Physics For Scientists And Engineers 6E - part 169

S E C T I O N 2 2 . 3 • Reversible and Irreversible Processes

673

22.3 Reversible and Irreversible Processes

In the next section we discuss a theoretical heat engine that is the most efficient possi-
ble. To understand its nature, we must first examine the meaning of reversible and ir-
reversible processes. In a

reversible process, the system undergoing the process can be

Example 22.2 Freezing Water

A certain refrigerator has a COP of 5.00. When the refrigera-
tor is running, its power input is 500 W. A sample of water of
mass 500 g and temperature 20.0°C is placed in the freezer
compartment.  How  long  does  it  take  to  freeze  the  water  to
ice at 0°C? Assume that all other parts of the refrigerator stay
at  the  same  temperature  and  there  is  no  leakage  of  energy
from  the  exterior,  so  that  the  operation  of  the  refrigerator
results only in energy being extracted from the water.

Solution Conceptualize  this  problem  by  realizing  that  en-
ergy  leaves  the  water,  reducing  its  temperature  and  then
freezing it into ice. The time interval required for this entire
process  is  related  to  the  rate  at  which  energy  is  withdrawn
from the water, which, in turn is related to the power input
of  the  refrigerator.  We  categorize  this  problem  as  one  in
which  we  will  need  to  combine  our  understanding  of  tem-
perature changes and phase changes from Chapter 20 with
our understanding of heat pumps from the current chapter.
To analyze the problem, we first find the amount of energy
that we must extract from 500 g of water at 20°C to turn it
into ice at 0°C. Using Equations 20.4 and 20.6,

2.08 ' 10

(0.500 kg)[(4 186 J/kg()C)(20.0)C) & 3.33 ' 10

J/kg]

! Q

! ! ! mc

%T & mL

! ! m

! c

T & L

Now we use Equation 22.4 to find out how much energy we
need to provide to the refrigerator to extract this much
energy from the water:

Using the power rating of the refrigerator, we find out
the time interval required for the freezing process to
occur:

To finalize this problem, note that this time interval is very
different  from  that  of  our  everyday  experience;  this  sug-
gests  the  difficulties  with  our  assumptions.  Only  a  small
part of the energy extracted from the refrigerator interior
in  a  given  time  interval  will  come  from  the  water.  Energy
must also be extracted from the container in which the wa-
ter  is  placed,  and  energy  that  continuously  leaks  into  the
interior from the exterior must be continuously extracted.
In reality, the time interval for the water to freeze is much
longer than 83.3 s.

83.3 s

! !

t !

4.17 ' 10

500 W

W ! 4.17 ' 10

COP !

! Q

W !

! Q

COP

2.08 ' 10

5.00

Quick Quiz 22.3

The energy entering an electric heater by electrical trans-

mission can be converted to internal energy with an efficiency of 100%. By what factor
does the cost of heating your home change when you replace your electric heating sys-
tem with an electric heat pump that has a COP of 4.00? Assume that the motor run-
ning the heat pump is 100% efficient. (a) 4.00 (b) 2.00 (c) 0.500 (d) 0.250

areas by burying the external coils deep in the ground. In this case, the energy is
extracted from the ground, which tends to be warmer than the air in the winter.

For a heat pump operating in the cooling mode, “what you gain” is energy

removed from the cold reservoir. The most effective refrigerator or air conditioner is
one that removes the greatest amount of energy from the cold reservoir in exchange
for the least amount of work. Thus, for these devices we define the COP in terms
of |Q

(22.4)

A good refrigerator should have a high COP, typically 5 or 6.

COP (cooling mode) !

returned to its initial conditions along the same path on a PV diagram, and every point
along this path is an equilibrium state. A process that does not satisfy these require-
ments is

irreversible.

All natural processes are known to be irreversible. From the endless number of ex-

amples that could be selected, let us examine the adiabatic free expansion of a gas,
which was already discussed in Section 20.6, and show that it cannot be reversible. Con-
sider a gas in a thermally insulated container, as shown in Figure 22.8. A membrane
separates the gas from a vacuum. When the membrane is punctured, the gas expands
freely into the vacuum. As a result of the puncture, the system has changed because it
occupies a greater volume after the expansion. Because the gas does not exert a force
through a displacement, it does no work on the surroundings as it expands. In addi-
tion, no energy is transferred to or from the gas by heat because the container is insu-
lated from its surroundings. Thus, in this adiabatic process, the system has changed but
the surroundings have not.

