle Wax

CONTENT:

This lab deals with heat transfer, combustion of candle wax, heat loss within a system, and physical/chemical changes that occur as a result of energy transformations. Transfer of heat is seen in weather studies (i.e., within thunderstorms the release of heat of condensation helps to power the cell), in absorption of solar radiation at the surface of the earth and re-release of that energy to the atmosphere (Greenhouse heating), and in biological systems when food is burned to provide energy for the organisms use.

Energy is an important consideration in all physical and chemical changes. This experiment has you look at the energy involved in the chemical change of candle wax. We will be using joules instead of calories (1 cal = 4.185 joules or 1 joule = 0.2390 calories.)

OBJECTIVE:

In this lab heat energy from a burning candle is directed to a Coke can filled with water. The elevation in the temperature of the water as the candle burns indicates heat is being transferred. Students should readily observe the principle of heat exchange. Students must attempt to design the procedure and equipment such that heat is neither gained nor lost to the environment. The student then must calculate the heat of combustion of candle wax (i.e., the amount of heat energy released per gram of wax consumed.)

Remember the principle of heat exchange: the heat lost by some object should be equal to the heat gained by the object to which the heat is transferred. Look over the general set-up. An aluminum Coke can filled with water is suspended over the burning candle. How can we determine the amount of heat lost by the candle as it burns and how can be eliminate heat loss to the environment or gain from the environment?

Make a drawing of the experimental setup. Try to incorporate into your design and experimental procedure factors that will eliminate heat loss/gain from the environment. (hint: Simple insulation of the system is not enough. Use your knowledge of heat transfer mechanisms, specific heats, temperature differences, etc. to design a system that will not product a high percent error.) After the lab group decides on the factors they wish to control and decides how to build their apparatus, the group should confirm what information they wish to collect and in what order.

PROCEDURE:

I. Attach candle to notecard base (with heated wax) and record mass of candle/card. Record.

II. Record mass of empty can (with rod inserted). Add 150 mL of very cold water. Record mass of can and water. Calculate amount of water used.

III. Record room temperature.

IV. Design insulating system. Remember that oxygen must get to the flame.

V. Prior to lighting candle, record temperature of water in can.

VI. With constant stirring as candle burns, heat water to a temperature as many degrees above room temperature as it was below room temperature prior to heating.

VII. Put out candle flame as gently as possible. Record final temperature (realize that the temperature may in fact continue to go up after the flame is put out. Calculate change in temperature.

VIII. Mass candle/card. Record. Calculate change in mass of candle.

Sample Data Table:
mass of candle and base before burning:
mass of candle and base after burning:
mass of empty can:
mass of can and water:
room temperature:
temperature of water before heating:
temperature of water after heating:

CALCULATIONS:

1. Using the specific heat of water, calculate the amount of heat transferred in both calories and joules.

2. Calculate the heat of combustion of candle wax in both cal/g and J/g.

3. Actual values for paraffin wax is about 42 kJ/g. Calculate percent error with your value.

Questions:

1. In what way did your insulation help control heat loss or gain?

2. Assuming some heat is always going to be gained from the room and that some heat will probably be lost, how can you have the two changes cancel each other out?

3. What purpose would be served by starting with cold water an equal number of degrees below room temperature as the number of degrees above room temperature finally reached?

4. To do the calculations, it was assumed that all the joules released by the burning was went totally into the water and that no extraneous joules of energy were absorbed by the water. What factors in this experiment could have validated those assumptions. What factors might have led to errors in the final results?

5. What further modifications in your procedure and experimental design would you now make to reduce error.

Problem:

A given birthday candle has a mass of 0.84 g. The mass remaining after 1 minute of burning = 0.77 g. What would be the burnout time for that candle. Explain why your answer might not be exactly the same as the actual burnout time.

Problem:

A 5.00 gram sample of motor oil is burned in a calorimeter. The calorimeter contained 720. mL of water initially at 21.4° C. After the oil was burned, the water temperature was measured at 33.9° C. What was the heat of combustion of the motor oil in J/g?

Extended Problem:

You are given a block of wax that has a mass of 150. grams (molar mass = 450). The melting point is 95.0° C, the specific heat of the solid is 0.80 J/g° C, the specific heat of the liquid is 1.2 J/g° C, the heat of fusion is

60. J/g. The was is heated on a stove that provides 6.0 E 3 J of heat per minute. The room temperature is 25 ° C.

a) how many joules of heat are required to heat the wax to melting?

b) how long will this take?

c) how many joules of heat are required to melt the wax and how long will this take?

d) how many joules of heat are required to heat the liquid wax to 245° C and how long will it take?

e) make a graph of the data. What do the slopes of the solid heating and liquid heating indicate about the relative specific heats?

f) at 245° C the wax will burst into flame and completely react with the air to produce CO₂ and H₂O. The heat of combustion is 1200 kJ/mole. How many joules of heat are produced with the wax burns? Is this enough to melt and heat to burning another block of wax of the same size?

