Baixe Respostas capítulo 19 Halliday e outras Exercícios em PDF para Física, somente na Docsity!
Chapter 19
The Second Law of Thermodynamics
Conceptual Problems
1 • Modern automobile gasoline engines have efficiencies of about 25%. About what percentage of the heat of combustion is not used for work but released as heat? ( a ) 25%, ( b ) 50%, ( c ) 75%, ( d ) 100%, ( e ) You cannot tell from the data given.
Determine the Concept The efficiency of a heat engine is the ratio of the work
done per cycle W to the heat absorbed from the high-temperature reservoir Q h.
The percentage of the heat of combustion (heat absorbed from the high- temperature reservoir) is the ratio of Q c to Q h. We can use the relationship between W, Q h , and Q c ( W = Q h − Q c) to find Q c/ Q h.
Use the definition of efficiency and the relationship between W, Q h , and Q c to obtain: h
c h
h c h
Q
Q
Q
Q Q
Q
W
Solving for Q c/ Q h yields:
h
c Q
Q
Substitute for ε to obtain: (^10). 25 0. 75
h
c (^) = − = Q
Q
and ( c ) is correct.
2 • If a heat engine does 100 kJ of work per cycle while releasing 400 kJ of heat, what is its efficiency? ( a ) 20%, ( b ) 25%, ( c ) 80%, ( d ) 400%, ( e ) You cannot tell from the data given.
Determine the Concept The efficiency of a heat engine is the ratio of the work
done per cycle W to the heat absorbed from the high-temperature reservoir Q h. We can use the relationship between W, Q h , and Q c ( W = Q h − Q c) to express the
efficiency of the heat engine in terms of Q c and W.
Use the definition of efficiency and the relationship between W, Q h , and Q c to obtain: (^) W
W Q Q
W
Q
W
h c 1 c
1868 Chapter 19
Substitute for Q c and W to obtain:
- 2
100 kJ
400 kJ 1
and ( a ) is correct.
3 • If the heat absorbed by a heat engine is 600 kJ per cycle, and it releases 480 kJ of heat each cycle, what is its efficiency? ( a ) 20%, ( b ) 80%, ( c ) 100%, ( d ) You cannot tell from the data given.
Determine the Concept The efficiency of a heat engine is the ratio of the work
done per cycle W^ to the heat absorbed from the high-temperature reservoir^ Q h. We can use the relationship between W, Q h , and Q c ( W = Q h − Q c) to express the
efficiency of the heat engine in terms of Q c and Q h.
Use the definition of efficiency and the relationship between W, Q h , and Q c to obtain: h
c h
h c h
Q
Q
Q
Q Q
Q
W
Substitute for Q c and Q h to obtain:
- 2 600 kJ
480 kJ
and ( a ) is correct.
4 • Explain what distinguishes a refrigerator from a ″heat pump.″
Determine the Concept The job of a refrigerator is to move heat from its cold
interior to the warmer kitchen environment. This process moves heat in a direction that is opposite its ″natural″ direction of flow, analogous to the use of a
water pump to pump water out of a boat. The term heat pump is used to describe devices, such as air conditioners, that are used to cool living and working spaces
in the summer and warm them in the winter.
5 • [SSM] An air conditioner’s COP is mathematically identical to that
of a refrigerator, that is, COPAC = COPref= c
Q
W
. However a heat pump’s COP is
defined differently, as COPhp = h
Q
W
. Explain clearly why the two COPs are
defined differently. Hint : Think of the end use of the three different devices.
Determine the Concept The COP is defined so as to be a measure of the
effectiveness of the device. For a refrigerator or air conditioner, the important
quantity is the heat drawn from the already colder interior, Q c. For a heat pump, the ideas is to focus on the heat drawn into the warm interior of the house, Q h.
1870 Chapter 19
Determine the Concept The efficiency of a Carnot cycle engine is given by
h
c C (^1) T
T
ε = − where T c and T h (in kelvins) are the temperatures of the cold and hot
reservoirs, respectively.
