Chapter 20:
Exercises: 3, 7, 11, 22, 28, 34
EOC: 40, 43, 46, 58
E3:
A gasoline engine takes in 1.80×104
and delivers 3800
gasoline with a heat of combustion of 4.60×10
Part A
4
of work per cycle. The heat is obtained by burning
.
What is the thermal efficiency?
ANSWER:
=
Part B
How much heat is discarded in each cycle?
ANSWER:
Part C
What mass of fuel is burned in each cycle?
ANSWER:
=
Part D
If the engine goes through 50.0 cycles per second, what is its power output in kilowatts?
ANSWER:
Part E
If the engine goes through 50.0 cycles per second, what is its power output in horsepower?
ANSWER:
E7:
Part A
What compression ratio must an Otto cycle have to achieve an ideal efficiency of 66.2
ANSWER:
=
if
= 1.45?
E11:
A window air-conditioner unit absorbs 9.80×104
same time period deposits 1.50×10
Part A
5
of heat per minute from the room being cooled and in the
of heat into the outside air.
What is the power consumption of the unit?
ANSWER:
Part B
What is the energy efficiency rating of the unit?
ANSWER:
=
E22:
Description: A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold
reservoir consisting of a large tub of ice and water. In five minutes of operation of the engine, the heat rejected
by the engine melts 0.0400 kg of...
A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir
consisting of a large tub of ice and water. In five minutes of operation of the engine, the heat rejected by the
engine melts 0.0400
Part A
of ice.
During this time, how much work
ANSWER:
=
is performed by the engine?
E28:
Description: You make tea with 0.250 kg of T water and let it cool to room temperature (20.0 degree(s) C)
before drinking it. (a) Calculate the entropy change of the water while it cools. (b) The cooling process is
essentially isothermal for the air in your...
You make tea with 0.250
Part A
of 86.0
water and let it cool to room temperature (20.0
) before drinking it.
Calculate the entropy change of the water while it cools.
ANSWER:
=
Part B
The cooling process is essentially isothermal for the air in your kitchen. Calculate the change in entropy of the
air while the tea cools, assuming that all the heat lost by the water goes into the air.
ANSWER:
=
Part C
What is the total entropy change of the system tea + air?
Express your answer using two significant figures.
ANSWER:
=
E34:
Description: A box is separated by a partition into two parts of equal volume. The left side of the box contains
500 molecules of nitrogen gas; the right side contains 100 molecules of oxygen gas. The two gases are at the
same temperature. The partition is...
A box is separated by a partition into two parts of equal volume. The left side of the box contains 500 molecules
of nitrogen gas; the right side contains 100 molecules of oxygen gas. The two gases are at the same temperature.
The partition is punctured, and equilibrium is eventually attained. Assume that the volume of the box is large
enough for each gas to undergo a free expansion and not change temperature.
Part A
On average, how many molecules of oxygen will there be in either half of the box?
Express your answer using two significant figures.
ANSWER:
molecules
Part B
On average, how many molecules of nitrogen will there be in either half of the box?
Express your answer using two significant figures.
ANSWER:
molecules
Part C
What is the change in entropy of the system when the partition is punctured?
ANSWER:
=
Part D
What is the probability that the molecules will be found in the same distribution as they were before the
partition was punctured, that is, 500 nitrogen molecules in the left half and 100 oxygen molecules in the right
half? Give the order of magnitude.
Express your answer as the order of magnitude. If your answer is
, so you would enter -73
ANSWER:
, then the order of magnitude is
20.40.
IDENTIFY: Use the ideal gas law to calculate p and V for each state. Use the first law and specific expressions for Q, W, and ∆U
for each process. Use Eq.(20.4) to calculate e. QH is the net heat flow into the gas.
SET UP: γ = 1.40
CV = R /(γ − 1) = 20.79 J/mol ⋅ K; C p = CV + R = 29.10 J/mol ⋅ K. The cycle is sketched in Figure 20.40.
T1 = 300 K
T2 = 600 K
T3 = 492 K
Figure 20.40
EXECUTE: (a) point 1
p1 = 1.00 atm = 1.013 × 105 Pa (given); pV = nRT ;
V1 =
nRT1 (0.350 mol)(8.3145 J/mol ⋅ K)(300 K)
=
= 8.62 × 10−3 m 3
p1
1.013 × 105 Pa
point 2
process 1 → 2 at constant volume so V2 = V1 = 8.62 × 10−3 m 3
pV = nRT and n, R, V constant implies p1 / T1 = p2 / T2
p2 = p1 (T2 / T1 ) = (1.00 atm)(600 K/300 K) = 2.00 atm = 2.03 × 105 Pa
point 3
Consider the process 3 → 1, since it is simpler than 2 → 3.
