Contents

Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency

Photograph of melting ice floes in the Arctic.

Figure 15.15 These ice floes melt during the Arctic summer. Some of them refreeze in the winter, but the second law of thermodynamics predicts that it would be extremely unlikely for the water molecules contained in these particular floes to reform the distinctive alligator-like shape they formed when the picture was taken in the summer of 2009. (credit: Patrick Kelley, U.S. Coast Guard, U.S. Geological Survey)

The second law of thermodynamics deals with the direction taken by spontaneous processes. Many processes occur spontaneously in one direction only—that is, they are irreversible, under a given set of conditions. Although irreversibility is seen in day-to-day life—a broken glass does not resume its original state, for instance—complete irreversibility is a statistical statement that cannot be seen during the lifetime of the universe. More precisely, an irreversible process is one that depends on path. If the process can go in only one direction, then the reverse path differs fundamentally and the process cannot be reversible. For example, as noted in the previous section, heat involves the transfer of energy from higher to lower temperature. A cold object in contact with a hot one never gets colder, transferring heat to the hot object and making it hotter. Furthermore, mechanical energy, such as kinetic energy, can be completely converted to thermal energy by friction, but the reverse is impossible. A hot stationary object never spontaneously cools off and starts moving. Yet another example is the expansion of a puff of gas introduced into one corner of a vacuum chamber. The gas expands to fill the chamber, but it never regroups in the corner. The random motion of the gas molecules could take them all back to the corner, but this is never observed to happen. (See Figure 15.16.)

Part a of the figure shows spontaneous heat transfers. A rectangular section is divided down the center, and then marked hot on the left end and cold on the right. Heat Q is shown to flow from the hot end to the cold end as shown by a bold arrow toward the right. Part b of the figure shows a car moving on a horizontal road toward the right with initial velocity v. The car brake is applied after some time. The final velocity v sub f is shown equal to zero. Heat is released by the car. Part c of the figure shows two parts. The first part shows a burst of gas let into a vacuum chamber using a sprayer. The molecules of gas are shown to move in a random manner shown as dashed zigzag arrows. The second part of the same diagram shows the next stage after the air burst is sprayed. The molecules of air are shown to be arranged in uniform distribution as shown by horizontal, parallel dashed curves in the medium.

Figure 15.16 Examples of one-way processes in nature. (a) Heat transfer occurs spontaneously from hot to cold and not from cold to hot. (b) The brakes of this car convert its kinetic energy to heat transfer to the environment. The reverse process is impossible. (c) The burst of gas let into this vacuum chamber quickly expands to uniformly fill every part of the chamber. The random motions of the gas molecules will never return them to the corner.

The fact that certain processes never occur suggests that there is a law forbidding them to occur. The first law of thermodynamics would allow them to occur—none of those processes violate conservation of energy. The law that forbids these processes is called the second law of thermodynamics. We shall see that the second law can be stated in many ways that may seem different, but which in fact are equivalent. Like all natural laws, the second law of thermodynamics gives insights into nature, and its several statements imply that it is broadly applicable, fundamentally affecting many apparently disparate processes.

The already familiar direction of heat transfer from hot to cold is the basis of our first version of the second law of thermodynamics.

The Second Law of Thermodynamics (first expression)

Heat transfer occurs spontaneously from higher- to lower-temperature bodies but never spontaneously in the reverse direction.

Another way of stating this: It is impossible for any process to have as its sole result heat transfer from a cooler to a hotter object.

Heat Engines

Now let us consider a device that uses heat transfer to do work. As noted in the previous section, such a device is called a heat engine, and one is shown schematically in Figure 15.17(b). Gasoline and diesel engines, jet engines, and steam turbines are all heat engines that do work by using part of the heat transfer from some source. Heat transfer from the hot object (or hot reservoir) is denoted as $Q_{h} Q_{h} size 12{Q rSub { size 8{h} } } {}$ , while heat transfer into the cold object (or cold reservoir) is $Q_{c} Q_{c} size 12{Q rSub { size 8{c} } } {}$ , and the work done by the engine is $W W size 12{W} {}$ . The temperatures of the hot and cold reservoirs are $T_{h} T_{h} size 12{T rSub { size 8{h} } } {}$ and $T_{c} T_{c} size 12{T rSub { size 8{c} } } {}$ , respectively.