Substituting numerical values for T c and T h yields: (^400) K^0.^25
300 K
εC = 1 − =
( b ) is correct.
11 •• The Carnot engine in Problem 10 is run in reverse as a refrigerator. Its COP is ( a ) 0.33, ( b ) 1.3, ( c ) 3.0 ( d ) 4.7.
Determine the Concept The coefficient of performance of a Carnot cycle engine
run in reverse as refrigerator is given by W
Q c COPref =. We can use the relationship
between W, Q c, and Q h to eliminate W from this expression and then use the
relationship, applicable only to a device operating in a Carnot cycle, h
c h
c T
T
Q
Q
= to
express the refrigerator’s COP in terms of T c and T h.
The coefficient of performance of a refrigerator is given by: (^) W
Q c COPref =
or, because W = Q h − Q c,
h c
c COPref Q Q
Q
Dividing the numerator and denominator of this fraction by Q c yields: 1
COP
c
h
ref −
Q
Q
For a device operating in a Carnot cycle: h
c h
c T
T
Q
Q
Substitute in the expression for COPref to obtain: 1
COP
c
h
ref,C −
T
T
The Second Law of Thermodynamics 1871
Substitute numerical values and evaluate COPref, C : 3.^0 1 300 K
400 K
COP (^) ref, C = −
( c ) is correct. 12 •• On a humid day, water vapor condenses on a cold surface. During condensation, the entropy of the water ( a ) increases, ( b ) remains constant, ( c ) decreases, ( d ) may decrease or remain unchanged. Explain your answer.
Determine the Concept When water vapor condenses, its entropy decreases (the liquid state is a more ordered state than is the vapor state) and the entropy of the
universe increases. ( a ) is correct.
13 •• An ideal gas is taken reversibly from an initial state P i , V i , T i to the final state P f, V f , T f. Two possible paths are (A) an isothermal expansion followed by an adiabatic compression and (B) an adiabatic compression followed by an isothermal expansion. For these two paths, ( a ) Δ E int A > Δ E int B , ( b ) Δ S A > Δ S B , ( c ) Δ S A < Δ S B , ( d ) None of the above.
Determine the Concept The two paths are shown on the PV diagram to the right. We can use the concept of a state function to choose from among the alternatives given as possible answers to the problem.
P
V V i V f
P f P i (^) i
f
B B
A A
T i
T f
( a ) Because E int is a state function and the initial and final states are the same for the two paths, Δ E (^) int, (^) A=Δ E int,B.
( b ) and ( c ) S , like E int , is a state function and its change when the system moves from one state to another depends only on the system’s initial and final states. It is not dependent on the process by which the change occurs. Thus Δ S (^) A (^) =Δ S B.
( d ) ( d ) is correct.
14 •• Figure 19-12 shows a thermodynamic cycle for an ideal gas on an ST diagram. Identify this cycle and sketch it on a PV diagram.
The Second Law of Thermodynamics 1873
On path b → c the entropy is given by: ⎟ ⎠
T
T
S Sb S Sb C V 1 b
The first and second derivatives,
dS dT and d^2 S dT^2 , give the slope
and concavity of the path. Calculate these derivatives assuming C V is constant. (For an ideal gas C V is a positive constant.):
V T 2
T
C
dT
dS (^) b
2 V 3
2 2 T
T
C
dT
d S b = −
These results tell us that, along path b → c, the slope of the path is positive and the slope decreases as T increases. The concavity of the path is negative for all T.
Following the same procedure on
path d → a gives:
T
T
S Sd C V 1 d
V T 2
T
C
dT
dS (^) d
2 V 3
2 2 T
T
C
dT
d S d = −
These results tell us that, along path d → a, the slope of the path is positive and the slope decreases as T increases. The concavity of the path is negative for all T.