Process 3 → 1 is at constant pressure so p3 = p1 = 1.00 atm = 1.013 × 105 Pa
pV = nRT and n, R, p constant implies V1 / T1 = V3 / T3
V3 = V1 (T3 / T1 ) = (8.62 × 10−3 m 3 )(492 K/300 K) = 14.1 × 10 −3 m 3
(b) process 1 → 2
constant volume ( ∆V = 0)
Q = nCV ∆T = (0.350 mol)(20.79 J/mol ⋅ K)(600 K − 300 K) = 2180 J
∆V = 0 and W = 0. Then ∆U = Q − W = 2180 J
process 2 → 3
Adiabatic means Q = 0.
∆U = nCV ∆T (any process), so
∆U = (0.350 mol)(20.79 J/mol ⋅ K)(492 K − 600 K) = −780 J
Then ∆U = Q − W gives W = Q − ∆U = +780 J. (It is correct for W to be positive since ∆V is positive.)
process 3 → 1
For constant pressure
W = p∆V = (1.013 × 105 Pa)(8.62 × 10−3 m3 − 14.1 × 10−3 m 3 ) = −560 J
or W = nR∆T = (0.350 mol)(8.3145 J/mol ⋅ K)(300 K − 492 K) = −560 J, which checks. (It is correct for W to be negative, since
∆V is negative for this process.)
Q = nC p ∆T = (0.350 mol)(29.10 J/mol ⋅ K)(300 K − 492 K) = −1960 J
∆U = Q − W = −1960 J − ( −560 K) = −1400 J
or ∆U = nCV ∆T = (0.350 mol)(20.79 J/mol ⋅ K)(300 K − 492 K) = −1400 J, which checks
(c) Wnet = W1→ 2 + W2 →3 + W3→1 = 0 + 780 J − 560 J = +220 J
(d) Qnet = Q1→ 2 + Q2 →3 + Q3→1 = 2180 J + 0 − 1960 J = +220 J
(e) e =
work output
W
220 J
=
=
= 0.101 = 10.1%.
heat energy input QH 2180 J
e(Carnot) = 1 − TC / TH = 1 − 300 K/600 K = 0.500.
EVALUATE:
For a cycle ∆U = 0, so by ∆U = Q − W it must be that Qnet = Wnet for a cycle. We can also check that ∆U net = 0:
∆U net = ∆U1→ 2 + ∆U 2 →3 + ∆U 3→1 = 2180 J − 1050 J − 1130 J = 0
e < e(Carnot), as it must.
20.43.
Tb = Tc and is equal to the maximum temperature. Use the ideal gas law to calculate Ta . Apply the appropriate
IDENTIFY:
expression to calculate Q for each process. e =
SET UP:
W
. ∆U = 0 for a complete cycle and for an isothermal process of an ideal gas.
QH
For helium, CV = 3R / 2 and C p = 5 R / 2. The maximum efficiency is for a Carnot cycle, and eCarnot = 1 − TC / TH .
EXECUTE: (a) Qin = Qab + Qbc . Qout = Qca . Tmax = Tb = Tc = 327°C = 600 K.
paVa pbVb
p
1
nRTb (2 moles)(8.31 J/mol ⋅ K)(600 K)
=
→ Ta = a Tb = (600 K) = 200 K. pbVb = nRTb → Vb =
=
= 0.0332 m 3 .
Ta
Tb
pb
pb
3
3.0 × 105 Pa
pbVb pcVc
p
3
=
→ Vc = Vb b = (0.0332 m3 )
= 0.0997 m 3 = Va .
Tb
Tc
pc
1
Qab = nCV ∆Tab = (2 mol)
Qbc = Wbc =
c
b
pdV =
c
b
3
( 8.31 J/mol ⋅ K ) (400 K) = 9.97 × 103 J
2
nRTb
V
dV = nRTb ln c = nRTb ln 3.
V
Vb
Qbc = (2.00 mol) ( 8.31 J/mol ⋅ K ) (600 K)ln 3 = 1.10 × 104 J. Qin = Qab + Qbc = 2.10 × 10 4 J.
Qout = Qca = nC p ∆Tca = (2.00 mol)
5
( 8.31 J/mol ⋅ K ) (400 K) = 1.66 × 104 J.
2
(b) Q = ∆U + W = 0 + W → W = Qin − Qout = 2.10 × 104 J − 1.66 × 104 J = 4.4 × 103 J.
4.4 × 103 J
= 0.21 = 21%.
2.10 × 104 J
T
(c) emax = eCarnot = 1 − C = 1 − 200 K = 0.67 = 67%
600 K
TH
EVALUATE: The thermal efficiency of this cycle is about one-third of the efficiency of a Carnot cycle that operates between the
same two temperatures.
e = W Qin =
20.46.
IDENTIFY: Use Eq.(20.4) to calculate e.
SET UP: The cycle is sketched in Figure 20.46.