Part a of the figure shows the spontaneous heat transfer from a hot system to a cold system. The hot reservoir at temperature T sub h is represented by a rectangular section in the top and the cold reservoir at temperature T sub c is shown as a rectangular section at the bottom. Heat is shown to flow from hot reservoir to cold reservoir as shown by a bold arrow pointing downward. Part b of the figure shows a heat engine represented as a circle. The hot reservoir at temperature T sub h is represented by a rectangular section at the top and a cold reservoir at temperature T sub c is shown as a rectangular section at the bottom. Heat Q sub h is transferred out of the hot reservoir, work W is the output equals Q sub h minus Q sub c, and heat Q sub c is the heat transferred into the cold reservoir. All these are shown using bold arrows.

Figure 15.17 (a) Heat transfer occurs spontaneously from a hot object to a cold one, consistent with the second law of thermodynamics. (b) A heat engine, represented here by a circle, uses part of the heat transfer to do work. The hot and cold objects are called the hot and cold reservoirs.

Q_{h} Q_{h} size 12{Q rSub { size 8{h} } } {}

is the heat transfer out of the hot reservoir,

W W size 12{W} {}

is the work output, and

Q_{c} Q_{c} size 12{Q rSub { size 8{c} } } {}

is the heat transfer into the cold reservoir.

Because the hot reservoir is heated externally, which is energy intensive, it is important that the work is done as efficiently as possible. In fact, we would like $W W size 12{W} {}$ to equal $Q_{h} Q_{h} size 12{Q rSub { size 8{h} } } {}$ , and for there to be no heat transfer to the environment ( $Q_{c} = 0 Q_{c} = 0 size 12{Q rSub { size 8{c} } =0} {}$ ). Unfortunately, this is impossible. The second law of thermodynamics also states, with regard to using heat transfer to do work (the second expression of the second law):

The Second Law of Thermodynamics (second expression)

It is impossible in any system for heat transfer from a reservoir to completely convert to work in a cyclical process in which the system returns to its initial state.

A cyclical process brings a system, such as the gas in a cylinder, back to its original state at the end of every cycle. Most heat engines, such as reciprocating piston engines and rotating turbines, use cyclical processes. The second law, just stated in its second form, clearly states that such engines cannot have perfect conversion of heat transfer into work done. Before going into the underlying reasons for the limits on converting heat transfer into work, we need to explore the relationships among $W W size 12{W} {}$ , $Q_{h} Q_{h} size 12{Q rSub { size 8{h} } } {}$ , and $Q_{c} Q_{c} size 12{Q rSub { size 8{c} } } {}$ , and to define the efficiency of a cyclical heat engine. As noted, a cyclical process brings the system back to its original condition at the end of every cycle. Such a system’s internal energy $U U$ is the same at the beginning and end of every cycle—that is, $Δ U = 0 Δ U = 0 size 12{ΔU=0} {}$ . The first law of thermodynamics states that

Δ U = Q - W, Δ U = Q - W, size 12{ΔU=Q - W} {}

15.22

where $Q Q size 12{Q} {}$ is the net heat transfer during the cycle ( $Q = Q_{h} - Q_{c} Q = Q_{h} - Q_{c} size 12{Q=Q rSub { size 8{h} } - Q rSub { size 8{c} } } {}$ ) and $W W size 12{W} {}$ is the net work done by the system. Since $Δ U = 0 Δ U = 0 size 12{ΔU=0} {}$ for a complete cycle, we have

0 = Q - W, 0 = Q - W, size 12{0=Q - W} {}

15.23

so that

W = Q . W = Q . size 12{W=Q} {}

15.24

Thus the net work done by the system equals the net heat transfer into the system, or

W = Q_{h} - Q_{c} (cyclical process), W = Q_{h} - Q_{c} size 12{W=Q rSub { size 8{h} }  - Q rSub { size 8{c} } } {}

15.25

just as shown schematically in Figure 15.17(b). The problem is that in all processes, there is some heat transfer $Q_{c} Q_{c} size 12{Q rSub { size 8{c} } } {}$ to the environment—and usually a very significant amount at that.

