15.11

Entropy

A Carnot engine has the maximum possible efficiency for its operating conditions because the processes occurring within it are reversible. Irreversible processes, such as friction, cause real engines to operate at less than maximum efficiency, for they reduce our ability to use heat to perform work. As an extreme example, imagine that a hot object is placed in thermal contact with a cold object, so heat flows spontaneously, and hence irreversibly, from hot to cold. Eventually both objects reach the same temperature, and TC = TH. A Carnot engine using these two objects as heat reservoirs is unable to do work, because the efficiency of the engine is zero []. In general, irreversible processes cause us to lose some, but not necessarily all, of the ability to perform work. This partial loss can be expressed in terms of a concept called entropy.
To introduce the idea of entropy we recall the relation (Equation 15.14) that applies to a Carnot engine. It is possible to rearrange this equation as , which focuses attention on the heat Q divided by the Kelvin temperature T. The quantity Q/T is called the change in the entropy :

| (15.18)

In this expression the temperature T must be in kelvins, and the subscript R refers to the word “reversible.” It can be shown that Equation 15.18 applies to any process in which heat enters (Q is positive) or leaves (Q is negative) a system reversibly at a constant temperature. Such is the case for the heat that flows into and out of the reservoirs of a Carnot engine. Equation 15.18 indicates that the SI unit for entropy is a joule per kelvin (J/K).
Entropy, like internal energy, is a function of the state or condition of the system. Only the state of a system determines the entropy S that a system has. Therefore, the change in entropy is equal to the entropy of the final state of the system minus the entropy of the initial state.
We can now describe what happens to the entropy of a Carnot engine. As the engine operates, the entropy of the hot reservoir decreases, since heat of magnitude |QH| departs reversibly at a Kelvin temperature TH. The corresponding change in the entropy is , where the minus sign is needed to indicate a decrease, since the symbol |QH| denotes only the magnitude of the heat. In contrast, the entropy of the cold reservoir increases by an amount , for the rejected heat reversibly enters the cold reservoir at a Kelvin temperature TC. The total change in entropy is

because according to Equation 15.14.
The fact that the total change in entropy is zero for a Carnot engine is a specific illustration of a general result. It can be proved that when any reversible process occurs, the change in the entropy of the universe is zero; for a reversible process. The word “universe” means that takes into account the entropy changes of all parts of the system and all parts of the environment. Reversible processes do not alter the total entropy of the universe. To be sure, the entropy of one part of the universe may change because of a reversible process, but if so, the entropy of another part changes in the opposite way by the same amount.
What happens to the entropy of the universe when an irreversible process occurs is more complex, because the expression does not apply directly. However, if a system changes irreversibly from an initial state to a final state, this expression can be used to calculate indirectly, as Figure 15-19 indicates. We imagine a hypothetical reversible process that causes the system to change between the same initial and final states and then find for this reversible process. The value obtained for also applies to the irreversible process that actually occurs, since only the nature of the initial and final states, and not the path between them, determines . Example 11 illustrates this indirect method and shows that spontaneous (irreversible) processes increase the entropy of the universe.

Figure 15-19 Although the relation applies to reversible processes, it can be used as part of an indirect procedure to find the entropy change for an irreversible process. This drawing illustrates the procedure discussed in the text.

| Example 11 | | | The Entropy of the Universe Increases

Figure 15-20 shows 1200 J of heat flowing spontaneously through a copper rod from a hot reservoir at 650 K to a cold reservoir at 350 K. Determine the amount by which this irreversible process changes the entropy of the universe, assuming that no other changes occur.

Figure 15-20 Heat flows spontaneously from a hot reservoir to a cold reservoir.

Reasoning The hot-to-cold heat flow is irreversible, so the relation is applied to a hypothetical process whereby the 1200 J of heat is taken reversibly from the hot reservoir and added reversibly to the cold reservoir.
Solution The total entropy change of the universe is the algebraic sum of the entropy changes for each reservoir:

The irreversible process causes the entropy of the universe to increase by 1.6 J/K.

Example 11 is a specific illustration of a general result: Any irreversible process increases the entropy of the universe. In other words, for an irreversible process. Reversible processes do not alter the entropy of the universe, whereas irreversible processes cause the entropy to increase. Therefore, the entropy of the universe continually increases, like time itself, and entropy is sometimes called “time’s arrow.” It can be shown that this behavior of the entropy of the universe provides a completely general statement of the second law of thermodynamics, which applies not only to heat flow but also to all kinds of other processes.

| The Second Law of Thermodynamics Stated in Terms of Entropy
| The total entropy of the universe does not change when a reversible process occurs () and increases when an irreversible process occurs ().

