3. The heat added to a system during a thermodynamic process depends on the path taken. Even with the same initial and fi nal states, two different thermo- dynamic paths can have different values of Q.
Properties of U, W, and Q for thermodynamic processes Properties of U, W, and Q for
thermodynamic processes
E X A M P L E 1 6 . 6 Analyzing a Cyclic Process
Consider the thermodynamic process described in Figure 16.21. The P–V diagram is similar to the one in Figure 16.20 except that now we begin and end at the same
point (the same state). The system in Figure 16.21 starts in state 1, moves as indicated to state 2, state 3, and state 4, and then moves back to state 1. This set of changes is called a cyclic process because the system returns to its initial state. Find the work done during this process.
RECOGNIZE T HE PRINCIPLE
We can use W P V (Eq. 16.6) to fi nd the work done along each portion of the path.
Adding those results gives the total work done for the entire process.
SK E TCH T HE PROBLEM
Figure 16.21 shows the P–V diagram.
IDENT IF Y T HE REL AT IONSHIPS
The volume does not change along the path from state 1 to state 2 or along the path from state 3 to state 4, so V 0 and the work done in these parts of the process is zero. Using W P V for the path from state 2 to state 3 gives
W235P DV5P31V32V12 Likewise, along the path from state 4 to state 1, we get
W415P11V12V32 Notice that W41 is negative because V1 V3.
SOLV E
Combining these results gives
Wtotal5W231W415P31V32V12 1P11V12V32 5 1P32P12 1V32V12
What does it mean?
From the geometry of Figure 16.21, Wtotal is just the area enclosed by the path in the P–V diagram.
Work Done by a Cyclic Process
Example 16.6 considered the work done in a cyclic process and showed that W is equal to the area enclosed by the path in the P–V diagram. The process analyzed in that example had an especially simple path, but the general result applies to all cyclic processes. The work done in any cyclic process equals the area enclosed by the path in the corresponding P–V diagram (Fig. 16.22). We’ll use this result when we analyze the performance of heat engines and other devices in Section 16.6.
CO N C E P T C H E C K 16.6 | Thermodynamic Processes
Figure 16.23 shows several thermodynamic processes. Which of these processes could be (a) an isothermal expansion, (b) an adiabatic compression, (c) an iso- baric expansion, and (d) an isobaric compression?
Figure 16.21 Example 16.6. A cyclic process is one that returns to its original state.
V1 P3
P1
V3
2 3
1 4
Wtotal area P
V
Figure 16.22 In any cyclic thermodynamic process, the work done by the system equals the area enclosed by the path in the P–V diagram.
P
Wtotal
V i
Figure 16.23 Concept Check 16.6.
Process 1 Process 3
P
V Process 2
P
V i
f
f
i
i f f i
P
V
Process 4 P
V
16.4 | THERMODYNAMIC PROCESSES 507
1 6 . 5 | R E V E R S I B L E A N D I R R E V E R S I B L E P R O C E S S E S A N D T H E S E CO N D L A W O F T H E R M O D Y N A M I C S
In our discussions of thermodynamic processes, we have focused so far on changes in the system. It is sometimes important to also consider what happens to the envi- ronment. Figure 16.24A shows an isothermal expansion of a dilute gas in which the system starts in an initial state i and moves to a fi nal state f. The piston moves to the right as the gas does some amount of work W 0, and during this time, it absorbs some heat Q 0 from the environment. This example shows a thermodynamically reversible process. We could imagine a reversed process in which the gas begins in state f and is isothermally compressed back to state i (Fig. 16.24B). During this compression, the gas does a negative amount of work (W 0) on the piston because an outside agent must push on the piston to compress the gas. At the same time, heat would fl ow out of the gas into the surroundings. The work done by the gas during the compression is equal in magnitude but opposite in sign compared with W during the expansion. Also, the amount of heat that fl ows out of the gas dur- ing compression is equal in magnitude to the heat that fl ows in during expansion.
Hence, both the system and the environment are brought back to precisely their original states, which is the key feature of a thermodynamically reversible process.
