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Chapter 10 - Thermodynamics

Thermodynamics
Chapter 10
Table of Contents
Section 1 Relationships Between Heat and Work
Section 2 The First Law of Thermodynamics
Section 3 The Second Law of Thermodynamics

Section 1 Relationships
Between Heat and WorkChapter 10
Objectives
• Recognize that a system can absorb or release
energy as heat in order for work to be done on or by
the system and that work done on or by a system can
result in the transfer of energy as heat.
• Compute the amount of work done during a
thermodynamic process.
• Distinguish between isovolumetric, isothermal, and
adiabatic thermodynamic processes.

Chapter 10
Heat, Work, and Internal Energy
• Heat and work are energy transferred to or from a
system. An object never has “heat” or “work” in it; it
has only internal energy.
Between Heat and Work

( )
work = pressure volume change
A F
W Fd Fd Ad P V
A A
W P V
   
= = = = ∆ ÷  ÷
   
= ∆
×
Chapter 10
Heat, Work, and Internal Energy, continued
• In thermodynamic systems, work is defined in terms
of pressure and volume change.

Chapter 10
• If the gas expands, as
shown in the figure, ∆V is
positive, and the work done
by the gas on the piston is
positive.
• If the gas is compressed,
∆V is negative, and the
work done by the gas on
the piston is negative. (In
other words, the piston
does work on the gas.)

Chapter 10
• When the gas volume remains constant, there is no
displacement and no work is done on or by the
system.
• Although the pressure can change during a process,
work is done only if the volume changes.

Chapter 10
Thermodynamic Processes
• An isovolumetric process is a thermodynamic
process that takes place at constant volume so that
no work is done on or by the system.
• An isothermal process is a thermodynamic process
that takes place at constant temperature.
• An adiabatic process is a thermodynamic process
during which no energy is transferred to or from the
system as heat.

Thermodynamic Processes
Chapter 10

Section 2 The First Law of
ThermodynamicsChapter 10
Objectives
• Illustrate how the first law of thermodynamics is a
statement of energy conservation.
• Calculate heat, work, and the change in internal
energy by applying the first law of thermodynamics.
• Apply the first law of thermodynamics to describe
cyclic processes.

Chapter 10
Energy Conservation
• If friction is taken into account, mechanical energy
is not conserved.
• Consider the example of a roller coaster:
– A steady decrease in the car’s total mechanical energy
occurs because of work being done against the friction
between the car’s axles and its bearings and between the
car’s wheels and the coaster track.
– If the internal energy for the roller coaster (the system) and
the energy dissipated to the surrounding air (the
environment) are taken into account, then the total energy
will be constant.
Thermodynamics

Energy Conservation
Chapter 10
Thermodynamics

Chapter 10
Energy Conservation
Thermodynamics

Chapter 10
Energy Conservation, continued
• The principle of energy conservation that takes into
account a system’s internal energy as well as work
and heat is called the first law of thermodynamics.
• The first law of thermodynamics can be expressed
mathematically as follows:
∆U = Q – W
Change in system’s internal energy = energy
transferred to or from system as heat – energy
transferred to or from system as work
Thermodynamics

Chapter 10
Signs of Q and W for a system
Thermodynamics

Chapter 10
Sample Problem
The First Law of Thermodynamics
A total of 135 J of work is done on a gaseous
refrigerant as it undergoes compression. If the
internal energy of the gas increases by 114 J during
the process, what is the total amount of energy
transferred as heat? Has energy been added to or
removed from the refrigerant as heat?
Thermodynamics

Chapter 10
Sample Problem, continued
1. Define
Given:
W = –135 J
∆U = 114 J
Thermodynamics
Tip: Work is done
on the gas, so work
(W) has a negative
value. The internal
energy increases
during the process,
so the change in
internal energy
(∆U) has a positive
value.
Diagram:
Unknown:
Q = ?

Chapter 10
2. Plan
Choose an equation or situation:
Apply the first law of thermodynamics using the values
for ∆U and W in order to find the value for Q.
∆U = Q – W
Thermodynamics
Rearrange the equation to isolate the unknown:
Q = ∆U + W

Chapter 10
3. Calculate
Substitute the values into the equation and solve:
Q = 114 J + (–135 J)
Q = –21 J
Thermodynamics
Tip: The sign for the value of Q is negative. This
indicates that energy is transferred as heat from
the refrigerant.

Chapter 10
4. Evaluate
Although the internal energy of the refrigerant
increases under compression, more energy is
added as work than can be accounted for by the
increase in the internal energy. This energy is
removed from the gas as heat, as indicated by the
minus sign preceding the value for Q.
Thermodynamics

First Law of Thermodynamics for Special
Processes
Chapter 10
Thermodynamics

Chapter 10
Cyclic Processes
• A cyclic process is a thermodynamic process in
which a system returns to the same conditions under
which it started.
• Examples include heat engines and refrigerators.
• In a cyclic process, the final and initial values of
internal energy are the same, and the change in
internal energy is zero.
∆Unet = 0 and Qnet = Wnet
Thermodynamics

Chapter 10
Cyclic Processes, continued
• A heat engine uses heat to do
mechanical work.
• A heat engine is able to do
work (b) by transferring energy
from a high-temperature
substance (the boiler) at Th (a)
to a substance at a lower
temperature (the air around the
engine) at Tc (c).
Thermodynamics
• The internal-combustion engine found in most
vehicles is an example of a heat engine.

