12.7

Heat and Temperature Change: Specific Heat Capacity

| Solids and Liquids

Greater amounts of heat are needed to raise the temperature of solids or liquids to higher values. A greater amount of heat is also required to raise the temperature of a greater mass of material. Similar comments apply when the temperature is lowered, except that heat must be removed. For limited temperature ranges, experiment shows that the heat Q is directly proportional to the change in temperature and to the mass m. These two proportionalities are expressed below in Equation 12.4, with the help of a proportionality constant c that is referred to as the specific heat capacity of the material.

| Heat Supplied or Removed in Changing the Temperature of a Substance
| The heat Q that must be supplied or removed to change the temperature of a substance of mass m by an amount is

| (12.4)

where c is the specific heat capacity of the substance.
Common Unit for Specific Heat Capacity: J/(kg·C°)

Solving Equation 12.4 for the specific heat capacity shows that , so the unit for specific heat capacity is J/(kg·C°). Table 12-2 reveals that the value of the specific heat capacity depends on the nature of the material. Examples 9, 10, and 11, Examples 9, 10, and 11 and Examples 9, 10, and 11 illustrate the use of Equation 12.4.

Table 12-2 Specific Heat Capacities* of Some Solids and Liquids

Substance | Specific Heat Capacity, c J/(kg·C°)
Solids
Aluminum |
Copper | 387
Glass | 840
Human body | 3500
(37 °C, average) |
Ice (−15 °C) |
Iron or steel | 452
Lead | 128
Silver | 235
Liquids
Benzene | 1740
Ethyl alcohol | 2450
Glycerin | 2410
Mercury | 139
Water (15 °C) | 4186

1) |  | Except as noted, the values are for 25 °C and 1 atm of pressure.

| Example 9 | | | A Hot Jogger

In a half hour, a 65-kg jogger can generate of heat. This heat is removed from the jogger’s body by a variety of means, including the body’s own temperature-regulating mechanisms. If the heat were not removed, how much would the body temperature increase?
Reasoning The increase in body temperature depends on the amount of heat Q generated by the jogger, her mass m, and the specific heat capacity c of the human body. Since numerical values are known for these three variables, we can determine the potential rise in temperature by using Equation 12.4.
Solution Table 12-2 gives the average specific heat capacity of the human body as 3500 J/(kg·C°). With this value, Equation 12.4 shows that

An increase in body temperature of 3.5 °C could be life-threatening. One way in which the jogger’s body prevents it from occurring is to remove excess heat by perspiring. In contrast, cats, such as the one in Figure 12-24, do not perspire but often pant to remove excess heat.

Figure 12-24 Cats, such as this Bengal tiger, often pant to get rid of excess heat. (PhotoDisc Blue/PhotoDisc, Inc./ Getty Images)

| Example 10 | | | Taking a Hot Shower

Cold water at a temperature of 15 °C enters a heater, and the resulting hot water has a temperature of 61 °C. A person uses 120 kg of hot water in taking a shower. (a) Find the energy needed to heat the water. (b) Assuming that the utility company charges $0.10 per kilowatt·hour for electrical energy, determine the cost of heating the water.
Reasoning The amount Q of heat needed to raise the water temperature can be found from the relation , since the specific heat capacity, mass, and temperature change of the water are known. To determine the cost of this energy, we multiply the cost per unit of energy ($0.10 per kilowatt·hour) by the amount of energy used, expressed in energy units of kilowatt·hours.
Solution (a) The amount of heat needed to heat the water is

| (12.4)

(b) The kilowatt·hour (kWh) is the unit of energy that utility companies use in your electric bill. To calculate the cost, we need to determine the number of joules in one kilowatt·hour. Recall that 1 kilowatt is 1000 watts (), 1 watt is 1 joule per second (; see Equation 6.10b), and 1 hour is equal to 3600 seconds (1 h = 3600 s). Thus,

The number of kilowatt·hours of energy used to heat the water is

At a cost of $0.10 per kWh, the bill for the heat is or 64 cents.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 11 | | | Heating a Swimming Pool

Figure 12-25 shows a swimming pool on a sunny day. If the water absorbs of heat from the sun, what is the change in the volume of the water?

Figure 12-25 When water absorbs heat Q from the sun, the water expands. (David Wall/Danita Delimont)

Reasoning As the water heats up, its volume increases. According to the relation (Equation 12.3), the change in volume depends on the change in temperature. The change in temperature, in turn, depends on the amount of heat Q absorbed by the water and on the mass m of water being heated, since (Equation 12.4). To evaluate the mass, we recognize that it depends on the density and the initial volume V0, since (Equation 11.1). These three relations will be used to determine the change in volume of the water.
Knowns and Unknowns The data for this problem are:

Description | Symbol | Value
Heat absorbed by water | Q |
Unknown Variable
Change in volume of water |  | ?

Modeling the Problem

| Step 1 Volume Thermal Expansion
| When the temperature of the water changes by an amount , the volume of the water changes by an amount , as given by Equation 12.3: , where V0 is the initial volume and is the coefficient of volume expansion for water. Both V0 and are unknown, however, so will be dealt with in Step 2 and V0 in Step 3.

| (12.3)

| Step 2 Heat and Change in Temperature
| When heat Q is supplied to the water, the water temperature changes by an amount . The relation between Q and is given by Equation 12.4 as , where c is the specific heat capacity and m is the mass of the water. Solving this equation for yields

which can be substituted into Equation 12.3, as shown at the right. The heat Q absorbed by the water is known, and the mass m will be considered in the next step.

| Step 3 Mass Density
| The mass m and initial volume V0 of the water are related to the mass density by Equation 11.1 as . Solving this equation for the mass yields

This expression for m can be substituted into Equation 1, as indicated in the right column.