For this process to be reversible, we need to be able to return the gas to its original

volume  and  temperature  without  changing  the  surroundings.  Imagine  that  we  try  to
reverse the process by compressing the gas to its original volume. To do so, we fit the
container  with  a  piston  and  use  an  engine  to  force  the  piston  inward.  During  this
process, the surroundings change because work is being done by an outside agent on
the  system.  In  addition,  the  system  changes  because  the  compression  increases  the
temperature of the gas. We can lower the temperature of the gas by allowing it to come
into contact with an external energy reservoir. Although this step returns the gas to its
original conditions, the surroundings are again affected because energy is being added
to the surroundings from the gas. If this energy could somehow be used to drive the
engine  that  compressed  the  gas,  then  the  net  energy  transfer  to  the  surroundings
would be zero. In this way, the system and its surroundings could be returned to their
initial  conditions,  and  we  could  identify  the  process  as  reversible.  However,  the
Kelvin–Planck statement of the second law specifies that the energy removed from the
gas to return the temperature to its original value cannot be completely converted to
mechanical energy in the form of the work done by the engine in compressing the gas.
Thus, we must conclude that the process is irreversible.

We could also argue that the adiabatic free expansion is irreversible by relying on

the  portion  of  the  definition  of  a  reversible  process  that  refers  to  equilibrium  states.
For  example,  during  the  expansion,  significant  variations  in  pressure  occur  through-
out the gas. Thus, there is no well-defined value of the pressure for the entire system at
any time between the initial and final states. In fact, the process cannot even be repre-
sented  as  a  path  on  a  PV diagram.  The  PV diagram  for  an  adiabatic  free  expansion
would  show  the  initial  and  final  conditions  as  points,  but  these  points  would  not  be
connected  by  a  path.  Thus,  because  the  intermediate  conditions  between  the  initial
and final states are not equilibrium states, the process is irreversible.

Although all real processes are irreversible, some are almost reversible. If a real

process occurs very slowly such that the system is always very nearly in an equilibrium
state, then the process can be approximated as being reversible. Suppose that a gas is
compressed isothermally in a piston–cylinder arrangement in which the gas is in ther-
mal contact with an energy reservoir, and we continuously transfer just enough energy
from the gas to the reservoir during the process to keep the temperature constant. For
example, imagine that the gas is compressed very slowly by dropping grains of sand
onto a frictionless piston, as shown in Figure 22.9. As each grain lands on the piston
and compresses the gas a bit, the system deviates from an equilibrium state, but is so
close to one that it achieves a new equilibrium state in a relatively short time interval.
Each grain added represents a change to a new equilibrium state but the differences
between states are so small that we can approximate the entire process as occurring
through continuous equilibrium states. We can reverse the process by slowly removing
grains from the piston.

A general characteristic of a reversible process is that no dissipative effects (such as

turbulence or friction) that convert mechanical energy to internal energy can be

674

C H A P T E R 2 2 • Heat Engines, Entropy, and the Second Law of Thermodynamics

Insulating

wall

Membrane

Vacuum

Gas at T

Figure 22.8 Adiabatic free

expansion of a gas.

Energy reservoir

Sand

Figure 22.9 A gas in thermal

contact with an energy reservoir is

compressed slowly as individual

grains of sand drop onto the

piston. The compression is

isothermal and reversible.

▲

PITFALL PREVENTION

22.2 All Real Processes

Are Irreversible

The reversible process is an ideal-
ization—all real processes on
Earth are irreversible.

S E C T I O N 2 2 . 4 • The Carnot Engine

675

present. Such effects can be impossible to eliminate completely. Hence, it is not
surprising that real processes in nature are irreversible.

22.4 The Carnot Engine

In 1824 a French engineer named Sadi Carnot described a theoretical engine, now
called a

Carnot engine, which is of great importance from both practical and theoreti-

cal viewpoints. He showed that a heat engine operating in an ideal, reversible cycle—
called a

Carnot cycle—between two energy reservoirs is the most efficient engine pos-

sible. Such an ideal engine establishes an upper limit on the efficiencies of all other
engines. That is, the net work done by a working substance taken through the Carnot
cycle is the greatest amount of work possible for a given amount of energy supplied to
the substance at the higher temperature.

Carnot’s theorem can be stated as follows:

No real heat engine operating between two energy reservoirs can be more efficient
than a Carnot engine operating between the same two reservoirs.

