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# 6.3: Entropy As a Macroscopic Quantity

• • Contributed by Benjamin Crowell
• Professor (Physics) at Fullerton College

# 5.3.1 Efficiency and grades of energy

Some forms of energy are more convenient than others in certain situations. You can't run a spring-powered mechanical clock on a battery, and you can't run a battery-powered clock with mechanical energy. However, there is no fundamental physical principle that prevents you from converting 100% of the electrical energy in a battery into mechanical energy or vice-versa. More efficient motors and generators are being designed every year. In general, the laws of physics permit perfectly efficient conversion within a broad class of forms of energy.

Heat is different. Friction tends to convert other forms of energy into heat even in the best lubricated machines. When we slide a book on a table, friction brings it to a stop and converts all its kinetic energy into heat, but we never observe the opposite process, in which a book spontaneously converts heat energy into mechanical energy and starts moving! Roughly speaking, heat is different because it is disorganized. Scrambling an egg is easy. Unscrambling it is harder.

We summarize these observations by saying that heat is a lower grade of energy than other forms such as mechanical energy.

Of course it is possible to convert heat into other forms of energy such as mechanical energy, and that is what a car engine does with the heat created by exploding the air-gasoline mixture. But a car engine is a tremendously inefficient device, and a great deal of the heat is simply wasted through the radiator and the exhaust. Engineers have never succeeded in creating a perfectly efficient device for converting heat energy into mechanical energy, and we now know that this is because of a deeper physical principle that is far more basic than the design of an engine. a / 1. The temperature difference between the hot and cold parts of the air can be used to extract mechanical energy, for example with a fan blade that spins because of the rising hot air currents. 2. If the temperature of the air is first allowed to become uniform, then no mechanical energy can be extracted. The same amount of heat energy is present, but it is no longer accessible for doing mechanical work.

## 5.3.2 Heat engines

Heat may be more useful in some forms than in others, i.e., there are different grades of heat energy. In figure a/1, the difference in temperature can be used to extract mechanical work with a fan blade. This principle is used in power plants, where steam is heated by burning oil or by nuclear reactions, and then allowed to expand through a turbine which has cooler steam on the other side. On a smaller scale, there is a Christmas toy, b, that consists of a small propeller spun by the hot air rising from a set of candles, very much like the setup shown in figure a.

In figure a/2, however, no mechanical work can be extracted because there is no difference in temperature. Although the air in a/2 has the same total amount of energy as the air in a/1, the heat in a/2 is a lower grade of energy, since none of it is accessible for doing mechanical work. b / A heat engine. Hot air from the candles rises through the fan blades, and makes the angels spin.

In general, we define a heat engine as any device that takes heat from a reservoir of hot matter, extracts some of the heat energy to do mechanical work, and expels a lesser amount of heat into a reservoir of cold matter. The efficiency of a heat engine equals the amount of useful work extracted, $$W$$, divided by the amount of energy we had to pay for in order to heat the hot reservoir. This latter amount of heat is the same as the amount of heat the engine extracts from the high-temperature reservoir, $$Q_H$$. (The letter $$Q$$ is the standard notation for a transfer of heat.) By conservation of energy, we have $$Q_H=W+Q_L$$, where $$Q_L$$ is the amount of heat expelled into the low-temperature reservoir, so the efficiency of a heat engine, $$W/Q_H$$, can be rewritten as

$\begin{equation*} \text{efficiency} = 1-\frac{Q_L}{Q_H} . \text{[efficiency of any heat engine]} \end{equation*}$

It turns out that there is a particular type of heat engine, the Carnot engine, which, although not 100% efficient, is more efficient than any other. The grade of heat energy in a system can thus be unambiguously defined in terms of the amount of heat energy in it that cannot be extracted even by a Carnot engine. How can we build the most efficient possible engine? Let's start with an unnecessarily inefficient engine like a car engine and see how it could be improved. The radiator and exhaust expel hot gases, which is a waste of heat energy. These gases are cooler than the exploded air-gas mixture inside the cylinder, but hotter than the air that surrounds the car. We could thus improve the engine's efficiency by adding an auxiliary heat engine to it, which would operate with the first engine's exhaust as its hot reservoir and the air as its cold reservoir. In general, any heat engine that expels heat at an intermediate temperature can be made more efficient by changing it so that it expels heat only at the temperature of the cold reservoir.