In the conversion of energy to work, we are always faced with the problem of getting less out than we put in. We define conversion efficiency $Eff Eff size 12{ ital "Eff"} {}$ to be the ratio of useful work output to the energy input (or, in other words, the ratio of what we get to what we spend). In that spirit, we define the efficiency of a heat engine to be its net work output $W W size 12{W} {}$ divided by heat transfer to the engine $Q_{h} Q_{h} size 12{Q rSub { size 8{h} } } {}$ ; that is,

Eff = \frac{W}{Q_{h}} . Eff = \frac{W}{Q_{h}} . size 12{ ital "Eff"= {  {W}  over  {Q rSub { size 8{h} } } } } {}

15.26

Since $W = Q_{h} - Q_{c} W = Q_{h} - Q_{c} size 12{W=Q rSub { size 8{h} } -Q rSub { size 8{c} } } {}$ in a cyclical process, we can also express this as

Eff = \frac{Q_{h} - Q_{c}}{Q_{h}} = 1 - \frac{Q_{c}}{Q_{h}} (cyclical process), Eff = \frac{Q_{h} - Q_{c}}{Q_{h}} = 1 - \frac{Q_{c}}{Q_{h}} size 12{ ital "Eff"= {  {Q rSub { size 8{h} }  - Q rSub { size 8{c} } }  over  {Q rSub { size 8{h} } } } =1 -  {  {Q rSub { size 8{c} } }  over  {Q rSub { size 8{h} } } } } {}

15.27

making it clear that an efficiency of 1, or 100%, is possible only if there is no heat transfer to the environment ( $Q_{c} = 0 Q_{c} = 0 size 12{Q rSub { size 8{c} } =0} {}$ ). Note that all $Q Q size 12{Q} {}$ s are positive. The direction of heat transfer is indicated by a plus or minus sign. For example, $Q_{c} Q_{c} size 12{Q rSub { size 8{c} } } {}$ is out of the system and so is preceded by a minus sign.

Example 15.3

Daily Work Done by a Coal-Fired Power Station, Its Efficiency and Carbon Dioxide Emissions

A coal-fired power station is a huge heat engine. It uses heat transfer from burning coal to do work to turn turbines, which are used to generate electricity. In a single day, a large coal power station has $2 . 50 \times 10^{14} J 2 . 50 \times 10^{14} size 12{2 "." "50" times "10" rSup { size 8{"14"} } J} {}$ of heat transfer from coal and $1 . 48 \times 10^{14} J 1 . 48 \times 10^{14} size 12{1 "." "48" times "10" rSup { size 8{"14"} } J} {}$ of heat transfer into the environment. (a) What is the work done by the power station? (b) What is the efficiency of the power station? (c) In the combustion process, the following chemical reaction occurs: _{$C + O_{2} \to {CO}_{2} C + O_{2} \to {CO}_{2} size 12{C+O rSub { size 8{2} } rightarrow "CO" rSub { size 8{2} } } {}$}. This implies that every 12 kg of coal puts 12 kg + 16 kg + 16 kg = 44 kg of carbon dioxide into the atmosphere. Assuming that 1 kg of coal can provide $2 . 5 \times 10^{6} J 2 . 5 \times 10^{6} size 12{2 "." 5 times "10" rSup { size 8{6} } J} {}$ of heat transfer upon combustion, how much _{${CO}_{2} {CO}_{2} size 12{"CO" rSub { size 8{2} } } {}$} is emitted per day by this power plant?

Strategy for (a)

We can use $W = Q_{h} - Q_{c} W = Q_{h} - Q_{c} size 12{W=Q rSub { size 8{h} } - Q rSub { size 8{c} } } {}$ to find the work output $W W size 12{W} {}$ , assuming a cyclical process is used in the power station. In this process, water is boiled under pressure to form high-temperature steam, which is used to run steam turbine-generators, and then condensed back to water to start the cycle again.