Refer to the Data Table below to answer questions:

Substance	Specific Heat	Heat of Fusion	Heat of Vaporization
	(J / g° C)	(J / g)	(J / g)
water (l)	4.185	334	2260
water (s)	2.06	334	2260
ethanol	2.45	109	879
aluminum	0.895	376	11371
copper	0.387	205	4726
silver	0.233	88	2300
granite	0.803	-	-

1. How much heat is required to melt a sample of 550 grams of copper?

2. How much heat is absorbed as a 95.0 gram sample of water is heated from 10.5° C to 48.2° C?

3. How much heat is absorbed by 25 grams of ethanol as it evaporates?

4. A 950. kg granite wall inside a solar house is used to absorb heat during the day. During the day it reaches a temperature of 24.0° C . At night, it will cool off to 17.0° C. How much heat will be released by the wall? How much heat would be released if the wall was made of 20. kg of aluminum filled with 180 kg of water and had the same temperature change?

CHARLES' LAW LABORATORY EXERCISE

This law states that as the temperature of a gas changes its volume will vary proportionally. The pressure acting on the system must remain constant as well as the amount of gas (air in this lab) must remain the same.

V₁/ T₁ = V₂/ T₂ V₂ = (V₁T₂) / T₁ V₂ = unknown, can be found by formula and experimentally

OBJECTIVES:
1. to observe the change in volume of a gas when its temp. is changed
2. to practice laboratory techniques involving gases
3. to perform a lab with a % error less than 2%

SAFETY PRECAUTIONS:

1. Handling extremely hot glassware and working with boiling water requires attention given to equipment and instructions.
2. Use of the Bunsen burner requires careful lighting and one lab group member assigned to watch the burner at all times.
3. Clean dry equipment will always provide better results and perform in a safe manner.

PROCEDURE:

1. Place stopper (with glass tubing already inserted) into the 250 mL Erlenmeyer flask firmly and as level as possible. Using tape mark the flask where the stopper extends.. This should be even around the neck of the flask.

2. Attach the universal clamp to the neck of the flask and also to the ring stand. The flask should extend down as far as possible in an 400/600 mL beaker sitting on wire gauze ring attached to the ring stand. Add water to the beaker so that it fills the beaker to approximately 1 inch below the lip of the flask. NO water should boil or splash out of the beaker. Use boiling chips.

3. Heat the empty flask using this water bath arrangement till the water has been boiling for 8-10 minutes. Record the temperature of the water. This should now also be the temperature of the air inside the flask.

4. Disconnect the clamp from the ring stand. Place your finger over the end of the glass tubing in the stopper and transfer the entire flask and clamp to the standing water in the sink. This must be done so that no air enters or escapes from the flask. Only remove your finger when the entire flask is submerged.

5. Keep the flask completely submerged for approximately 8-10 minutes. Record the temp of the water in the sink. This should be the temperature of the water and air inside the flask.

6. The most critical step in attempting to reduce the % error comes at this point. Continue to submerge the end of the flask with the stopper but raise the base of the flask. Look inside the flask and attempt to have the water level inside the flask correspond to the water level of the water in the sink. This sounds more difficult that it really is. By keeping the water levels inside and out of the flask exactly the same the atmospheric pressure inside and outside the flask will be the same (held constant) and we can eliminate it from our calculations.

7. Place your finger over the end of the glass tubing (maintaining the same level of water inside and out) and then lift out the entire flask and clamp. Remove and dry the clamp with paper towels. Carefully pour the water inside the flask into a dry graduated cylinder and record the amount.

8. Fill the flask with water up to the mark made earlier (the neck where the stopper had extended). Again pour this water into a dry graduated cylinder (this may take several full cylinders, just accumulate the volume). The final volume (V₂) can be found by using the formula or by simply subtracting the amount left in the flask after cooling from the original total amount (V).

