15.10

Refrigerators, Air Conditioners, and Heat Pumps

The natural tendency of heat is to flow from hot to cold, as indicated by the second law of thermodynamics. However, if work is used, heat can be made to flow from cold to hot, against its natural tendency. Refrigerators, air conditioners, and heat pumps are, in fact, devices that do just that. As Figure 15-14 illustrates, these devices use work () to extract heat () from the cold reservoir and deposit heat () into the hot reservoir. Generally speaking, such a process is called a refrigeration process. A comparison of this drawing with Figure 15-13 shows that the directions of the arrows symbolizing heat and work in a refrigeration process are opposite to those in an engine process. Nonetheless, energy is conserved during a refrigeration process, just as it is in an engine process, so . Moreover, if the process occurs reversibly, we have ideal devices that are called Carnot refrigerators, Carnot air conditioners, and Carnot heat pumps. For these ideal devices, the relation (Equation 15.14) applies, just as it does for a Carnot engine.

Figure 15-14 In the refrigeration process on the left, work |W| is used to remove heat |QC| from the cold reservoir and deposit heat |QH| into the hot reservoir. Compare this with the engine process on the right.

The physics of refrigerators. In a refrigerator, the interior of the unit is the cold reservoir, while the warmer exterior is the hot reservoir. As Figure 15-15 illustrates, the refrigerator takes heat from the food inside and deposits it into the kitchen, along with the energy needed to do the work of making the heat flow from cold to hot. For this reason, the outside surfaces (usually the sides and back) of most refrigerators are warm to the touch while the units operate. Thus, a refrigerator warms the kitchen.

Figure 15-15 A refrigerator.

The physics of air conditioners. An air conditioner is like a refrigerator, except that the room itself is the cold reservoir and the outdoors is the hot reservoir. Figure 15-16 shows a window unit, which cools a room by removing heat and depositing it outside, along with the work used to make the heat flow from cold to hot. Conceptual Example 9 considers a common misconception about refrigerators and air conditioners.

Figure 15-16 A window air conditioner removes heat from a room, which is the cold reservoir, and deposits heat outdoors, which is the hot reservoir.

| Conceptual Example 9 | | | You Can’t Beat the Second Law of Thermodynamics

Is it possible to cool your kitchen by leaving the refrigerator door open or cool your bedroom by putting a window air conditioner on the floor by the bed? Reasoning and Solution The heat () removed from the air directly in front of the open refrigerator is deposited back into the kitchen at the rear of the unit. Moreover, according to the second law, work () is needed to move that heat from cold to hot, and the energy from this work is also deposited into the kitchen as additional heat. Thus, the open refrigerator puts into the kitchen an amount of heat , which is more than it removes. Rather than cooling the kitchen, the open refrigerator warms it up. Putting a window air conditioner on the floor to cool your bedroom is similarly a no-win game. The heat pumped out the back of the air conditioner and into the bedroom is greater than the heat pulled into the front of the unit. Consequently, the air conditioner actually warms the bedroom.
Related Homework: Problem 63

Need more practice?

Interactive LearningWare15.2
The water in a glass on a table has a mass of 0.25 kg (about eight liquid ounces). It has the same temperature as the room, which is 27 °C. A Carnot refrigerator is operating in this room. The water is put in the refrigerator, the inside of which is maintained at 3 °C. How much work is done by the refrigerator in cooling the water from 27 to 3 °C?
Related Homework: Problem 86

The quality of a refrigerator or air conditioner is rated according to its coefficient of performance. Such appliances perform well when they remove a relatively large amount of heat () from a cold reservoir by using as small an amount of work () as possible. Therefore, the coefficient of performance is defined as the ratio of |QC| to |W|, and the greater this ratio is, the better the performance is:

| (15.16)

Commercially available refrigerators and air conditioners have coefficients of performance in the range 2 to 6, depending on the temperatures involved. The coefficients of performance for these real devices are less than those for ideal, or Carnot, refrigerators and air conditioners.

The physics of heat pumps.

In a sense, refrigerators and air conditioners operate like pumps. They pump heat “uphill” from a lower temperature to a higher temperature, just as a water pump forces water uphill from a lower elevation to a higher elevation. It would be appropriate to call them heat pumps. However, the name “heat pump” is reserved for the device illustrated in Figure 15-17, which is a home heating appliance. The heat pump uses work |W| to make heat |QC| from the wintry outdoors (the cold reservoir) flow up the temperature “hill” into a warm house (the hot reservoir). According to the conservation of energy, the heat pump deposits inside the house an amount of heat . The air conditioner and the heat pump do closely related jobs. The air conditioner refrigerates the inside of the house and heats up the outdoors, while the heat pump refrigerates the outdoors and heats up the inside. These jobs are so closely related that most heat pump systems serve in a dual capacity, being equipped with a switch that converts them from heaters in the winter into air conditioners in the summer.