Combustion Engines
Chapter 10
Thermodynamics

Chapter 10
The Steps of a Gasoline Engine Cycle
Thermodynamics

Refrigeration
Chapter 10
Thermodynamics

Chapter 10
The Steps of a Refrigeration Cycle
Thermodynamics

Section 3 The Second Law of
ThermodynamicsChapter 10
Objectives
• Recognize why the second law of thermodynamics
requires two bodies at different temperatures for work
to be done.
• Calculate the efficiency of a heat engine.
• Relate the disorder of a system to its ability to do
work or transfer energy as heat.

Chapter 10
Efficiency of Heat Engines
• The second law of thermodynamics can be stated
as follows:
No cyclic process that converts heat entirely
into work is possible.
• As seen in the last section, Wnet = Qnet = Qh – Qc.
– According to the second law of thermodynamics,
W can never be equal to Qh in a cyclic process.
– In other words, some energy must always be
transferred as heat to the system’s surroundings
(Qc > 0).
Thermodynamics

Chapter 10
Efficiency of Heat Engines, continued
• A measure of how well an engine operates is given
by the engine’s efficiency (eff ).
Thermodynamics
• Because of the second law of thermodynamics, the
efficiency of a real engine is always less than 1.
eff =
Wnet
Qh
=
Qh – Qc
Qh
= 1−
Qc
Qh

Chapter 10
Sample Problem
Heat-Engine Efficiency
Find the efficiency of a gasoline engine that, during
one cycle, receives 204 J of energy from combustion
and loses 153 J as heat to the exhaust.
Thermodynamics
1. Define
Given: Diagram:
Qh = 204 J
Qc = 153 J
Unknown
eff = ?

Chapter 10
2. Plan
Choose an equation or situation: The efficiency of
a heat engine is the ratio of the work done by the
engine to the energy transferred to it as heat.
Thermodynamics
eff =
Wnet
Qh
= 1−
Qc
Qh

Chapter 10
3. Calculate
Substitute the values into the equation and
solve:
Thermodynamics
eff = 1−
Qc
Qh
= 1−
153 J
204 J
eff = 0.250
4. Evaluate
Only 25 percent of the energy added as heat is used
by the engine to do work. As expected, the efficiency
is less than 1.0.

Chapter 10
Entropy
• In thermodynamics, a system left to itself tends to go
from a state with a very ordered set of energies to
one in which there is less order.
• The measure of a system’s disorder or randomness
is called the entropy of the system. The greater the
entropy of a system is, the greater the system’s
disorder.
• The entropy of a system tends to increase.
Thermodynamics

Chapter 10
Entropy, continued
• The second law of thermodynamics can also be
expressed in terms of entropy change:
The entropy of the universe increases in all
natural processes.
Thermodynamics

Chapter 10
Energy Changes Produced by a Refrigerator
Freezing Water
Thermodynamics
Because of the refrigerator’s less-than-perfect efficiency, the entropy of
the outside air molecules increases more than the entropy of the freezing
water decreases.

Entropy of the Universe
Chapter 10
Thermodynamics

Multiple Choice
1. If there is no change in the internal energy of a gas,
even though energy is transferred to the gas as heat
and work, what is the thermodynamic process that
the gas undergoes called?
A. adiabatic
B. isothermal
C. isovolumetric
D. isobaric
Chapter 10

Multiple Choice, continued
2. To calculate the efficiency of a heat engine, which
thermodynamic property does not need to be known?
F. the energy transferred as heat to the engine
G. the energy transferred as heat from the engine
H. the change in the internal energy of the engine
J. the work done by the engine
Chapter 10

3. In which of the following processes is no work done?
A. Water is boiled in a pressure cooker.
B. A refrigerator is used to freeze water.
C. An automobile engine operates for several
minutes.
D. A tire is inflated with an air pump.
Chapter 10

4. A thermodynamic process occurs in which the
entropy of a system decreases. From the second law
of thermodynamics, what can you conclude about the
entropy change of the environment?
F. The entropy of the environment decreases.
G. The entropy of the environment increases.
H. The entropy of the environment remains
unchanged.
J. There is not enough information to state what
happens to the environment’s entropy.
Chapter 10

Use the passage and diagrams to answer questions 5–8.
A system consists of steam within the confines of a steam engine,
whose cylinder and piston are shown in the figures below.
Chapter 10
5. Which of the figures
describes a situation in
which ∆U < 0, Q < 0, and
W = 0?
A. (a)
B. (b)
C. (c)
D. (d)

Chapter 10
which ∆U > 0, Q = 0, and
W < 0?
F. (a)
G. (b)
H. (c)
J. (d)