An ST diagram for the Otto cycle is shown to the right.
S
T
a (^) b
d c
17 •• [SSM] Sketch an SV diagram of the Carnot cycle for an ideal gas.
Determine the Concept Referring to Figure 19-8, process 1→2 is an isothermal expansion. In this process heat is added to the system and the entropy and volume increase. Process 2→3 is adiabatic, so S is constant as V increases. Process 3→ 4 is an isothermal compression in which S decreases and V also decreases. Finally, process 4→1 is adiabatic, that is, isentropic, and S is constant while V decreases.
1874 Chapter 19
During the isothermal expansion (from
point 1 to point 2) the work done by
the gas equals the heat added to the gas. The change in entropy of the gas
from point 1 (where the temperature is T 1 ) to an arbitrary point on the curve is
given by:
T 1
Q
Δ S =
For an isothermal expansion, the work done by the gas, and thus the
heat added to the gas, are given by:
1
1 ln^ V
V
Q W nRT
Substituting for Q yields: ⎟⎟ ⎠
1
ln V
V
S nR
Since S = S 1 +Δ S , we have: ⎟⎟ ⎠
1
1 ln^ V
V
S S nR
The graph of S as a function of V for an
isothermal expansion shown to the right was plotted using a spreadsheet
program. This graph establishes the curvature of the 1→2 and 3→4 paths
for the SV graph.
V
S
An SV graph for the Carnot cycle (see Figure 19-8) is shown to the right.
V
S
1
(^2 )
4
18 •• Sketch an SV diagram of the Otto cycle. (The Otto cycle is discussed in Section 19-1.)
1876 Chapter 19
disorder, it is not true that order cannot be brought forth from disorder. What is
required is an agent doing work – for example, your friend – on the system in
order to reduce the level of chaos and bring about order. His cleanup efforts will be rewarded with an orderly system after a sufficient time for him to complete the
task. It is true that order will not come about from the disordered chaos of his room – unless he applies some elbow grease.
Estimation and Approximation
21 • Estimate the change in COP of your electric food freezer when it is removed from your kitchen to its new location in your basement, which is 8°C cooler than your kitchen.
Picture the Problem We can use the definition of the coefficient of performance
to express the ratio of the coefficient of performance in your basement to the coefficient of performance in the kitchen. If we further assume that the freezer operates in a Carnot cycle, then we can use the proportion Q h (^) Q c= T h T c to
express the ratio of the coefficients of performance in terms of the temperatures in
the kitchen, basement, and freezer.
The ratio of the coefficients of performance in the basement and
kitchen is given by:
c,kit
c,kit
c,basement
c,basement
kit
basement COP
COP
W
Q
W
Q
Because W = Q h − Q cfor a heat
engine or refrigerator:
h,kit c, kit
c,kit
h,basement c,basement
c,basement
kit
basement COP
COP
Q Q
Q
Q Q
Q
Divide the numerators and denominators by Q c,basement and Q c,kit
and simplify to obtain:
COP
COP
c,basement
h,basement
c,kit
h,kit
c,kit
h,kit
c,basement
h,basement
kit
basement
Q
Q
Q
Q
Q
Q
Q
Q
The Second Law of Thermodynamics 1877
If we assume that the freezer unit
operates in a Carnot cycle, then
c
h c
h T
T
Q
Q
= and our expression for the
ratio of the COPs becomes:
COP
COP
c,basement
h,basement
c,kit
h,kit
kit
basement −
T
T
T
T
Assuming that the temperature in your kitchen is 20°C and that the
temperature of the interior of your freezer is − 5 °C, substitute numerical
values and evaluate the ratio of the coefficients of performance:
268 K
285 K
268 K
293 K
COP
COP
kit
basement (^) = −
or an increase of 47 % in the performance of the freezer!
22 •• Estimate the probability that all the molecules in your bedroom are located in the (open) closet which accounts for about 10% of the total volume of the room.