When an irreversible process occurs and the entropy of the universe increases, the energy available for doing work decreases, as the next example illustrates.

| Example 12 | | | Energy Unavailable for Doing Work

Suppose that 1200 J of heat is used as input for an engine under two different conditions. In Figure 15-21a the heat is supplied by a hot reservoir whose temperature is 650 K. In part b of the drawing, the heat flows irreversibly through a copper rod into a second reservoir whose temperature is 350 K and then enters the engine. In either case, a 150-K reservoir is used as the cold reservoir. For each case, determine the maximum amount of work that can be obtained from the 1200 J of heat.

Figure 15-21 Heat in the amount of is used as input for an engine under two different conditions in parts a and b.

Reasoning According to Equation 15.11, the work () obtained from the engine is the product of its efficiency e and the input heat (), or . For a given input heat, the maximum amount of work is obtained when the efficiency is a maximum—that is, when the engine is a Carnot engine. The efficiency of a Carnot engine is given by Equation 15.15 as . Therefore, the efficiency may be determined from the Kelvin temperatures of the hot and cold reservoirs.
Solution

When the 1200 J of input heat is taken from the 350-K reservoir instead of the 650-K reservoir, the efficiency of the Carnot engine is smaller. As a result, less work (680 J versus 920 J) can be extracted from the input heat.

Example 12 shows that 240 J less work (920 J – 680 J) can be performed when the input heat is obtained from the hot reservoir with the lower temperature. In other words, the irreversible process of heat flow through the copper rod causes energy to become unavailable for doing work in the amount of . Example 11 shows that this irreversible process simultaneously causes the entropy of the universe to increase by an amount . These values for and are in fact related. If you multiply by 150 K, which is the lowest Kelvin temperature in Example 12, you obtain . This illustrates the following general result:

| (15.19)

where T0 is the Kelvin temperature of the coldest heat reservoir. Since irreversible processes cause the entropy of the universe to increase, they cause energy to be degraded, in the sense that part of the energy becomes unavailable for the performance of work. In contrast, there is no penalty when reversible processes occur, because for them , and there is no loss of work.
Entropy can also be interpreted in terms of order and disorder. As an example, consider a block of ice (Figure 15-22) with each of its H2O molecules fixed rigidly in place in a highly structured and ordered arrangement. In comparison, the puddle of water into which the ice melts is disordered and unorganized, because the molecules in a liquid are free to move from place to place. Heat is required to melt the ice and produce the disorder. Moreover, heat flow into a system increases the entropy of the system, according to . We associate an increase in entropy, then, with an increase in disorder. Conversely, we associate a decrease in entropy with a decrease in disorder or a greater degree of order. Example 13 illustrates an order-to-disorder change and the increase of entropy that accompanies it.

Figure 15-22 A block of ice is an example of an ordered system relative to a puddle of water.

| Example 13 | | | Order to Disorder

Find the change in entropy that results when a 2.3-kg block of ice melts slowly (reversibly) at 273 K (0 °C). Reasoning Since the phase change occurs reversibly at a constant temperature, the change in entropy can be found by using Equation 15.18, , where Q is the heat absorbed by the melting ice. This heat can be determined by using the relation (Equation 12.5), where m is the mass and is the latent heat of fusion of water (see Table 12.3).
Solution Using Equation 15.18 and Equation 12.5, we find that the change in entropy is

a result that is positive, since the ice absorbs heat as it melts.

Figure 15-23 shows another order-to-disorder change that can be described in terms of entropy.

Figure 15-23 With the aid of explosives, demolition experts caused the Kingdome in Seattle, Washington, to go from the ordered state (lower entropy) shown in the top photograph to the disordered state (higher entropy) shown in the bottom photograph. (Anthony Bolante/Reuters/Landov LLC)

Check Your Understanding 4

Two equal amounts of water are mixed together in an insulated container. The initial temperatures of the water are different, but the mixture reaches a uniform temperature. Do the energy and the entropy of the water increase, decrease, or remain constant as a result of the mixing process? (The answer is given at the end of the book.)

| Energy of the Water | Entropy of the Water
a. | Increases | Increases
b. | Decreases | Decreases
c. | Remains constant | Decreases
d. | Remains constant | Increases
e. | Remains constant | Remains constant

Background: This question focuses on energy, entropy, and the important idea of reversible versus irreversible processes.
For similar questions (including calculational counterparts), consult Self-Assessment Test 15.2. This test is described next.

Test your understanding of the central ideas presented in Sections 15.7–15.11:

· | The Second Law of Thermodynamics
· | Refrigerators, Air Conditioners, and Heat Pumps
· | Heat Engines and the Carnot Engine
· | Entropy

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