An example of an irreversible process is sketched in Figure 16.25. Here a hockey puck is initially sliding to the right on a horizontal surface. There is some friction between the puck and the surface, causing the puck to come to a stop as all its initial kinetic energy is converted to heat energy. This heat energy fl ows from the region where the puck is in contact with the horizontal surface into its surroundings (Fig.
16.25A). Suppose we now try to reverse this process by arranging for heat to fl ow from the surroundings into the puck, perhaps with a torch as sketched in Figure 16.25B. Intuition tells us that this will not cause the puck to begin moving! Of course, we could start the puck in motion again by striking it with a hockey stick, but for a process to be thermodynamically reversible, it must be possible to return both the system (the puck) and its surroundings (the rest of the universe) back to their original states. There is nothing we could do to reverse the process in Figure 16.25; it is simply not possible to take the puck and the rest of the universe back to precisely their original states. This is the essential feature of an irreversible process.
And, as in Figure 16.25, irreversible processes usually involve friction.
The distinction between reversible and irreversible processes is the subject of the second law of thermodynamics. There are several different ways of stating the second law of thermodynamics; all are equivalent, but because they emphasize dif- ferent aspects of the second law, it is useful to discuss more than just one of them.
Perhaps the simplest version is given as follows.
Second law of thermodynamics
Heat fl ows spontaneously from a warm body to a colder one. It is not possible for heat to fl ow spontaneously from a cold body to a warmer one.
Reversible versus irreversible processes Reversible versus irreversible processes
Second law of thermodynamics Second law of thermodynamics
Figure 16.25 A A hockey puck sliding across a fl oor with friction is an example of a thermodynami- cally irreversible process. B Heat- ing a hockey puck will not cause it to move back to its original position!
Torch
v v 0
Q
Q A
B
Figure 16.24 Schematic of a thermodynamically reversible process. A Isothermal expansion from state i to state f.
B Isothermal compression from state f back to state i.
Isothermal expansion Isothermal compression
Q 0 Q 0
W 0 W 0
V V P P
i
f i
f
A B
This statement certainly seems obvious. The part about heat fl owing from a warm object to a colder one is implied by the zeroth law of thermodynamics. Two crucial aspects of the second law go beyond the zeroth law, however. The fi rst is the notion of a spontaneous process. Such a process will essentially take place “on its own,” without work done by any added or external forces. For example, the fl ow of heat from the surroundings into the ice cube in Figure 16.8 is a spontaneous process. The other crucial part of the second law of thermodynamics is the state- ment that heat cannot fl ow spontaneously from a cold object to a warmer one. This statement leads to many important results, including the impossibility of building a perpetual motion machine.
1 6 . 6 | H E AT E N G I N E S A N D O T H E R T H E R M O D Y N A M I C D E V I C E S
One of the discoverers of the second law of thermodynamics was an engineer named Sadi Carnot (1796–1832), who was interested in how the science of thermodynam- ics could be applied to practical devices such as steam engines and other marvels of the Industrial Revolution that were being developed around that time. Carnot’s engineering approach leads to a rather different way of thinking about the second law of thermodynamics.
Imagine you are an engineer and you have been asked to invent the world’s best heat engine. A heat engine takes heat energy, as might be generated by a coal- burning furnace, and converts this energy into work. For example, this work might be in the form of a rotating shaft as would be useful in a steamboat, locomotive, or automobile. With that in mind, we can draw a heat engine in the highly sche- matic form shown in Figure 16.26A. An amount of heat QH is extracted from the furnace and fed into some sort of mechanical device, which then does an amount of work W. It is your engineering challenge to design the mechanical device. The principle of conservation of energy and the fi rst law of thermodynamics tell us that the work done cannot be larger than the energy input QH, but you might hope to design and patent a device that reaches this limit. The world’s best heat engine would perform an amount of work W QH, and from the principle of conservation of energy, you would be absolutely guaranteed that no one could design an engine better than that one.