Picture the Problem The probability that all the molecules in your bedroom are
located in the (open) closet is given by
N
V
V
p (^) ⎟⎟ ⎠
1
(^2) where N is the number of air
molecules in your bedroom and V 1 and V 2 are the volumes of your bedroom and
closet, respectively. We can use the ideal-gas law to find the number of molecules N. We’ll assume that the volume of your room is about 50 m^3 and that the
temperature of the air is 20°C.
If the original volume of the air in
your bedroom is V 1 , the probability p of finding the N molecules,
normally in your bedroom, confined to your closet whose volume is V 2 is
given by:
N
V
V
p (^) ⎟⎟ ⎠
1
2
or, because V 2 (^) = 101 V 1 , N p (^) ⎟ ⎠
Use the ideal-gas law to express N : kT
PV
N =
Substitute numerical values and
evaluate N :
- 252 10 molecules
1. 381 10 J/K 293 K
- 325 kPa 50 m
27
23
3
= ×
×
N = −
The Second Law of Thermodynamics 1879
Picture the Problem If we assume that the temperature on the inside of the refrigerator is 0°C (273 K) and the room temperature to be about 30°C (303 K), then the refrigerator must be able to maintain a temperature difference of about 30 K. We can use the definition of the COP of a refrigerator and the relationship between the temperatures of the hot and cold reservoir and Q h and Q c to find an upper limit on the COP of a household refrigerator. In ( b ) we can solve the definition of COP for Q c and differentiate the resulting equation with respect to time to estimate the rate at which heat is being drawn from the refrigerator compartment.
( a ) Using its definition, express the COP of a household refrigerator: (^) W
Q c COP = (1)
Apply conservation of energy to the refrigerator to obtain:
W + Q c = Q h⇒ W = Q h − Q c
Substitute for W and simplify to obtain: 1
COP
c
h c h
c −
Q
Q Q Q
Q
Assume, for the sake of finding the upper limit on the COP, that the refrigerator is a Carnot refrigerator and relate the temperatures of the hot and cold reservoirs to Q h and Q c:
c
h c
h T
T
Q
Q
Substitute for c
h Q
Q
to obtain: 1
COP
c
h max −
T
T
Substitute numerical values and evaluate COP (^) max : 9.^1 1 273 K
303 K
COPmax = −
( b ) Solve equation (1) for Q c: (^) Q c (^) = W ( COP) (2)
Differentiate equation (2) with respect to time to obtain: (^ )^ dt
dW dt
dQ c (^) = COP
Substitute numerical values and
evaluate dt
dQ c :
c (^) = ( 9.1)( 600 J/s) = 5. 5 kW dt
dQ
1880 Chapter 19
25 •• [SSM] The average temperature of the surface of the Sun is about 5400 K, the average temperature of the surface of Earth is about 290 K. The solar constant (the intensity of sunlight reaching Earth’s atmosphere) is about 1.37 kW/m^2. ( a ) Estimate the total power of the sunlight hitting Earth. ( b ) Estimate the net rate at which Earth’s entropy is increasing due to this solar radiation.
Picture the Problem We can use the definition of intensity to find the total power of sunlight hitting the earth and the definition of the change in entropy to find the changes in the entropy of Earth and the Sun resulting from the radiation from the Sun.
( a ) Using its definition, express the intensity of the Sun’s radiation on Earth in terms of the power P delivered to Earth and Earth’s cross sectional area A :
A
P
I =
Solve for P and substitute for A to obtain:
P = IA = I π R^2 where R is the radius of Earth.