Solution for (a)

Work output is given by:

W = Q_{h} - Q_{c} . W = Q_{h} - Q_{c} . size 12{W=Q rSub { size 8{h} }  - Q rSub { size 8{c} } } {}

15.28

Substituting the given values:

\begin{array}{l} W & = & 2.50 \times 10^{14} J - 1 . 48 \times 10^{14} J \\ = & 1.02 \times 10^{14} J . \end{array} \begin{array}{l} W & = & 2.50 \times 10^{14} \\ = 1.02 \times 10^{14} \end{array} alignl { stack {
 size 12{W=2 "." "50"´"10" rSup { size 8{"14"} } " J" +- 1 "." "48"´"10" rSup { size 8{"14"} } " J"}  {} #
=1 "." "02"´"10" rSup { size 8{"14"} } " J" "."  {}
} } {}

15.29

Strategy for (b)

The efficiency can be calculated with $Eff = \frac{W}{Q_{h}} Eff = \frac{W}{Q_{h}} size 12{ ital "Eff"= { {W} over {Q rSub { size 8{h} } } } } {}$ since $Q_{h} Q_{h} size 12{Q rSub { size 8{h} } } {}$ is given and work $W W size 12{W} {}$ was found in the first part of this example.

Solution for (b)

Efficiency is given by: $Eff = \frac{W}{Q_{h}} Eff = \frac{W}{Q_{h}} size 12{ ital "Eff"= { {W} over {Q rSub { size 8{h} } } } } {}$ . The work $W W$ was just found to be ${1.02 × 10}^{14} J {1.02 × 10}^{14}$ , and $Q_{h} Q_{h} size 12{Q rSub { size 8{h} } } {}$ is given, so the efficiency is

\begin{array}{l} Eff & = & \frac{1 . 02 \times 10^{14} J}{2 . 50 \times 10^{14} J} \\ = & 0.408, or 40 . 8% \end{array} \begin{array}{l} Eff & = & \frac{1 . 02 \times 10^{14}}{2 . 50 \times 10^{14}} \\ = 0.408, or 40 . 8% \end{array} alignl { stack {
 size 12{ ital "Eff"= {  {1 "." "02" times "10" rSup { size 8{"14"} } J}  over  {2 "." "50" times "10" rSup { size 8{"14"} } J} } }  {} #
=0 "." "408"", or ""40" "." 8% {}
} } {}

15.30

Strategy for (c)

The daily consumption of coal is calculated using the information that each day there is $2.50 \times 10^{14} J 2.50 \times 10^{14} size 12{2 "." "50"´"10" rSup { size 8{"14"} } " J"} {}$ of heat transfer from coal. In the combustion process, we have _{$C + O_{2} \to {CO}_{2} C + O_{2} \to {CO}_{2} size 12{C+O rSub { size 8{2} } rightarrow "CO" rSub { size 8{2} } } {}$}. So every 12 kg of coal puts 12 kg + 16 kg + 16 kg = 44 kg of _{${CO}_{2} {CO}_{2} size 12{"CO" rSub { size 8{2} } } {}$} into the atmosphere.

Solution for (c)

The daily coal consumption is

\frac{2.50 \times 10^{14} J}{2.50 \times 10^{6} J/kg} = 1 . 0 \times 10^{8} kg. \frac{2.50 \times 10^{14}}{2.50 \times 10^{6}} = 1 . 0 \times 10^{8} size 12{ {  {2 "." "50"´"10" rSup { size 8{"14"} } " J"}  over  {2 "." "50"´"10" rSup { size 8{6} } " J/kg"} } =1 "." 0´"10" rSup { size 8{7} } " J/kg"} {}

15.31

Assuming that the coal is pure and that all the coal goes toward producing carbon dioxide, the carbon dioxide produced per day is

1.0 \times 10^{8} kg coal \times \frac{{44 kg CO}_{2}}{12 kg coal} = 3 . 7 \times 10^{8} {kg CO}_{2} . 1.0 \times 10^{8} . size 12{1 "." 0´"10" rSup { size 8{7} } " kg coal"´ {  {"44 kg CO" rSub { size 8{2} } }  over  {"12 kg coal"} } =3 "." 7´"10" rSup { size 8{7} } " kg CO" rSub { size 8{2} } } {}

15.32

This is 370,000 metric tons of _{${CO}_{2} {CO}_{2} size 12{"CO" rSub { size 8{2} } } {}$} produced every day.