Chapter 10
which ∆U < 0, Q = 0, and
W > 0?
A. (a)
B. (b)
C. (c)
D. (d)

Chapter 10
which ∆U > 0, Q > 0, and
W = 0?
F. (a)
G. (b)
H. (c)
J. (d)

9. A power plant has a power output of 1055 MW and
operates with an efficiency of 0.330. Excess energy is
carried away as heat from the plant to a nearby river.
How much energy is transferred away from the power
plant as heat?
A. 0.348 × 109
J/s
B. 0.520 × 109
J/s
C. 0.707 × 109
J/s
D. 2.14 × 109
J/s
Chapter 10

10. How much work must be done by air pumped into a
tire if the tire’s volume increases from 0.031 m3
to
0.041 m3
and the net, constant pressure of the air is
300.0 kPa?
F. 3.0 × 102
J
G. 3.0 × 103
J
H. 3.0 × 104
J
J. 3.0 × 105
J
Chapter 10

Use the passage below to answer questions 11–12.
An air conditioner is left running on a table in the middle
of the room, so none of the air that passes through the
air conditioner is transferred to outside the room.
11. Does passing air through the air conditioner affect
the temperature of the room? (Ignore the thermal
effects of the motor running the compressor.)
Chapter 10
Short Response

11. Does passing air through the air conditioner affect
the temperature of the room? (Ignore the thermal
effects of the motor running the compressor.)
Answer: No, because the energy removed from the
cooled air is returned to the room.
Chapter 10
Short Response

12. Taking the compressor motor into account, what
would happen to the temperature of the room?
Chapter 10
Short Response, continued

12. Taking the compressor motor into account, what
would happen to the temperature of the room?
Answer: The temperature increases.
Chapter 10

13. If 1600 J of energy are transferred as heat to an
engine and 1200 J are transferred as heat away
from the engine to the surrounding air, what is the
efficiency of the engine?
Chapter 10

13. If 1600 J of energy are transferred as heat to an
engine and 1200 J are transferred as heat away
from the engine to the surrounding air, what is the
efficiency of the engine?
Answer: 0.25
Chapter 10

14. How do the temperature of combustion and the
temperatures of coolant and exhaust affect the
efficiency of automobile engines?
Chapter 10
Extended Response

14. How do the temperature of combustion and the
temperatures of coolant and exhaust affect the
efficiency of automobile engines?
Answer: The greater the temperature difference is, the
greater is the amount of energy transferred as heat. For
efficiency to increase, the heat transferred between the
combustion reaction and the engine (Qh) should be
made to increase, whereas the energy given up as
waste heat to the coolant and exhaust (Qc) should be
made to decrease.
Chapter 10
Extended Response

Use the information below to answer questions 15–18. In each
problem, show all of your work.
A steam shovel raises 450.0 kg of dirt a vertical distance of 8.6 m.
The steam shovel’s engine provides 2.00 × 105
J of energy as heat
for the steam shovel to lift the dirt.
Chapter 10
Extended Response, continued
15. How much work is done by
the steam shovel in lifting the
dirt?

J of energy as heat
Chapter 10
15. How much work is done by
the steam shovel in lifting the
dirt?
Answer: 3.8 × 104
J

J of energy as heat
Chapter 10
16. What is the efficiency of the
steam shovel?

J of energy as heat
Chapter 10
16. What is the efficiency of the
steam shovel?
Answer: 0.19

J of energy as heat
Chapter 10
17. Assuming there is no change
in the internal energy of the
steam shovel’s engine, how
much energy is given up by
the shovel as waste heat?

J of energy as heat
Chapter 10
17. Assuming there is no change
in the internal energy of the
steam shovel’s engine, how
much energy is given up by
the shovel as waste heat?
Answer: 1.62 × 105
J

J of energy as heat
Chapter 10
18. Suppose the internal energy
of the steam shovel’s engine
increases by 5.0 × 103
J. How
much energy is given up now
as waste heat?

J of energy as heat
Chapter 10
18. Suppose the internal energy
of the steam shovel’s engine
increases by 5.0 × 103
J. How
much energy is given up now
as waste heat?
Answer: 1.57 × 105
J

19. One way to look at heat and work is to think of
energy transferred as heat as a “disorganized” form
of energy and energy transferred as work as an
“organized” form. Use this interpretation to show
that the increased order obtained by freezing
water is less than the total disorder that results
from the freezer used to form the ice.
Chapter 10

19. One way to look at heat and work is to think of
energy transferred as heat as a “disorganized” form
of energy and energy transferred as work as an
“organized” form. Use this interpretation to show
that the increased order obtained by freezing
water is less than the total disorder that results
from the freezer used to form the ice.
Chapter 10
Answer: Disorganized energy is removed from water to
form ice, but a greater amount of organized energy
must become disorganized to operate the freezer.

Chapter 10
Entropy
Thermodynamics

Chapter 10
Energy Changes Produced by a Refrigerator
Freezing Water
Thermodynamics

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