Substitute numerical values and evaluate P :
1. 746 10 W 1. 75 10 W
- 37 kW/m 6. 37 10 m 17 17
2 6 2
= × = ×
P =π ×
( b ) Express the rate at which Earth’s entropy S Earth changes due to the flow of solar radiation: Earth
Earth T
P
dt
dS
Substitute numerical values and
evaluate dt
dS Earth :
- 02 10 J/K s
290 K
1. 746 10 W
14
17 Earth
= × ⋅
×
dt
dS
26 •• A 1.0-L box contains N molecules of an ideal gas, and the positions of the molecules are observed 100 times per second. Calculate the average time it should take before we observe all N molecules in the left half of the box if N is equal to ( a ) 10, ( b ) 100, ( c ) 1000, and ( d ) 1.0 mole. ( e ) The best vacuums that have been created to date have pressures of about 10 –12^ torr. If a vacuum chamber has the same volume as the box, how long will a physicist have to wait before all of the gas molecules in the vacuum chamber occupy only the left half of it? Compare that to the expected lifetime of the universe, which is about 10 10 years.
Picture the Problem If you had one molecule in a box, it would have a 50% chance of being on one side or the other. We don’t care which side the molecules are on as long as they all are on one side, so with one molecule you have a 100% chance of it being on one side or the other. With two molecules, there are four
1882 Chapter 19
Substituting for x yields:
10 y
3.156 10 s
1 y 2100 s
23
23
10
1 7
10
×
t ≈ (^) −
( e ) Solve the ideal gas law for the number of molecules N in the gas: (^) kT
PV
N =
Assuming the gas to be at room temperature (300 K), substitute numerical values and evaluate N :
- 22 10 molecules
1. 381 10 J/K 300 K
10 torr 133. 32 Pa/torr 1. 0 L
7
23
12
= ×
×
− N
Evaluate t for N = 3.22× 107
molecules: ( 1 )
- 2210
2100 s
7 −
× t =
To evaluate
- 22107 2 ×^ let
- 22107 10 x^ = 2 × and take the logarithm of both sides of the equation to obtain:
( 3. 22 × 107 ) ln 2 = x ln 10 ⇒ x ≈ 107
Substituting for x yields:
10 y
3.156 10 s
1 y 2100 s
7
7
10
1 7
10
×
t = (^) − ×
Express the ratio of this waiting time to the lifetime of the universe t universe:
7 7 10 10
10
universe
10 y
10 y = ≈ t
t
or
universe t ≈ 1010 7 t
Heat Engines and Refrigerators
27 • [SSM] A heat engine with 20.0% efficiency does 0.100 kJ of work during each cycle. ( a ) How much heat is absorbed from the hot reservoir during each cycle? ( b ) How much heat is released to the cold reservoir during each cycle?
Picture the Problem ( a ) The efficiency of the engine is defined to be
ε = W Q hwhere W is the work done per cycle and Q h is the heat absorbed from
the hot reservoir during each cycle. ( b ) Because, from conservation of energy, Q h (^) = W + Q c, we can express the efficiency of the engine in terms of the heat Q c
The Second Law of Thermodynamics 1883
released to the cold reservoir during each cycle.
( a ) Q h absorbed from the hot reservoir during each cycle is given by:
500 J
100 J
h =ε = =
W
Q
( b ) Use Q h (^) = W + Q cto obtain: (^) Q c = Q h− W = 500 J− 100 J= 400 J
28 • A heat engine absorbs 0.400 kJ of heat from the hot reservoir and does 0.120 kJ of work during each cycle. ( a ) What is its efficiency? ( b ) How much heat is released to the cold reservoir during each cycle?
Picture the Problem ( a ) The efficiency of the engine is defined to be
ε = W Q hwhere W is the work done per cycle and Q h is the heat absorbed from
the hot reservoir during each cycle. ( b ) We can apply conservation of energy to the engine to obtain Q h (^) = W + Q cand solve this equation for the heat Q c released to the cold reservoir during each cycle.