Similarly, any heat engine that absorbs some energy at an intermediate temperature can be made more efficient by adding an auxiliary heat engine to it which will operate between the hot reservoir and this intermediate temperature.

Based on these arguments, we define a Carnot engine as a heat engine that absorbs heat only from the hot reservoir and expels it only into the cold reservoir. Figures d-g show a realization of a Carnot engine using a piston in a cylinder filled with a monoatomic ideal gas. This gas, known as the working fluid, is separate from, but exchanges energy with, the hot and cold reservoirs. As proved on page 325, this particular Carnot engine has an efficiency given by

$\begin{equation*} \text{efficiency} = 1 - \frac{T_L}{T_H} , \text{[efficiency of a Carnot engine]} \end{equation*}$

where $$T_L$$ is the temperature of the cold reservoir and $$T_H$$ is the temperature of the hot reservoir.

Even if you do not wish to dig into the details of the proof, the basic reason for the temperature dependence is not so hard to understand. Useful mechanical work is done on strokes d and e, in which the gas expands. The motion of the piston is in the same direction as the gas's force on the piston, so positive work is done on the piston. d / The beginning of the first expansion stroke, in which the working gas is kept in thermal equilibrium with the hot reservoir. e / The beginning of the second expansion stroke, in which the working gas is thermally insulated. The working gas cools because it is doing work on the piston and thus losing energy.

In strokes f and g, however, the gas does negative work on the piston. We would like to avoid this negative work, but we must design the engine to perform a complete cycle. Luckily the pressures during the compression strokes are lower than the ones during the expansion strokes, so the engine doesn't undo all its work with every cycle. The ratios of the pressures are in proportion to the ratios of the temperatures, so if $$T_L$$ is 20% of $$T_H$$, the engine is 80% efficient. f / The beginning of the first compression stroke. The working gas begins the stroke at the same temperature as the cold reservoir, and remains in thermal contact with it the whole time. The engine does negative work. g / The beginning of the second compression stroke, in which mechanical work is absorbed, heating the working gas back up to $$T_H$$.

We have already proved that any engine that is not a Carnot engine is less than optimally efficient, and it is also true that all Carnot engines operating between a given pair of temperatures $$T_H$$ and $$T_L$$ have the same efficiency. (This can be proved by the methods of section 5.4.) Thus a Carnot engine is the most efficient possible heat engine.

## 5.3.3 Entropy h / Entropy can be understood using the metaphor of a water wheel. Letting the water levels equalize is like letting the entropy maximize. Taking water from the high side and putting it into the low side increases the entropy. Water levels in this metaphor correspond to temperatures in the actual definition of entropy.

We would like to have some numerical way of measuring the grade of energy in a system. We want this quantity, called entropy, to have the following two properties:

(1) Entropy is additive. When we combine two systems and consider them as one, the entropy of the combined system equals the sum of the entropies of the two original systems. (Quantities like mass and energy also have this property.)

(2) The entropy of a system is not changed by operating a Carnot engine within it.

It turns out to be simpler and more useful to define changes in entropy than absolute entropies. Suppose as an example that a system contains some hot matter and some cold matter. It has a relatively high grade of energy because a heat engine could be used to extract mechanical work from it. But if we allow the hot and cold parts to equilibrate at some lukewarm temperature, the grade of energy has gotten worse. Thus putting heat into a hotter area is more useful than putting it into a cold area. Motivated by these considerations, we define a change in entropy as follows:

$\begin{multline*} \Delta S = \frac{Q}{T} \\ {\text{[change in entropy when adding}}\ \\ {\text{heat Q to matter at temperature T;}} \ {\text{\Delta S is negative if heat is taken out]}} \end{multline*}$

A system with a higher grade of energy has a lower entropy.

Since changes in entropy are defined by an additive quantity (heat) divided by a non-additive one (temperature), entropy is additive.