That is the problem Carnot tackled. Although he was not able to construct the heat engine in Figure 16.26A, Carnot was able to show that such an engine is impossible! He showed that it is simply not possible to construct a device whose sole effect is to convert heat energy into work, that is, to convert a given amount of heat energy completely into mechanical energy. Instead, all heat engines must be diagrammed as in Figure 16.26B. Here we again have a mechanical device that takes an amount of heat energy QH from a hot reservoir and does an amount of
16.6 | HEAT ENGINES AND OTHER THERMODYNAMIC DEVICES 509 W
QC QH
QH W Imaginary heat engine
(NOT possible) W
QH
(From furnace) Hot reservoir, TH
Cold reservoir, TC Real heat engines all
expel some heat to a cold reservoir.
A B
Heat engine
Heat engine
Figure 16.26 Schematic diagrams for a heat engine.
A A hypothetical engine, which the laws of thermodynamics tell us is not possible. B All real heat engines must expel some heat to a cold reservoir.
work W. However, Carnot showed that all real heat engines must also expel some heat energy QC to a colder reservoir.
Applying conservation of energy to the heat engine in Figure 16.26B, we have W5QH2QC (16.17)
Your goal as an engine designer is get the most work (the largest W) possible out of the engine, so we defi ne the effi ciency e of the engine as
e5 W
QH (16.18) Combining Equation 16.18 with Equation 16.17 gives
e5 W
QH 5 QH2QC
QH
e512 QC
QH (16.19) When comparing heat engines, a larger value of e means that the engine converts a larger fraction of the input energy QH into work.
Carnot’s Engine
To maximize the effi ciency e, one must make QC as small as possible, which makes W as large as possible. Carnot considered this problem for a heat engine that makes use of the reversible compression and expansion of an ideal gas. Because Carnot’s engine uses only reversible processes, it is an example of a reversible heat engine.
He was able to prove that all reversible heat engines have the same effi ciency as his particular design. Carnot also showed that for a hot reservoir at temperature TH (providing the heat input QH) and a cold reservoir at temperature TC, the heat expelled to the cold reservoir satisfi es
QC QH5 TC
TH (16.20) Here the temperatures TH and TC must be measured in Kelvin units (not degrees Celsius). Combining Equation 16.20 with the defi nition of effi ciency (Eq. 16.19) leads to
e512 TC
TH (16.21) Equation 16.21 is an amazing result: it applies to all reversible engines, no mat- ter how they are constructed. Carnot did not stop here, but discovered one more crucial point: no engine can have an effi ciency that is better than the effi ciency of a reversible engine. In practice, all heat engines will always be irreversible to some extent, so Carnot’s result sets an absolute limit on the effi ciency of all real heat engines. Carnot’s results are the basis for an alternative statement of the second law of thermodynamics.
Second law of thermodynamics (alternative form) The effi ciency of a reversible heat engine is given by
e512 TC TH
No heat engine can have a greater effi ciency than this.
We have now seen two statements of the second law, this one involving the effi - ciency of heat engines and another (in Section 16.5) involving the direction of heat fl ow. Although it is not obvious, these two statements are completely equivalent.
Given either one of them as a starting point, the other one can be derived!
Effi ciency of a reversible heat engine Effi ciency of a reversible heat engine
Second law of thermodynamics:
formulation involving heat engines Second law of thermodynamics:
formulation involving heat engines
16.6 | HEAT ENGINES AND OTHER THERMODYNAMIC DEVICES 511
E X A M P L E 1 6 . 7 Effi ciency of a Steam Engine
Steamboats were used extensively in the 1800s to carry people and goods (Fig. 16.27). These boats used heat engines: a hot reservoir contained steam (TH⬇ 200°C) that was heated by burning wood or coal, and the cold reservoir was provided by the air or river water (TC⬇ 10°C). The engine turned a shaft that drove the boat’s paddle wheels. Suppose one of these steam engines does W 10,000 J of work each second. If it is a reversible heat engine, how much heat QH does it absorb from the hot reservoir in 1 s?
RECOGNIZE T HE PRINCIPLE
The effi ciency of any heat engine is defi ned as (Eq. 16.18) e5 W
QH (1)
For a reversible heat engine, Carnot showed that e is also equal to e512 TC
TH (2)
We can use these two relations to fi nd the heat absorbed from the hot reservoir QH in terms of the given quantities W, TH, and TC.