( a ) The efficiency of the heat engine is given by:
400 J
120 J
h
Q
W
ε
( b ) Apply conservation of energy to the engine to obtain:
Q h (^) = W + Q c⇒ Q (^) c= Q h− W
Substitute numerical values and evaluate Q c:
Q c = 400 J− 120 J= 280 J
29 • A heat engine absorbs 100 J of heat from the hot reservoir and releases 60 J of heat to the cold reservoir during each cycle. ( a ) What is its efficiency? ( b ) If each cycle takes 0.50 s, find the power output of this engine.
Picture the Problem We can use its definition to find the efficiency of the engine and the definition of power to find its power output.
( a ) The efficiency of the heat engine is given by: (^) h
c h
h c h Q
Q
Q
Q Q
Q
W
Substitute numerical values and evaluate ε:
100 J
60 J
ε= 1 − =
The Second Law of Thermodynamics 1885
pressure is again 1.00 atm. It is then compressed at constant pressure to its original state. All the steps are quasi-static and reversible. ( a ) Show this cycle on a PV diagram. For each step of the cycle, find the work done by the gas, the heat absorbed by the gas, and the change in the internal energy of the gas. ( b ) Find the efficiency of the cycle.
Picture the Problem To find the heat added during each step we need to find the temperatures in states 1, 2, 3, and 4. We can then find the work done on the gas during each process from the area under each straight-line segment and the heat that enters the system from Q = C V Δ T and Q = C P Δ T .We can use the 1st^ law of thermodynamics to find the change in internal energy for each step of the cycle. Finally, we can find the efficiency of the cycle from the work done each cycle and the heat that enters the system each cycle.
( a ) The cycle is shown to the right:
Apply the ideal-gas law to state 1 to find T 1 :
( )( )
( )
300 K
mol K
L atm 1.00mol 8.206 10
1.00atm 24.6L 2
1 1 1 = ⎟ ⎠
×
nR −
PV
T
The pressure doubles while the volume remains constant between states 1 and 2. Hence:
T 2 = 2 T 1 = 600 K
The volume doubles while the pressure remains constant between states 2 and 3. Hence:
T 3 = 2 T 2 = 1200 K
The pressure is halved while the volume remains constant between states 3 and 4. Hence:
T 4 = 21 T 3 = 600 K
1886 Chapter 19
For path 1 → 2 :
W 12 = P Δ V 12 = 0
and
( 600 K 300 K) 3. 74 kJ molK
J
12 VΔ^1223 Δ 12 23 8.314 ⎟ − =
Q = C T = R T =
The change in the internal energy of the system as it goes from state 1 to state 2 is given by the 1st^ law of thermodynamics:
Δ E int = Q in+ W on
Because W 12 (^) = 0 : (^) Δ E int, 12 = Q 12 = 3. 74 kJ
For path 2 → 3 :
( )( ) 4. 99 kJ L atm
101.325J
on 23 Δ^23 2.00atm 49.2L 24.6L ⎟= − ⎠
W = − W =− P V =− −
( 1200 K 600 K) 12. 5 kJ molK
J
23 PΔ^2325 Δ 23 25 8.314 ⎟ − =
Q = C T = R T =
Apply Δ E (^) int= Q in+ W onto obtain: (^) Δ E int, 23 = 12. 5 kJ− 4. 99 kJ= 7. 5 kJ
For path 3 → 4 :
W 34 = P Δ V 34 = 0
and
( 600 K 1200 K) 7. 48 kJ molK
J
34 Δ^ int, 34 VΔ 34 23 Δ 34 23 8.314 ⎟ − = − ⎠
Q = E = C T = R T =
Apply Δ E int = Q in+ W onto obtain: (^) Δ E int, 34 =− 7. 48 kJ+ 0 = − 7. 48 kJ
For path 4 → 1 :
( )( ) 2. 49 kJ L atm
101.325J
on 41 Δ^41 1.00atm 24.6L^49.^2 L ⎟= ⎠
W = − W =− P V =− −
and
( 300 K 600 K) 6. 24 kJ molK
J
41 PΔ^4125 Δ 41 25 8.314 ⎟ − = −
Q = C T = R T =
