13.5

Concepts & Calculations

Heat transfer by conduction is governed by Equation 13.1, as we have seen. Our first example illustrates a familiar application of this relation in the kitchen. It also gives us the opportunity to review the idea of latent heat of vaporization, which Section 12.8 discusses.

| Concepts & Calculations Example 9 | | | Boiling Water

Two pots are identical, except that in one case the flat bottom is aluminum and in the other it is copper. Each pot contains the same amount of boiling water and sits on a heating element that has a temperature of 155 °C. In the aluminum-bottom pot, the water boils away completely in 360 s. How long does it take the water in the copper-bottom pot to boil away completely?
Concept Questions and Answers Is the heat needed to boil away the water in the aluminum-bottom pot less than, greater than, or the same as the heat needed in the copper-bottom pot?

Answer The heat Q needed is the same in each case. When water boils, it changes from the liquid phase to the vapor phase. The heat needed to make the water boil away is Q = mLv, according to Equation 12.5, where m is the mass of the water and Lv is the latent heat of vaporization for water. Since the amount of water in each pot is the same, the mass of water is the same in each case. Moreover, the latent heat is a characteristic of water and, therefore, is also the same in each case.

One of the factors in Equation 13.1 that influences the amount of heat conducted through the bottom of each pot is the temperature difference between the upper and lower surfaces of the pot’s bottom. Is this temperature difference for the aluminum-bottom pot less than, greater than, or the same as that for the copper-bottom pot?

Answer For each pot the temperature difference is the same. At the upper surface of each pot bottom the temperature is 100.0 °C, because water boils at 100.0 °C under normal conditions of atmospheric pressure. The temperature remains at 100.0 °C until all the water is gone. For each pot the temperature at the lower surface of the pot bottom is 155 °C, the temperature of the heating element. Therefore, for each pot.

Is the time required to boil away the water completely in the copper-bottom pot less than, greater than, or the same as that required for the aluminum-bottom pot?

Answer The time is less for the copper-bottom pot. The factors that influence the amount of heat conducted in a given time are the thermal conductivity, the area of the bottom, the temperature difference across the bottom, and the thickness of the bottom. All of these factors are the same for each pot except for the thermal conductivity, which is greater for copper (Cu) than for aluminum (Al) (see Table 13-1). The greater the thermal conductivity, the greater the heat that is conducted in a given time, other things being equal. Therefore, less time is required to boil the water away using the copper-bottom pot.

Solution Applying Equation 13.1 to the conduction of heat into both pots and using Equation 12.5 to express the heat needed to boil away the water, we have

In these two equations the area A, the temperature difference , the thickness L of the pot bottom, the mass m of the water, and the latent heat of vaporization of water Lv have the same values. Therefore, we can set the two heats equal and obtain

Solving for tCu and taking values for the thermal conductivities from Table 13-1, we find

As expected, the boil-away time is less for the copper-bottom pot.

Heat transfer by conduction is only one way in which heat gets from place to place. Heat transfer by radiation is another way, and it is governed by the Stefan–Boltzmann law of radiation, as Section 13.3 discusses. Concepts & Calculations Example 10 deals with a case in which heat loss by radiation leads to freezing of water. It stresses the importance of the area from which the radiation occurs and also provides a review of the idea of latent heat of fusion, which Section 12.8 discusses.

| Concepts & Calculations Example 10 | | | Freezing Water

One half of a kilogram of liquid water at 273 K (0 °C) is placed outside on a day when the temperature is 261 K (− 12 °C). Assume that heat is lost from the water only by means of radiation and that the emissivity of the radiating surface is 0.60. How long does it take for the water to freeze into ice at 0 °C when the surface area from which the radiation occurs is (a) 0.035 m2 (as it could be in a cup) and (b) 1.5 m2 (as it could be if the water were spilled out to form a thin sheet)?
Concept Questions and Answers In case (a) is the heat that must be removed to freeze the water less than, greater than, or the same as in case (b)?

Answer The heat that must be removed is the same in both cases. When water freezes, it changes from the liquid phase to the solid phase. The heat that must be removed to make the water freeze is , according to Equation 12.5, where m is the mass of the water and Lf is the latent heat of fusion for water. The mass is the same in both cases and so is Lf, since it is a characteristic of the water.

The loss of heat by radiation depends on the temperature of the radiating object. Does the temperature of the water change as the water freezes?

Answer No. The temperature of the water does not change as the freezing process takes place. The heat removed serves only to change the water from the liquid to the solid phase, as Section 12.8 discusses. Only after all the water has frozen does the temperature of the ice begin to fall below 0 °C.

The water both loses and gains heat by radiation. How, then, can heat transfer by radiation lead to freezing of the water?

Answer The water freezes because it loses more heat by radiation than it gains. The gain occurs because the environment radiates heat and the water absorbs it. But the temperature of the environment is less than the temperature of the water. As a result, the environmental radiation is insufficient to offset completely the loss of heat due to radiation from the water.

Will it take longer for the water to freeze in case (a) when the area is smaller or in case (b) when the area is larger?

Answer It will take longer when the area is smaller. This is because the amount of energy radiated in a given time is proportional to the area from which the radiation occurs. A smaller area means that less energy is radiated per second, so more time will be required to freeze the water by removing heat via radiation.

Solution We use Equation 13.3 to take into account that the water both gains and loses heat via radiation. This expression gives the net power lost, the net power being the net heat divided by the time. Thus, we have

Using Equation 12.5 to express the heat Q as and taking the latent heat of fusion for water from Table 12-3 (), we find

a. | Smaller area

b. | Larger area

As expected, the freezing time is longer when the area is smaller.

At the end of the problem set for this chapter, you will find homework problems that contain both conceptual and quantitative parts. These problems are grouped under the heading Concepts & Calculations.

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