SK E TCH T HE PROBLEM
Figure 16.27 shows a picture of a real steamboat, but the diagram of a heat engine in Figure 16.26B is more useful for analyzing the operation of the engine.
IDENT IF Y T HE REL AT IONSHIPS Combining Equations (1) and (2) gives
W
QH512 TC TH Rearranging to solve for QH, we have
QH5 W
12 1TC/TH2 (3)
SOLV E
When using Equation (3), we must be careful to use values of the temperature on the Kelvin scale; Carnot’s result in Equation (2) assumes the temperatures are measured on this scale. Converting the given values of TH and TC gives TH (200 273) K 473 K and TC (10 273) K 283 K. Inserting these values of the reservoir tem- peratures and the given value of W into Equation (3), we fi nd
QH5 W
12 1TC/TH2 5 10,000 J
12 1283 K/473 K2 5 25,000 J
Figure 16.27 Example 16.7.
Steamboats in the 1800s used heat engines to drive their paddle wheels.
© The stern-wheeler ‘Cochan’ on the Colorado River, 1890 (b&w photo), American photographer, (19th century)/Private Collection, Peter Neward American Pictures/The Bridgeman Art Library
The result in Equation 16.21 shows that the effi ciency of an engine depends on the temperatures of the reservoirs. If the temperature of the cold reservoir is zero (TC 0 K), the effi ciency e 1. In this case, we also have QC 0, and it would thus seem that we have created the hypothetical heat engine in Figure 16.26A—a heat engine that expels no heat—which we said was impossible. One way out of this apparent contradiction is to hypothesize that it is impossible to lower the tempera- ture of a reservoir to T 0 K, the temperature called absolute zero. We’ll see this hypothesis again in Section 16.8 when we discuss the third law of thermodynamics.
Insight 16.1
EFFICIENCY OF A DIESEL ENGINE A diesel engine is similar to a gasoline internal combustion engine (Example 16.8). One difference is that a gasoline engine ignites the fuel mixture with a spark from a spark plug, whereas a diesel engine ignites the fuel mixture purely “by compression” (without a spark). The compression of the fuel mixture is therefore much greater in a diesel engine, which leads to a higher temperature in the hot reservoir.
According to Equation 16.21 and Example 16.8, this higher tempera- ture gives a higher theoretical limit on the effi ciency of a diesel engine.
E X A M P L E 1 6 . 8 Effi ciency of a Modern Automobile Engine
The internal combustion engine found in most cars is a type of heat engine (Fig.
16.28). A hot reservoir created by exploding gasoline provides heat energy, causing the gas in a piston to expand and do work as it rotates the driveshaft of the engine. Some heat is then expelled to a cold reservoir (the air) via the radiator and the exhaust.
What is the maximum possible effi ciency of such an engine?
RECOGNIZE T HE PRINCIPLE
A real internal combustion engine is irreversible, so its effi ciency will always be less than that of a reversible engine. This irreversibility is caused by friction between the moving parts inside the engine, the failure to “capture” and use all the heat energy created by combustion, and other factors. Nevertheless, an upper limit on the effi - ciency is set by the reversible case (Eq. 16.21)
e512 TC
TH (1)
To apply Equation (1) to an internal combustion engine, we need to know the tem- peratures of the two reservoirs TC and TH; these values are not given, so we need to use some common sense and knowledge about this type of engine to fi nd approximate values.
SK E TCH T HE PROBLEM
The schematic heat engine in Figure 16.26B describes the problem. The corresponding hot and cold reservoirs for a real automobile engine are shown in Figure 16.28.
IDENT IF Y T HE REL AT IONSHIPS
The cold reservoir (the radiator and exhaust) is at air temperature, so we estimate TC⬇ 300 K. For the hot reservoir, we can use reference data (from the Internet or else- where) to fi nd that the temperature of exploding gas in a piston is about TH⬇ 600 K.
SOLV E
Inserting our estimates for the reservoir temperatures into Equation (1) gives e512 TC
TH512 300 K
600 K5 0.5 (2)
What have we learned?