Example 11: Entropy isn't changed by a Carnot engine.

The efficiency of a heat engine is defined by

$\begin{equation*} \text{efficiency} = 1 - Q_L/ Q_H , \end{equation*}$

and the efficiency of a Carnot engine is

$\begin{equation*} \text{efficiency} = 1 - T_L/ T_H , \end{equation*}$

so for a Carnot engine we have $$Q_L/ Q_H = T_L/ T_H$$, which can be rewritten as $$Q_L/ T_{L} = Q_{H}/ T_H$$. The entropy lost by the hot reservoir is therefore the same as the entropy gained by the cold one.

Example 12: Entropy increases in heat conduction.

When a hot object gives up energy to a cold one, conservation of energy tells us that the amount of heat lost by the hot object is the same as the amount of heat gained by the cold one. The change in entropy is $$- Q/ T_{H}+ Q/ T_L$$, which is positive because $$T_L\lt T_H$$.

Example 13: Entropy is increased by a non-Carnot engine.

The efficiency of a non-Carnot engine is less than 1 - $$T_L/ T_H$$, so $$Q_L/ Q_{H} > T_{L}/ T_H$$ and $$Q_L/ T_{L} > Q_{H}/ T_H$$. This means that the entropy increase in the cold reservoir is greater than the entropy decrease in the hot reservoir.

Example 14: A book sliding to a stop

A book slides across a table and comes to a stop. Once it stops, all its kinetic energy has been transformed into heat. As the book and table heat up, their entropies both increase, so the total entropy increases as well.

All of these examples involved closed systems, and in all of them the total entropy either increased or stayed the same. It never decreased. Here are two examples of schemes for decreasing the entropy of a closed system, with explanations of why they don't work.

Example 15: Using a refrigerator to decrease entropy?

$$\triangleright$$ A refrigerator takes heat from a cold area and dumps it into a hot area. (1) Does this lead to a net decrease in the entropy of a closed system? (2) Could you make a Carnot engine more efficient by running a refrigerator to cool its low-temperature reservoir and eject heat into its high-temperature reservoir?

$$\triangleright$$ (1) No. The heat that comes off of the radiator coils is a great deal more than the heat the fridge removes from inside; the difference is what it costs to run your fridge. The heat radiated from the coils is so much more than the heat removed from the inside that the increase in the entropy of the air in the room is greater than the decrease of the entropy inside the fridge. The most efficient refrigerator is actually a Carnot engine running in reverse, which leads to neither an increase nor a decrease in entropy.

(2) No. The most efficient refrigerator is a reversed Carnot engine. You will not achieve anything by running one Carnot engine in reverse and another forward. They will just cancel each other out.

Example 16: Maxwell's demon

$$\triangleright$$ Maxwell imagined a pair of rooms, their air being initially in thermal equilibrium, having a partition across the middle with a tiny door. A miniscule demon is posted at the door with a little ping-pong paddle, and his duty is to try to build up faster-moving air molecules in room B and slower moving ones in room A. For instance, when a fast molecule is headed through the door, going from A to B, he lets it by, but when a slower than average molecule tries the same thing, he hits it back into room A. Would this decrease the total entropy of the pair of rooms?

$$\triangleright$$ No. The demon needs to eat, and we can think of his body as a little heat engine, and his metabolism is less efficient than a Carnot engine, so he ends up increasing the entropy rather than decreasing it.

Observations such as these lead to the following hypothesis, known as the second law of thermodynamics:

The entropy of a closed system always increases, or at best stays the same: $$\Delta S\ge0$$.

At present our arguments to support this statement may seem less than convincing, since they have so much to do with obscure facts about heat engines. In the following section we will find a more satisfying and fundamental explanation for the continual increase in entropy. To emphasize the fundamental and universal nature of the second law, here are a few exotic examples.

Example 17: Entropy and evolution

A favorite argument of many creationists who don't believe in evolution is that evolution would violate the second law of thermodynamics: the death and decay of a living thing releases heat (as when a compost heap gets hot) and lessens the amount of energy available for doing useful work, while the reverse process, the emergence of life from nonliving matter, would require a decrease in entropy. Their argument is faulty, since the second law only applies to closed systems, and the earth is not a closed system. The earth is continuously receiving energy from the sun.