To argue the validity of this theorem, imagine two heat engines operating between

the same energy reservoirs. One is a Carnot engine with efficiency e

, and the other is

an engine with efficiency e, where we assume that e # e

. The more efficient engine is

used to drive the Carnot engine as a Carnot refrigerator. The output by work of the
more efficient engine is matched to the input by work of the Carnot refrigerator. For
the combination of the engine and refrigerator, no exchange by work with the sur-
roundings occurs. Because we have assumed that the engine is more efficient than the
refrigerator, the net result of the combination is a transfer of energy from the cold to
the hot reservoir without work being done on the combination. According to the
Clausius statement of the second law, this is impossible. Hence, the assumption that
e # e

must be false.

All real engines are less efficient than the Carnot engine

because they do not operate through a reversible cycle. The efficiency of a real
engine is further reduced by such practical difficulties as friction and energy losses by
conduction.

To describe the Carnot cycle taking place between temperatures T

and T

, we

assume that the working substance is an ideal gas contained in a cylinder fitted with a
movable piston at one end. The cylinder’s walls and the piston are thermally noncon-
ducting. Four stages of the Carnot cycle are shown in Figure 22.10, and the PV diagram
for the cycle is shown in Figure 22.11. The Carnot cycle consists of two adiabatic
processes and two isothermal processes, all reversible:

1. Process A : B (Fig. 22.10a) is an isothermal expansion at temperature T

. The gas

is placed in thermal contact with an energy reservoir at temperature T

. During the

expansion, the gas absorbs energy

! from the reservoir through the base of the

cylinder and does work W

in raising the piston.

2. In process B : C (Fig. 22.10b), the base of the cylinder is replaced by a thermally

nonconducting wall, and the gas expands adiabatically—that is, no energy enters or
leaves the system by heat. During the expansion, the temperature of the gas
decreases from T

to T

and the gas does work W

in raising the piston.

3. In process C : D (Fig. 22.10c), the gas is placed in thermal contact with an energy

reservoir at temperature T

and is compressed isothermally at temperature T

. Dur-

ing this time, the gas expels energy

! to the reservoir, and the work done by the

piston on the gas is W

4. In the final process D : A (Fig. 22.10d), the base of the cylinder is replaced by a

nonconducting wall, and the gas is compressed adiabatically. The temperature of
the gas increases to T

, and the work done by the piston on the gas is W

Sadi Carnot

French engineer (1796–1832)

Carnot was the first to show the

quantitative relationship between

work and heat. In 1824 he

published his only work—

Reflections on the Motive Power

of Heat—which reviewed the

industrial, political, and economic

importance of the steam engine.

In it, he defined work as “weight

lifted through a height.”
(J.-L. Charmet/Science Photo
Library/Photo Researchers, Inc.)

▲

PITFALL PREVENTION

22.3 Don’t Shop for a

Carnot Engine

The Carnot engine is an idealiza-
tion—do  not  expect  a  Carnot
engine to be developed for com-
mercial  use.  We  explore  the
Carnot  engine  only  for  theoreti-
cal considerations.

The net work done in this reversible, cyclic process is equal to the area enclosed by

the path ABCDA in Figure 22.11. As we demonstrated in Section 22.1, because the
change in internal energy is zero, the net work W

eng

done by the gas in one cycle

equals the net energy transferred into the system,

! " !Q

!. The thermal efficiency

of the engine is given by Equation 22.2:

In Example 22.3, we show that for a Carnot cycle

(22.5)

! Q

e !

eng

! Q

! " ! Q

! Q

1 "

! Q

676

C H A P T E R 2 2 • Heat Engines, Entropy, and the Second Law of Thermodynamics

Cycle

Energy reservoir at T

→ D

Isothermal

compression

→ C

Adiabatic

expansion

Q = 0

(b)

Q = 0

(d)

Energy reservoir at T

(a)

→ B

Isothermal

expansion

→ A

Adiabatic

compression

Active Figure 22.10 The Carnot cycle. (a) In process A : B, the gas expands

isothermally while in contact with a reservoir at T

. (b) In process B : C, the gas

expands adiabatically (Q ! 0). (c) In process C : D, the gas is compressed

isothermally while in contact with a reservoir at T

. (d) In process D : A, the gas

is compressed adiabatically. The arrows on the piston indicate the direction of its

motion during each process.

At the Active Figures link

at http://www.pse6.com, you

can observe the motion of the

piston in the Carnot cycle while

you also observe the cycle on

the PV diagram of Figure 22.11.

eng

Active Figure 22.11 PV diagram

for the Carnot cycle. The net work

done W

eng

equals the net energy

transferred into the Carnot engine

in one cycle,

! " !Q

!. Note that

int

0 for the cycle.

At the Active Figures link

at http://www.pse6.com, you

can observe the Carnot cycle

on the PV diagram while you

also observe the motion of the

piston in Figure 22.10.

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