Even in the best possible case, an automobile engine can only convert about half the chemical energy in the gasoline into useful work. This limit on the effi ciency could be improved by increasing the temperature of the hot reservoir (the tem- perature in the piston), which can be accomplished by using a suitable blend of gasoline with a higher octane rating. Such high-performance engines and fuels are more expensive than typical engines and fuels, however.
In a real engine, a substantial amount of energy is lost to friction. In a typical car engine, only about 25% of the chemical energy in the gasoline is converted into work.
Figure 16.28 Example 16.8. A real automobile engine is a type of heat engine following the general schematic form in Figure 16.26B.
The laws of thermodynamics place rigorous limits on the effi ciency of all real heat engines.
Fuel intake
Combustion chamber
Piston
Crankshaft Spark plug
Exhaust Valve and spring
QC QH
What does it mean?
Because 10,000 J of work were done by the engine, the energy expelled to the cold reservoir was QC QH W 15,000 J. Most of the energy extracted from the hot reservoir was expelled to the cold reservoir and was thus “wasted.” Also, when analyzing the operation of a heat engine, temperature values should always be expressed in Kelvin units.
Perpetual Motion and the Second Law of Thermodynamics
The construction of a perpetual motion machine has been the quest of many inventors. Various laws of physics imply that such machines are impossible. One popular perpetual motion machine design sketched in Figure 16.29 makes use of the hypothetical (and impossible) heat engine in Figure 16.26A. The machine in Figure 16.29 takes some heat energy Q from a hot reservoir, which here is the atmosphere, and converts it all to an amount of work W. This work is then used to power a car. As the car moves, it loses energy to friction due to air drag, the wheels, and so on. This loss causes all the energy W to be converted to heat energy,
which returns to the atmosphere. The net effect is that the car has moved, whereas the energy in the atmosphere is back to its original value; hence, the process can be repeated, and the car can move on forever.
Carnot’s work and the second law of thermodynamics tell us that the perpet- ual motion machine in Figure 16.29 is impossible because a real heat engine must always expel some heat energy to a cold reservoir. Hence, all the heat energy Q extracted from the atmosphere cannot later be returned to the atmosphere. In this way, thermodynamics rules out many hypothetical perpetual motion designs. In fact, the second law along with conservation of energy principles rule out all per- petual motion machines.
The Carnot Cycle
We have so far represented heat engines in a highly schematic form as in the block diagram in Figure 16.26B and have ignored how one might construct a real engine.
Carnot showed how to use an ideal gas and piston to make a reversible heat engine.
Figure 16.30A shows the operation of Carnot’s engine as a path in the P–V plane.
The corresponding changes of the gas and piston are shown in Figure 16.30B, where we show only the mechanical part of the heat engine; the thermal reservoirs could be air or water held at temperatures TH and TC.
The system begins in state 1 in the P–V diagram. It then moves progressively to states 2, 3, and 4 and fi nally back to state 1.
Change from state 1 to state 2: The system is placed in contact with the hot reservoir, during which time the gas absorbs an amount of heat QH, which causes the gas to expand to state 2. Because the system is in contact with the hot reservoir, this process is an isothermal expansion at temperature TH. Change from state 2 to state 3: The system is isolated from its surroundings
(detached from the hot reservoir) and allowed to expand to state 3. Because no heat is absorbed from or expelled to the environment, this process is an adiabatic expansion.
16.6 | HEAT ENGINES AND OTHER THERMODYNAMIC DEVICES 513 Figure 16.29 Hypotheti- cal (and impossible) design of a perpetual motion machine. This machine is not possible because it violates the second law of thermo- dynamics.
W Q Q
Hot reservoir (atmosphere)
Machine
IMPOSSIBLE
Energy lost to air drag (friction) Heat energy returns
to the atmosphere.
Work powers a car.
V P
QH
QH
QC
QC CARNOT
CYCLE W Work done by engine
TC 1
2
3 4
Isothermal expansion Adiabatic expansion
Adiabatic compression Isothermal compression
1 2 2 3
1 4 4 3
Q 0
Q 0
A B
TH
Figure 16.30 An example of a thermodynamically reversible heat engine. This engine uses a cyclic process called a Carnot cycle. A The work done during a Carnot cycle is equal to the area enclosed in the P–V diagram. B The four steps in the Carnot cycle.