SAMPLE DATA TABLE

Temp. of water bath = T₁       ___________º C
Temp. of water in sink = T₂   ___________º C
Volume of water in flask after cooling    __________ mL
Total volume of water in flask (V₁)          ___________mL
Final volume of air in flask (by subtraction: = V₂) _______________mL

CALCULATIONS:

1. Find V₂ using Charles' Law:

2. Perform graphing example as shown in class: value of absolute value found: ___________º C

3. Perform the following percentage calculations:

PERCENTAGE ERROR:
% error = |experimental – theoretical result|/ theoretical result x 100 =

PERCENTAGE DIFFERENCE: (for V₂)
% difference = |difference of two values| / larger of two values x 100 =

QUESTIONS:

1. Explain what your lab group could have done to reduce your error.
2. Why/how did we eliminate the need to measure the atmospheric pressure inside the flask after cooling.
3. Find examples of Charles' Law in our daily lives.

Answer Space:

Energy Content of Foods

Energy content is an important property of food. The energy your body needs for running, talking, and thinking comes from the food you eat. Energy content is the amount of heat produced by the burning of 1 gram of a substance, and is measured in joules per gram (J/g).

You can determine energy content by burning a portion of food and capturing the heat released to a known mass of water in a calorimeter. If you measure the initial and final temperatures, the energy released can be calculated using the equation

Q = Dt•m•c_p

where Q = heat energy absorbed (in J), Dt = change in temperature (in °C), m = mass (in g), and c_p= specific heat capacity (4.185 J/g°C for water). Dividing the resulting energy value by grams of food burned gives the energy content (in J/g).

PROCEDURE

1. Obtain and wear goggles/aprons.

2. Get a sample of food and a food holder. Find and record the initial mass of the food sample and food holder. CAUTION: Do not eat or drink in the laboratory.

4. Set up the apparatus. Place aluminum foil on the table top to catch any spills and also to reflect heat upward.

· Determine and record the mass of an empty can.

• Place about 50.0 mL of cold water into the can.

• Determine and record the mass of the can plus water.

• Use a rod to suspend the can about 2.5 cm (1") above the food sample.

• Use a utility clamp and stopper to suspend the thermometer in the water. The thermometer should not touch the bottom of the can.

5. Record the initial temperature of the water.

6. Remove the food sample from under the can and use a wooden splint to light it. Quickly place the burning food sample directly under the center of the can. Allow the water to be heated until the food sample stops burning. CAUTION: Keep hair and clothing away from an open flame.

7. Stir the water until the temperature stops rising. Record this final temperature.

8. Determine the final mass of the food sample and food holder.

9. Repeat the procedure for a second food sample. Use a new 50.0 mL portion of cold water.

10. When you are done, place burned food, used matches, and partly-burned wooden splints in the container supplied by the teacher.

DATA

Sample 1 Sample 2

            Food used                                                     _______                      _______
Mass of food and holder (initial)                                 ______ g                     ______g
Mass of food and holder (final)                                  ______ g                     ______g
Mass of empty can                                                    ______ g                     ______g
Mass of can plus water                                              ______ g                     ______g
Initial water temperature                                             ______ °C                   ______°C
Final water temperature                                             ______ °C                   ______ °C

PROCESSING THE DATA

1. Calculate change in water temperature, Dt, for each sample, by subtracting the initial temperature from the final temperature (Dt = t_final – t_initial).

2. Calculate the mass (in g) of the water heated for each sample. Subtract the mass of the empty can from the mass of the can plus water.

3. Use the results of Steps 1 and 2 to determine the heat energy gained by the water (in J). Use the equation Q = Dt•m•c_p

where Q = heat absorbed (in J), Dt = change in temperature (in °C), m = mass of the water heated (in g), and c_p = specific heat capacity (4.185 J/g°C for water).

4. Calculate the mass (in g) of each food sample burned. Subtract the final mass from the initial mass.

5. Use the results of Steps 3 and 4 to calculate the energy content (in J/g) of each food sample.

6. Record your results and the results of other groups below. Share data with other groups.

Food Type Food Type Food Type Food Type

            ____________               ____________            ____________            ____________
_________ J/g                _________ J/g             _________ J/g             _________J/g
_________ J/g                _________ J/g             _________ J/g             _________J/g
_________ J/g                _________ J/g             _________ J/g             _________J/g
_________ J/g                _________ J/g             _________ J/g             _________J/g
_________ J/g                _________ J/g             _________ J/g             _________J/g
_________ J/g                _________ J/g             _________ J/g             _________J/g

Avg. _________ J/g _________ J/g _________ J/g _________ J/g

7. Which of the foods has the greatest energy content? Why do these foods have the greatest energy content?

Heat of Fusion

During melting, heat is absorbed by the melting solid. In this experiment, you will determine how much heat is needed to melt 1 g of ice. Heat has units of joules (J). The heat used to melt the ice will come from the cooling of warm water and will be measured with a calorimeter. A calorimeter is an insulated container fitted with a device for measuring temperature.