Example 18: The heat death of the universe

Living things have low entropy: to demonstrate this fact, observe how a compost pile releases heat, which then equilibrates with the cooler environment. We never observe dead things to leap back to life after sucking some heat energy out of their environments! The only reason life was able to evolve on earth was that the earth was not a closed system: it got energy from the sun, which presumably gained more entropy than the earth lost.

Victorian philosophers spent a lot of time worrying about the heat death of the universe: eventually the universe would have to become a high-entropy, lukewarm soup, with no life or organized motion of any kind. Fortunately (?), we now know a great many other things that will make the universe inhospitable to life long before its entropy is maximized. Life on earth, for instance, will end when the sun evolves into a giant star and vaporizes our planet.

Any process that could destroy heat (or convert it into nothing but mechanical work) would lead to a reduction in entropy. Black holes are supermassive stars whose gravity is so strong that nothing, not even light, can escape from them once it gets within a boundary known as the event horizon. Black holes are commonly observed to suck hot gas into them. Does this lead to a reduction in the entropy of the universe? Of course one could argue that the entropy is still there inside the black hole, but being able to “hide” entropy there amounts to the same thing as being able to destroy entropy.

The physicist Steven Hawking was bothered by this question, and finally realized that although the actual stuff that enters a black hole is lost forever, the black hole will gradually lose energy in the form of light emitted from just outside the event horizon. This light ends up reintroducing the original entropy back into the universe at large.

### Discussion Questions

◊ In this discussion question, you'll think about a car engine in terms of thermodynamics. Note that an internal combustion engine doesn't fit very well into the theoretical straightjacket of a heat engine. For instance, a heat engine has a high-temperature heat reservoir at a single well-defined temperature, $$T_H$$. In a typical car engine, however, there are several very different temperatures you could imagine using for $$T_H$$: the temperature of the engine block ($$\sim100°\text{C}$$), the walls of the cylinder ($$\sim250°\text{C}$$), or the temperature of the exploding air-gas mixture ($$\sim1000°\text{C}$$, with significant changes over a four-stroke cycle). Let's use $$T_H\sim1000°\text{C}$$.

Burning gas supplies heat energy $$Q_H$$ to your car's engine. The engine does mechanical work $$W$$, but also expels heat $$Q_L$$ into the environment through the radiator and the exhaust. Conservation of energy gives

$\begin{equation*} Q_H = Q_L+W , \end{equation*}$

and the relative proportions of $$Q_L$$ and $$W$$ are usually about 90% to 10%. (Actually it depends quite a bit on the type of car, the driving conditions, etc.)

(1) $$Q_L$$ is obviously undesirable: you pay for it, but all it does is heat the neighborhood. Suppose that engineers do a really good job of getting rid of the effects that create $$Q_L$$, such as friction. Could $$Q_L$$ ever be reduced to zero, at least theoretically?

(2) A gallon of gas releases about 140 MJ of heat $$Q_H$$ when burned. Estimate the change in entropy of the universe due to running a typical car engine and burning one gallon of gas. (You'll have to estimate how hot the environment is. For the sake of argument, assume that the work done by the engine, $$W$$, remains in the form of mechanical energy, although in reality it probably ends up being changed into heat when you step on the brakes.) Is your result consistent with the second law of thermodynamics?

(3) What would happen if you redid the calculation in #2, but assumed $$Q_L=0$$? Is this consistent with your answer to #1?

◊ When we run the Carnot engine in figures d-g, there are four parts of the universe that undergo changes in their physical states: the hot reservoir, the cold reservoir, the working gas, and the outside world to which the shaft is connected in order to do physical work. Over one full cycle, discuss which of these parts gain entropy, which ones lose entropy, and which ones keep the same entropy. During which of the four strokes do these changes occur?

## Contributors

Benjamin Crowell (Fullerton College). Conceptual Physics is copyrighted with a CC-BY-SA license.