Discussion

If all the work output is converted to electricity in a period of one day, the average power output is 1180 MW (this is left to you as an end-of-chapter problem). This value is about the size of a large-scale conventional power plant. The efficiency found is acceptably close to the value of 42% given for coal power stations. It means that fully 59.2% of the energy is heat transfer to the environment, which usually results in warming lakes, rivers, or the ocean near the power station, and is implicated in a warming planet generally. While the laws of thermodynamics limit the efficiency of such plants—including plants fired by nuclear fuel, oil, and natural gas—the heat transfer to the environment could be, and sometimes is, used for heating homes or for industrial processes. The generally low cost of energy has not made it economical to make better use of the waste heat transfer from most heat engines. Coal-fired power plants produce the greatest amount of _{${CO}_{2} {CO}_{2} size 12{"CO" rSub { size 8{2} } } {}$} per unit energy output (compared to natural gas or oil), making coal the least efficient fossil fuel.

With the information given in Example 15.3, we can find characteristics such as the efficiency of a heat engine without any knowledge of how the heat engine operates, but looking further into the mechanism of the engine will give us greater insight. Figure 15.18 illustrates the operation of the common four-stroke gasoline engine. The four steps shown complete this heat engine’s cycle, bringing the gasoline-air mixture back to its original condition.

The Otto cycle shown in Figure 15.19(a) is used in four-stroke internal combustion engines, although in fact the true Otto cycle paths do not correspond exactly to the strokes of the engine.

The adiabatic process AB corresponds to the nearly adiabatic compression stroke of the gasoline engine. In both cases, work is done on the system (the gas mixture in the cylinder), increasing its temperature and pressure. Along path BC of the Otto cycle, heat transfer $Q_{h} Q_{h} size 12{Q rSub { size 8{h} } } {}$ into the gas occurs at constant volume, causing a further increase in pressure and temperature. This process corresponds to burning fuel in an internal combustion engine, and takes place so rapidly that the volume is nearly constant. Path CD in the Otto cycle is an adiabatic expansion that does work on the outside world, just as the power stroke of an internal combustion engine does in its nearly adiabatic expansion. The work done by the system along path CD is greater than the work done on the system along path AB, because the pressure is greater, and so there is a net work output. Along path DA in the Otto cycle, heat transfer $Q_{c} Q_{c} size 12{Q rSub { size 8{c} } } {}$ from the gas at constant volume reduces its temperature and pressure, returning it to its original state. In an internal combustion engine, this process corresponds to the exhaust of hot gases and the intake of an air-gasoline mixture at a considerably lower temperature. In both cases, heat transfer into the environment occurs along this final path.

The net work done by a cyclical process is the area inside the closed path on a $PV PV size 12{ ital "PV"} {}$ diagram, such as that inside path ABCDA in Figure 15.19. Note that in every imaginable cyclical process, it is absolutely necessary for heat transfer from the system to occur in order to get a net work output. In the Otto cycle, heat transfer occurs along path DA. If no heat transfer occurs, then the return path is the same, and the net work output is zero. The lower the temperature on the path AB, the less work has to be done to compress the gas. The area inside the closed path is then greater, and so the engine does more work and is thus more efficient. Similarly, the higher the temperature along path CD, the more work output there is. (See Figure 15.20.) So efficiency is related to the temperatures of the hot and cold reservoirs. In the next section, we shall see what the absolute limit to the efficiency of a heat engine is, and how it is related to temperature.