OBJECTIVE: Determine the heat of fusion of ice (in J/g)

PROCEDURE:

1. Use a balance to measure the mass of the calorimeter. Record this mass in your data table. Using the large flask, obtain some hot water from the hot water faucet at the teacher’s desk and mix with tap water until the temperature of the water is about 30 to 35 °C.

2. Use a 100-mL graduated cylinder to measure out 100.0 mL of this water into the calorimeter. Measure the mass of the calorimeter and the 100.0 mL of warm water. Record this value in the data table.

3. Place the thermometer into the warm water inside the cup to warm the instrument to the temperature of the water. The thermometer must be in the warm water for at least 15 seconds before ice is added.

4. Obtain several pieces of ice out of the cooler. Drain the ice cubes while allowing them to still have a wet surface. Add the ice cubes to the warm water.

5. Gently stir the contents of the cup as the ice melts. The temperature will stop dropping and level off when the ice has all melted. End your data collection when the temperature stops dropping.

6. Measure and record the mass of the calorimeter and water (original water + ice melt).

DATA

mass of calorimeter                                 ______ g
mass of calorimeter and warm water        ______ g
mass of warm water                               _______g
mass of calorimeter and all water/ice      ______ g
mass of ice added                                  _______g

initial water temperature (maximum) _______°C
final water temperature (minimum) _______°C

PROCESSING THE DATA

1. Calculate the change in water temperature, Dt (t_max – t_min).

2. Calculate the heat (in J) lost by the water melting the ice using the equation

Q = m • Δt • 4.185 J/g°C

where Q = heat (in joules), m = mass of warm water (in g), and Δt = change in temperature (in °C)

3. Calculate the heat needed to melt 1 g of ice (J/g).

4. An accepted value for the heat of fusion of ice is 334 J/g. Calculate your percent error:

5. What assumption did we make about heat lost by the water in the calorimeter as compared to heat gained by the melting ice?

EXTENSION

Design an experiment to find out if an ice cube taken from a freezer and immediately placed into a calorimeter needs the same amount of energy per gram for melting as does an ice cube that has been outside the freezer for ten minutes.

Thermochemistry is the study of heat changes that take place in a change of state or a chemical reaction - heat energy is either absorbed or released. If a process releases energy in the form of heat, the process is called exothermic. A process that absorbs heat is called endothermic.

Heat is defined as the energy transferred from one object to another due to a difference in temperature. We do not observe or measure heat directly - we measure the temperature change that accompanies heat transfer. In a chemical reaction it is often not possible to measure the temperature of the reactants or products themselves. Instead, we measure the temperature change of the surroundings.

The difference between the system and the surroundings is a key concept in thermochemistry. The system consists of the reactants and the products of the reaction. The solvent, the container, the atmosphere about the reaction (in other words, the rest of the universe) are considered the surroundings. Heat may be transferred from the system to the surroundings (the temperature of the surroundings will increase) or from the surroundings to the system (the temperature of the surroundings will decrease).

When a system releases heat to the surroundings during a reaction, the temperature of the surroundings increases and the reaction container feels warm to the touch. This is an exothermic reaction. Heat flows out of the system. An example of an exothermic reaction is the combustion of propane (C₃H₈) in a barbecue grill to produce carbon dioxide, water, and heat. Note that heat appear in the product side of the equation:

C₃H₈ (g) + 5 O₂ (g) à 3CO₂ (g) + 4H₂O (g) + heat

When a system absorbs heat from the surroundings during a reaction, the temperature of the surroundings decreases and the reaction container feels cold to the touch. This in an endothermic reaction. Heat flows into the system. A familiar example of an endothermic process is the melting of ice. Solid water (ice) needs heat energy to break the forces holding the molecules together in the solid state.

The heat absorbed by water in turning it to steam has been used to break apart the forces (i.e., hydrogen bonding) between water molecules in the liquid phase. The amount of heat that must be absorbed to vaporize a specific quantity of liquid (usually one gram or one mole) is called the heat of vaporization. In a similar manner, heat is also required to melt ice. The amount of heat that must be absorbed to melt a specific quantity of solid is called the heat of fusion.

The amount of heat transferred in these processes depends on the difference in the energy stored in each substance. This stored energy is called the heat content or enthalpy, and is represented by the symbol H.. The enthalpy change ( ΔH) for a physical process or a chemical reaction is defined as the heat change that occurs at a constant pressure. This is convenient, because most of the reactions that are carried out in the lab are in flasks or containers that are open to the atmosphere – that is, they take place at a constant pressure equal to the barometric pressure.