Figure 15-17 In a heat pump the cold reservoir is the wintry outdoors, and the hot reservoir is the inside of the house.

Heat pumps are popular for home heating in today’s energy-conscious world, and it is easy to understand why. Suppose 1000 J of energy is available for home heating. Figure 15-18 shows that a conventional electric heating system uses this 1000 J to heat a coil of wire, just as in a toaster. A fan blows air across the hot coil, and forced convection carries the 1000 J of heat into the house. In contrast, the heat pump in Figure 15-17 does not use the 1000 J directly as heat. Instead, it uses it to do the work |W| of pumping heat |QC| from the cooler outdoors into the warmer house and, in so doing, delivers an amount of energy . With , this becomes , so that the heat pump delivers more than 1000 J of heat, whereas the conventional electric heating system delivers only 1000 J. The next example shows how the basic relations and are used with heat pumps.

Figure 15-18 This conventional electric heating system is delivering 1000 J of heat to the living room.

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 10 | | | A Heat Pump

An ideal, or Carnot, heat pump is used to heat a house to a temperature of 294 K (21 °C). How much work must the pump do to deliver 3350 J of heat into the house on a day when the outdoor temperature is 273 K (0 °C) and on another day when the outdoor temperature is 252 K (−21 °C)?
Reasoning The conservation of energy dictates that the heat delivered into the house (the hot reservoir) equals the energy from the work done by the heat pump plus the energy in the form of heat taken from the cold outdoors (the cold reservoir). The heat delivered into the house is given, so that we can use energy conservation to determine the work, provided that we can obtain a value for the heat taken from the outdoors. We will obtain this value by using Equation 15.14, which relates the ratio of the magnitudes of the heats for the cold and hot reservoirs to the ratio of the reservoir temperatures (in kelvins).
Knowns and Unknowns We have the following data:

Description | Symbol | Value | Comment
Temperature of hot reservoir (interior of house) | TH | 294 K | Temperature in kelvins must be used.
Temperature of cold reservoir (outdoors) | TC | 273 K or 252 K | Temperature in kelvins must be used.
Magnitude of heat delivered into house | |QH| | 3350 J |
Unknown Variable
Magnitude of work done by pump | |W| | ? |

Modeling the Problem

| Step 1 The Conservation of Energy
| The energy-conservation principle requires that (Equation 15.12), where |QH|, |W|, and |QC| are, respectively, the magnitudes of the heat delivered into the house (the hot reservoir), the work done by the heat pump, and the heat taken from the cold outdoors (the cold reservoir). Solving for |W| gives Equation 1 at the right. In this result, we have a value for |QH| but not for |QC|. Step 2 deals with this missing information.

| (1)

| Step 2 Heat and Thermodynamic Temperature
| To obtain a value for |QC|, we will use the information given about the temperatures TH for the hot reservoir and TC for the cold reservoir. According to Equation 15.14, , which can be solved for |QC|:

At the right, we substitute this result into Equation 15.1.

Solution Combining the results of each step algebraically, we find that

It follows that the magnitude of the work for the two given outdoor temperatures is

More work must be done when the outdoor temperature is lower, because the heat is pumped up a greater temperature “hill.”
Related Homework: Problem 58

Problem solving insight When applying Equation 15.14 () to heat pumps, refrigerators, or air conditioners, be sure the temperatures TC and TH are expressed in kelvins; degrees Celsius or degrees Fahrenheit will not do.

It is also possible to specify a coefficient of performance for heat pumps. However, unlike refrigerators and air conditioners, the job of a heat pump is to heat, not to cool. As a result, the coefficient of performance of a heat pump is the ratio of the magnitude of the heat |QH| delivered into the house to the magnitude of the work |W| required to deliver it:

| (15.17)

The coefficient of performance depends on the indoor and outdoor temperatures. Commercial units have coefficients of about 3 to 4 under favorable conditions.

Check Your Understanding 3

Each drawing represents a hypothetical heat engine or a hypothetical heat pump and shows the corresponding heats and work. Only one is allowed in nature. Which is it? (The answer is given at the end of the book).

Background: The operation of any device must obey both the first and second laws of thermodynamics. One is a statement about the conservation of energy, and the other about the direction of heat flow.
For similar questions (including calculational counterparts), consult Self-Assessment Test 15.2. This test is described at the end of Section 15.11.

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