Solution Algebraically combining the results of the three steps, we have
Note that the initial volume V0 of the water is eliminated algebraically, so it does not appear in the final expression for . Taking values of from Table 12-1, from Table 12-2, and from Table 12-1, we have that

Related Homework: Problems 46, 48

| Gases

As we will see in Section 15.6, the value of the specific heat capacity depends on whether the pressure or volume is held constant while energy in the form of heat is added to or removed from a substance. The distinction between constant pressure and constant volume is usually not important for solids and liquids but is significant for gases. As we will see in Section 15.6, a greater value for the specific heat capacity is obtained for a gas at constant pressure than for a gas at constant volume.

| Heat Units Other Than the Joule

There are three heat units other than the joule in common use. One kilocalorie (1 kcal) was defined historically as the amount of heat needed to raise the temperature of one kilogram of water by one Celsius degree.* With , and , the equation shows that such a definition is equivalent to a specific heat capacity for water of . Similarly, one calorie (1 cal) was defined as the amount of heat needed to raise the temperature of one gram of water by one Celsius degree, which yields a value of . (Nutritionists use the word “Calorie,” with a capital C, to specify the energy content of foods; this use is unfortunate, since 1 Calorie = 1000 calories = 1 kcal.) The British thermal unit (Btu) is the other commonly used heat unit and was defined historically as the amount of heat needed to raise the temperature of one pound of water by one Fahrenheit degree.
It was not until the time of James Joule (1818–1889) that the relationship between energy in the form of work (in units of joules) and energy in the form of heat (in units of kilocalories) was firmly established. Joule’s experiments revealed that the performance of mechanical work, like rubbing your hands together, can make the temperature of a substance rise, just as the absorption of heat can. His experiments and those of later workers have shown that

Because of its historical significance, this conversion factor is known as the mechanical equivalent of heat.

| Calorimetry

In Section 6.8 we encountered the principle of conservation of energy, which states that energy can be neither created nor destroyed, but can only be converted from one form to another. There we dealt with kinetic and potential energies. In this chapter we have expanded our concept of energy to include heat, which is energy that flows from a higher-temperature object to a lower-temperature object because of the difference in temperature. No matter what its form, whether kinetic energy, potential energy, or heat, energy can be neither created nor destroyed. This fact governs the way objects at different temperatures come to an equilibrium temperature when they are placed in contact. If there is no heat loss to the external surroundings, the heat lost by the hotter objects equals the heat gained by the cooler ones, a process that is consistent with the conservation of energy. Just this kind of process occurs within a thermos. A perfect thermos would prevent any heat from leaking out or in. However, energy in the form of heat can flow between materials inside the thermos to the extent that they have different temperatures; for example, between ice cubes and warm tea. The transfer of energy continues until a common temperature is reached at thermal equilibrium.
The kind of heat transfer that occurs within a thermos of iced tea also occurs within a calorimeter, which is the experimental apparatus used in a technique known as calorimetry. Figure 12-26 shows that, like a thermos, a calorimeter is essentially an insulated container. It can be used to determine the specific heat capacity of a substance, as the next example illustrates.

Figure 12-26 A calorimeter can be used to measure the specific heat capacity of an unknown material.

| Example 12 | | | Measuring the Specific Heat Capacity

The calorimeter cup in Figure 12-26 is made from 0.15 kg of aluminum and contains 0.20 kg of water. Initially, the water and the cup have a common temperature of 18.0 °C. A 0.040-kg mass of unknown material is heated to a temperature of 97.0 °C and then added to the water. The temperature of the water, the cup, and the unknown material is 22.0 °C after thermal equilibrium is reestablished. Ignoring the small amount of heat gained by the thermometer, find the specific heat capacity of the unknown material.
Reasoning Since energy is conserved and there is negligible heat flow between the calorimeter and the outside surroundings, the heat gained by the cold water and the aluminum cup as they warm up is equal to the heat lost by the unknown material as it cools down. Each quantity of heat can be calculated using the relation , where, in calorimetry, the change in temperature is always the higher temperature minus the lower temperature. The equation “Heat gained = Heat lost” contains a single unknown quantity, the desired specific heat capacity.

Problem solving insight In the equation “Heat gained = Heat lost,” both sides must have the same algebraic sign. Therefore, when calculating heat contributions, always write any temperature changes as the higher minus the lower temperature.

Solution

The changes in temperature for the three substances are , and . Table 12-2 contains values for the specific heat capacities of aluminum and water. Substituting these data into the equation above, we find that

Check Your Understanding 3

Consider a mass m of a material and a change in its temperature. Various possibilities for these variables are listed in the table below. Rank these possibilities in descending order (largest first), according to how much heat is needed to bring about the change in temperature. (The answer is given at the end of the book.)

| m(kg) |
(a) | 2.0 | 15
(b) | 1.5 | 40
(c) | 3.0 | 25
(d) | 2.5 | 20

Background: The concepts of heat and specific heat capacity lie at the heart of this question.
For similar questions (including calculational counterparts), consult Self-Assessment Test 12.2. This test is described at the end of Section 12.10.

Need more practice?

Interactive LearningWare12.2
Water is moving with a speed of 5.00 m/s just before it passes over the top of a waterfall. At the bottom, 5.00 m below, the water flows away with a speed of 3.00 m/s. What is the largest amount by which the temperature of the water at the bottom could exceed the temperature of the water at the top?
Related Homework: Problem 89

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.