Equation 1 shows the equality between the change in enthalpy (ΔH) of a system and the amount of heat transferred, symbolized by q_p, for a reaction carried out at a constant pressure.

ΔH = q_p Equation 1

The amount of heat (q_p) transferred to a substance or object depends on three factors: the mass (m) of the object, its specific heat (c_p) and the resulting temperature change (Δt)

q_p = m • Δt • c_p Equation 2

The specific heat (c_p) of a substance reflects its ability to absorb heat energy and is defined as the amount of heat needed to raise the temperature of one gram of a substance by one degree Celsius. The specific heat of water is equal to 4.185 J/gºC. In most laboratory situations, the temperature change is measured not for the system itself (the reactants and products) but for the surroundings (the solution and reaction vessel). The amount of heat released by the system must be equal to the amount of heat absorbed by the surroundings. The sign convention in Equation 3 reveals that the heat change occurs in the opposite direction.

q₍_system) = - q_{(surroundings)}

For an exothermic reaction, the heat released by the system results in a temperature increase for the surroundings (Δt is positive) and the heat absorbed by the surroundings will be a positive quantity. The heat released by the system must have the reverse sign – it must be a negative quantity. According to this convention, the enthalpy change for an exothermic reaction is always a negative value. For an endothermic reaction, in contrast, the heat absorbed by the system results in a temperature decrease for the surroundings (Δt) is negative) and the heat released by the soundings will be a negative quantity. The heat absorbed by the system must have the opposite sign - it must be a positive quantity. According to this convention, the enthalpy change for an endothermic reaction is always a positive value.

The energy or enthalpy change associated with the process of a solute dissolving in a solvent is called the heat of solution (ΔH_soln). In the case of an ionic compound dissolving in water, the overall energy change is the net result of two processes - the energy required to break the attractive forces (ionic bonds) between the ions in the crystal lattice, and the energy released when the dissociated (free) ions form ion-dipole attractive forces with the water molecules.

Heats of solution and other enthalpy changes are generally measured in an insulated vessel called a calorimeter that reduces or prevents heat loss to the atmosphere outside the reaction vessel. The process of a solute dissolving in water may either release heat into the aqueous solution or absorb heat from the solution, the amount of heat exchange between the calorimeter and the outside surroundings should be minimal. When using a calorimeter, the reagents being studied are mixed directly in the calorimeter and the temperature is recorded both before and after the reaction ahs occurred. The amount of heat change occurring in the calorimeter may be calculated using the equation: q_p = m • Δt • c_p The specific heat of the solution is generally assumed to be the same as that of the water, namely 4.185 J/gºC.

The mass of grams of solute is usually the independent variable and will be varied in different trials. The temperature change that is produced depends on the mass of the solute and is thus the dependent variable in a calorimetry experiment.

Calorimeter Constant

All chemical reactions involve changes in energy as a result of bond breaking and bond formation during the chemical change. This energy change is an important parameter when studying chemical reactions and is normally measured in an insulated vessel called a calorimeter. When using a calorimeter, the chemical reagents being studied are mixed directly in the calorimeter with the temperature recorded both before and after the reaction. If the mass of the material is also recorded, then the change in the energy occurring in the calorimeter can be calculated by the relationship:

Δq = m • Δt • c_p

where Δq is the change in energy, m is the mass of the solution, Δt is the temperature change of the solution, and c_p is the specific heat of the solution. Since most reactions are carried out in dilute water solutions, the specific heat of the solution is assumed to be the same as that of water (1.00 cal/g°C or 4.185 J/g°C).

The underlying principle utilized in calorimetry is the law of Conservation of Energy. The basic premise of this principle is that “energy can neither be created nor destroyed but may be converted from one form to another.” In other words, the heat lost (or gained) by the chemical reaction is equal to the heat gained (or lost) by the solution in the calorimeter.