The figure shows four diagrams a, b, c, and d representing four stages of a four stroke gasoline engine. The construction of the engine has the base chamber whose cross section is in the shape of a square with flat corners, the top portion of the chamber is extended into a cylindrical section. The cylindrical section ends in the upper section with two valves, an inlet and an outlet. The cylindrical section has a movable cylinder with a piston attached to it. The piston is connected to the crank shaft in the base gas chamber. There is a spark plug on the top most part of the cylinder between the two valves. The four parts of the diagram show various stages of this four stoke engine. Part a of the diagram shows the air fuel mixture enters through the inlet valve in the upper section of the engine. The outlet valve is shown to be closed. The air and fuel is shown to exert a pressure on the piston acting downward. This force is shown to move the rotating crank shaft in clockwise direction in the gas chamber. This is the intake stroke. Part b of the diagram shows the compression stroke. Both the inlet and outlet valves are closed. The air and fuel mixture is shown to be compressed. The piston is shown to rise up as shown by a vertically pointing arrow. The piston is at the edge of the cylinder near the valves. The crankshaft in the gas chamber has shown to complete one complete cycle of rotation in the gas chamber. Part c of the diagram shows the power stroke. It has two parts, first the ignition stroke. This shows ignition of the fuel in the cylinder and pressure buildup in the region. Then in the second part the piston is shown to descend down the cylinder moving the crankshaft in the gas chamber in the clockwise direction. Part d of the figure shows the exhaust stroke. The piston expels the hot gas by moving upward and the gas is expelled through the exhaust valve.

Figure 15.18 In the four-stroke internal combustion gasoline engine, heat transfer into work takes place in the cyclical process shown here. The piston is connected to a rotating crankshaft, which both takes work out of and does work on the gas in the cylinder. (a) Air is mixed with fuel during the intake stroke. (b) During the compression stroke, the air-fuel mixture is rapidly compressed in a nearly adiabatic process, as the piston rises with the valves closed. Work is done on the gas. (c) The power stroke has two distinct parts. First, the air-fuel mixture is ignited, converting chemical potential energy into thermal energy almost instantaneously, which leads to a great increase in pressure. Then the piston descends, and the gas does work by exerting a force through a distance in a nearly adiabatic process. (d) The exhaust stroke expels the hot gas to prepare the engine for another cycle, starting again with the intake stroke.

Part a of the figure shows a graph of pressure P versus volume V for an Otto cycle. The pressure P is along the Y axis and the volume V is along the X axis. The graph shows a complete cycle A B C D. The path begins at point A; the curve rises upward from point A to point B along the direction of the negative X axis. This is marked as an adiabatic process. Then the curve rises vertically up from point B to point C in a direction perpendicular to the X axis. Then the curve moves smoothly down to point D along the direction of the positive X axis. This is also marked as an adiabatic process. The last part of the curve drops vertically down from point D back to point A. The path A B is slightly lower than path C D. Heat Q sub h is shown to enter the system as shown by a bold arrow to the curve B C. Heat Q sub c is shown to leave the system as shown by a bold arrow near D A. The area inside A B C D is shaded, and the area of the shaded region is shown proportional to the work W. Part b of the diagram shows an internal combustion engine represented as a circle. The hot reservoir is a rectangular section at the top of the circle shown at temperature T sub h. A cold reservoir is shown as a rectangular section in the bottom part of the circle at temperature T sub c. Heat Q sub h is shown to enter the heat engine, as shown by a bold arrow. Work W is produced as output, shown to leave the system, and the remaining heat Q sub c is returned back to the cold reservoir as shown by bold arrow toward it.

Figure 15.19

PV PV size 12{ ital "PV"} {}

diagram for a simplified Otto cycle, analogous to that employed in an internal combustion engine. Point A corresponds to the start of the compression stroke of an internal combustion engine. Paths AB and CD are adiabatic and correspond to the compression and power strokes of an internal combustion engine, respectively. Paths BC and DA are isochoric and accomplish similar results to the ignition and exhaust-intake portions, respectively, of the internal combustion engine’s cycle. Work is done on the gas along path AB, but more work is done by the gas along path CD, so that there is a net work output.

Figure 15.20 This Otto cycle produces a greater work output than the one in Figure 15.19, because the starting temperature of path CD is higher and the starting temperature of path AB is lower. The area inside the loop is greater, corresponding to greater net work output.