In this experiment we are going to simply mix hot and cold water, determine the change in the temperature, and compare the total amount of energy lost with the total amount of energy gained. If the Law of Conservation of Energy is valid, then

Δq_hot_{water
=} Δq_cold_water

where Δq_hot_water is the energy lost by the hot water and the Δq_cold_water is the energy gained by the cold water. Unfortunately, no system is perfect and some of the energy transferred is absorbed by the calorimeter. Therefore, the actual relationship for the Law of Conservation of Energy in this experiment should be:

Δq_hot_{water
=} Δq_cold_water + Δq_calorimeter

The purpose of this experiment is to verify the Law of Conservation of Energy and to determine the amount of energy absorbed or lost by the calorimeter called the calorimeter constant. Calorimeter constants are unique to each calorimeter and must be determined experimentally. they are usually express in calories or Joules per degree Celsius. That is, calorimeter constants represent the amount of energy absorbed or lost by the calorimeter for every one degree change in the temperature. The equation to be used:

calorimeter constant = | Δq _{hot water} - Δq _{cold water}|

_______________________

Δt

The change in energy of both the hot and cold water is calculated from the mass of the water, the specific heat of water, and the difference in the temperature of the water in each calorimeter before and after mixing. By measuring these values, the calorimeter constant can be determined and utilized in each subsequent experiment which uses the calorimeters.

Procedure:

P1. Label the two calorimeters “cold water” and “warm water”. Mass each. Add 8 mL of the cold tap water to the calorimeter labeled for cold water and find its mass. Add 8 mL of the warm tap water to the calorimeter labeled for the warm water and find its mass.

P2. Stir the water in the calorimeter frequently and when the temperature has been constant over several five-second intervals, record the temperature in each calorimeter to the nearest 0.1 °C. Pour the cold water into the warm water and record the resulting temperature of the mixture.

P3. Thoroughly dry each of the calorimeters, re-weigh, and repeat the experiment another time.

Calculations and Questions:

Q1a. Calculate the mass of the water in each calorimeter.

Q1b. Calculate the changes in temperature, Δt_cold and Δt_warm of the water in each calorimeter by subtracting the final temperature, t_final from the initial temperature of both the cold and warm water.

Δt_cold = t_cold - t_final

Δt_warm = t_warm - t_final

Q2. Calculate the energy gained by the cold water by multiplying the mass of the cold water (Q1a) by the specific heat capacity of water (c_p = 1.00 cal/g°C), by the change in temperature of the cold water (Q1b).

Δq _{cold water
=} m_cold • Δt_cold • c_p

Q3. Repeat for the warm water.

Δq _{hot water
=} m_warm •Δt_warm • c_p

Q4a. Calculate the absolute difference between the energy lost by the warm water (Q2), and the energy gained by the cold water (Q3).

Q4b. Calculate the calorimeter constant in cal/°C by dividing the absolute difference in energy (Q4a) by the temperature change in the warm water (Q1b).

calorimeter constant = | Δq_warm - Δq_cold|

_________________

Δt_warm

Q4c. Repeat the experiment and calculate the average calorimeter constant in

cal/ °C. Record your average value on the chalkboard. Then repeat the experiment with Styrofoam cups and record your values on the board.

Q5. The Law of Conservation of Energy states that “energy can neither be created nor destroyed but may be converted from one form to another”. According to your calculations, has energy been conserved? That is, is the heat lost by the warm water equal to the heat gained by the cold water? Explain your answer in terms of your calculated results.

Tips:

1) It is important that the calorimeter be completely dry form one trial to the next. A dry calorimeter should have almost exactly the same mass at the start of each trial.

2) Covers are not needed for the pink micro-scale calorimeters. For this experiment they will not be used on the Styrofoam cups since equilibrium is established in just a few seconds.

Heat of Solution

The energy change associated with the process in which a solute dissolves in a solvent is called the heat of solution. This energy change in the net result of two processes, the energy required to break the solute-solute bonds, the crystal lattice energy, and the energy released when the solute particles bond with the solvent molecules, the heat of hydration. For example, the heat of solution for KCl in water can be considered the sum of the two heat effects.

The crystal lattice energy of KCl, the energy necessary to break apart the KCl crystal lattice and form free ions is represented by equation 1:

[1] KCl (s) à K¹⁺ (g) + Cl¹^- (g) ΔH = + 167.6 kcal

The heat of hydration of KCl, the energy released when the free ions are hydrated, is represented by equation 2.

[2] K¹⁺ (g) + Cl¹^- (g) à K¹⁺ (aq) + Cl^1- (aq) ΔH = -163.5 kcal

In this example, the overall reaction is endothermic, and the heat of solution is positive since more energy is required in step 1 than is released in step 2.

Overall reaction: KCl (s) à K¹⁺ (aq) + Cl¹^- (aq) ΔH = + 4.1 kcal

The purpose of this experiment is to determine the heat of solution for some selected compounds from the energy change associated with their solution process.

Procedure:

P1. Mass a small, clean dry massing pan to the nearest 0.01 g. Add 1.00 to 1.25 grams of the solid to be studied to the massing pan and re-mass to the nearest 0.01 g.

P2. Mass a clean, dry calorimeter to the nearest 0.01 g. Add roughly 15 mL of tap water to the calorimeter and re-mass. Stir and record the initial temperature of the water in the calorimeter to the nearest 0.1 °C.

P3. Add the solid to be studied to the calorimeter. Stir the mixture until the solid is completely dissolved and the temperature remains constant over several readings. Record the final temperature.

P4. Rinse and dry both the thermometer and calorimeter, repeating the experiment with the same solid.

P5. Repeat experiment with the second solid.

Questions and Calculations:

Q1a. Calculate the mass of the solid, m_solid, by subtracting the mass of the empty massing pan from the mass of massing pan plus solid.

Q1b. Calculate the number of moles, n, of your assigned solid dissolved in each of your trials by dividing the mass of solid dissolved, m_solid, by its formula mass in g/mol.

Q1c. Calculate the mass of the water, m_water, by subtracting the mass of the empty calorimeter from the mass of the calorimeter plus water.

Q1d. Calculate the total mass of the reactants, m_total, by adding the mass of the water, m_water, to the mass of dissolved slat, m_solid.

Q2. Calculate the change in temperature, Δt, by subtracting the initial temperature of the water, t_initial, from the final temperature of the solution, t_final.

Q3a. Calculate the heat required or liberated when your assigned solid dissolved in water, Δq_water, from the total mass of water plus salt (Q1b), the specific heat capacity for water

(c_p = 1.00 cal/g°C), and the change in temperature (Q2).

Δq_water = m_total• Δt • c_p

Q3b. Calculate the energy absorbed or released by the calorimeter, Δq_cal, using the calorimeter constant and the change in temperature of the water (Q2)

Δq_cal = calorimeter constant • Δt

Q3c. Calculate the total energy absorbed or released by the solution process, Δq_soln, from the sum of the change in energy of the water (Q3a), plus the change in energy of the calorimeter (Q3b).

Δq_soln = Δq_water + Δq_cal

Q4a. Calculate the heat of solution, ΔH_soln , in cal/mol by dividing the amount of heat liberated or absorbed when the salt dissolves (Q3c) by the number of moles of salt dissolved (Q1a).

ΔH_soln = Δq_soln / n

Q4b. Convert each of the heats of reaction in cal/mol into kcal/mol by dividing each ΔH_soln value by 1000 cal/kcal.

Q5. Calculate the average heat of solution, ΔH_avg, in kcal/mol by averaging the heats of solution (Q4b) for all your trials.

Q6. Calculate the percent error between your average heat of solution, ΔH_avg, and the theoretical value, ΔH_theor.

% error = | ΔH_avg - ΔH_theor | • 100

________________

ΔH_theor

Theoretical heat of solution, ΔH_theor (kcal/mol):NaOH = -10.6 exothermic,

NH₄NO₃ = 6.1 endothermic, KNO₃ = 8.0 endothermic

Data: Heat of Solution Lab Name: _____________________

Formula of solid studied: _____________________ Formula mass: ____________ g/mol

		Trial #1	Trial #2	Trial #3
P1	mass of empty massing pan
P2	mass of pan + solid
Q1a	mass of solid, m_solid
Q1b	# moles of solid, n
P2	mass of empty calorimeter
P2	mass of calorimeter + water
Q1c	mass of water, m_water
Q1d	total mass reactants, m_total
P2	initial temp. water, t_initial
P3	final temp. of water, t_final
Q2	change in temp., Δt
Q3a	energy lost/gained by water, Δq_water
Q3b	energy absorbed/released by calorimeter, Δq_cal
Q3c	energy absorbed/released by dissolving salt, Δq_soln
Q4b	Heat of solution, ΔH_soln
Q5	average heat of solution, ΔH_avg
Q6	theoretical heat of solution, ΔH_theor
Q6	percent error

As an ionic compound dissolves in water, the entropy increases. The ions in the solid phase are arranged in a regular geometric pattern. For example, sodium chloride crystals have a cubic structure. In water, however, the ions are separated from one another and are scattered randomly throughout their container giving the solution a much higher entropy.

Specific Heat of a Substance

When a hot object is added to cold water in a calorimeter, the heat lost by the hot object,

Δq _{hot object
is} equal to the heat gained by the cold water, Δq _{cold water}and the calorimeter,

Δq _calorimenter as predicted by the Law of Conservation of Energy.

Δq _{hot
object =} Δq _{cold water} + Δq _calorimeter

[ m • Δt • c_p ]_{hot object} = (calorimeter constant • Δt)_calorimeter + [ m • Δt • c_p ]_{cold water}

The purpose of this experiment is to compare the energy lost by a hot object with the energy gained by cold water and a calorimeter and to determine the specific heat of the hot object.

Procedure:

P1. Heat about 300 mL of tap water in a 400 mL beaker using a hot, blue flame. Mass the test tube and then fill it about one-half full of the material to be studied, and re-mass to the nearest 0.01 g. Place the test tube containing the material in the warm water. As soon as the water boils, reduce the flame so that the water barely boils.

P2. Mass a dry calorimeter, add 15 mL of cold tap water, and re-mass. Record the temperature of both the water in the calorimeter and the temperature of the material in the test tube to the nearest 0.1 °C. The temperature of the material in the test tube can be measured by pressing the thermometer against the material until the temperature is constant over several five-second intervals.

P3. When the temperature of the material is constant, quickly pour the hot material into the cold water in the calorimeter. Stir the mixture and record the temperature to the nearest

0.1°C when it is constant over several five-second intervals.

P4. Drain the water from the material studied and return it to the collection point designated by the teacher. Dry the calorimeter and repeat the experiment with another material.

Calculations and Questions:

Q1a. Calculate the mass of the material in the test tube.

Q1b. Calculate the mass of the water in the calorimeter.

Q1c. Calculate the change in the temperature of both the water and calorimeter, Δt_water, by subtracting the initial temperature of the cold water from the final temperature of the water after the hot material was added to it, t_final.

Q1d. Calculate the temperature change of the hot material in the test tube, Δt_material, by subtracting the initial temperature of the material while it was still in the boiling water from the final temperature of the mixture.

Q2a. Calculate the change in the energy of the water, Δq_water , from its mass (Q1b), the specific heat of water (1.00 cal/g°C), and the change in its temperature (Q1c).

Δq _water = m_water • Δt_water • c_{p water}

Q2b. Calculate the change in the energy of the calorimeter (Δq _cal) from the calorimeter constant and the change in temperature of the water (Q1c).

Δq _cal = calorimeter constant • Δt_water

Q2c. Calculate the total energy lost by the hot material, Δq _material, from the energy gained by both the cold water (Q2a) and the calorimeter (Q2b).

Δq _material = Δq _water + Δq _cal

Q3. Calculate the specific heat of the material studied, c_{p material}, from the energy lost by the hot material (Q2c), the mass of the material (Q1a), and the temperature change of the material (Q2d).

Δq _material = m_material • Δt_material • c_{p material} (solve for c_p)

Note: if multiple trials of the same material are used, average the results you have for the different trials at this time.

Q4. Calculate the percent error by comparing your value for the material (Q3 or average of all Q3s) with the accepted value found of the chalkboard.

% error = | c_p - c_p _theoretical | x 100

_______________

c_p _theoretical

Procedure 2

Using the final specific heat of the metal found above, repeat the experiment replacing the water in the calorimeter with ethylene glycol (antifreeze) and then a 50/50 mixture of ethylene glycol and water (prepared for you by the teacher). Solve for the specific heat of the ethylene glycol (which is known to be 0.571 cal/g°C) and for the mixture (which is assumed to be

0.785 cal/g°C, the difference between 1.00 and 0.571 since the mixture was based on 50% by mass of water and ethylene glycol). Complete the calculations for this section by doing % error calculations.

Analysis Questions:

Why use boiling water as the method for heating the metal samples each time? Why not just hold the samples over a burner?
Using the three liquids, list their calculated specific heat capacities, in order from lowest to highest. Next give the change in temperatures experienced by each liquid. Assuming the metal samples transferred the same amount of heat to the three liquids (can this be checked?), is the ranking of the different changes in temperature correct?
How did the heat lost by the metal samples in each experiment compare? Give values and explain their validity.
Which of the liquids would be the most effective coolant (during the summer) in a car’s radiator. Why? Which liquid is the most effective coolant for year round usage? Why?
Explain how specific heat capacities be used to identify an unknown metal.
Given a true 50/50 mixture of the ethylene glycol and water, calculate the mass of the mixture (given the density of pure ethylene glycol as 1.1157 g/cm³)

Write-up Procedure:

Include all collected data in a table (use the P# and Q#s to help identify) including all appropriate units and significant figures.
Show all calculations required. Include the appropriate formula, fill in all data with units, and solve. Circle the final answer to all calculations (remember units and sig figs).
Answer all analysis questions fully.