CV = 5 R / 2
for an ideal gas C p = CV + R = 7 R / 2
Figure 20.46
SET UP: Calculate Q and W for each process.
process 1 → 2
∆V = 0 implies W = 0 & ∆V = 0 implies Q = nCV ∆T = nCV (T2 − T1 )
But pV = nRT and V constant says p1V = nRT1 and p2V = nRT2 .
Thus ( p2 − p1 )V = nR (T2 − T1 ); V ∆p = nR∆T (true when V is constant).
Then Q = nCV ∆T = nCV (V ∆p / nR ) = (CV /R )V ∆p = (CV /R)V0 (2 p0 − p0 ) = (CV /R) p0V0 . Q > 0; heat is absorbed by the gas.)
process 2 → 3
∆p = 0 so W = p∆V = p (V3 − V2 ) = 2 p0 (2V0 − V0 ) = 2 p0V0 (W is positive since V increases.)
∆p = 0 implies Q = nC p ∆T = nC p (T2 − T1 )
But pV = nRT and p constant says pV1 = nRT1 and pV2 = nRT2 .
Thus p (V2 − V1 ) = nR (T2 − T1 ); p∆V = nR∆T (true when p is constant).
Then Q = nC p ∆T = nC p ( p∆V/nR ) = (C p /R) p∆V = (C p /R )2 p0 (2V0 − V0 ) = (C p /R )2 p0V0 . (Q > 0; heat is absorbed by the gas.)
process 3 → 4
∆V = 0 implies W = 0
∆V = 0 so
Q = nCV ∆T = nCV (V ∆p/nR ) = (CV /R)(2V0 )( p0 − 2 p0 ) = −2(CV /R ) p0V0
(Q < 0 so heat is rejected by the gas.)
process 4 → 1
∆p = 0 so W = p∆V = p (V1 − V4 ) = p0 (V0 − 2V0 ) = − p0V0 (W is negative since V decreases)
∆p = 0 so Q = nC p ∆T = nC p ( p∆V/nR ) = (C p /R) p∆V = (C p /R ) p0 (V0 − 2V0 ) = −(C p /R ) p0V0 (Q < 0 so heat is rejected by the gas.)
total work performed by the gas during the cycle:
Wtot = W1→ 2 + W2 →3 + W3→ 4 + W4 →1 = 0 + 2 p0V0 + 0 − p0V0 = p0V0
(Note that Wtot equals the area enclosed by the cycle in the pV-diagram.)
total heat absorbed by the gas during the cycle (QH ): Heat is absorbed in processes 1 → 2 and 2 → 3.
QH = Q1→ 2 + Q2 →3 =
C
C + 2C p
CV
p0V0 + 2 p p0V0 = V
p0V0
R
R
R
But C p = CV + R so QH =
CV + 2(CV + R )
3CV + 2 R
p0V0 =
p0V0 .
R
R
total heat rejected by the gas during the cycle (QC ): Heat is rejected in processes 3 → 4 and 4 → 1.
QC = Q3→ 4 + Q4 →1 = −2
C
2CV + C p
CV
p0V0 − p p0V0 = −
p0V0
R
R
R
But C p = CV + R so QC = −
efficiency
e=
2CV + (CV + R )
3CV + R
p0V0 = −
p0V0 .
R
R
W
p0V0
R
R
2
=
=
=
= . e = 0.105 = 10.5%
QH ([3CV + 2 R ]/ R ) ( p0V0 ) 3CV + 2 R 3(5 R/2) + 2 R 19
EVALUATE: As a check on the calculations note that QC + QH = −
3CV + R
3CV + 2 R
p0V0 +
p0V0 = p0V0 = W , as it should.
R
R
20.58.
IDENTIFY and SET UP: First use the methods of Chapter 17 to calculate the final temperature T of the system.
EXECUTE: 0.600 kg of water (cools from 45.0°C to T )
Q = mc∆T = (0.600 kg)(4190 J/kg ⋅ K)(T − 45.0°C) = (2514 J/K)T − 1.1313 × 105 J
0.0500 kg of ice (warms to 0°C, melts, and water warms from 0°C to T )
Q = mcice (0°C − ( −15.0°C)) + mLf + mcwater (T − 0°C)
Q = 0.0500 kg (2100 J/kg ⋅ K)(15.0°C) + 334 × 103 J/kg + (4190 J/kg ⋅ K)(T − 0°C)
Q = 1575 J + 1.67 × 104 J + (209.5 J/K)T = 1.828 × 104 J + (209.5 J/K)T
Qsystem = 0 gives (2514 J/K)T − 1.1313 × 105 J + 1.828 × 104 J + (209.5 J/K)T = 0
(2.724 × 103 J/K)T = 9.485 × 104 J
T = (9.485 × 104 J)/(2.724 × 103 J/K) = 34.83°C = 308 K
EVALUATE: The final temperature must lie between –15.0°C and 45.0°C. A final temperature of 34.8°C is consistent with only
liquid water being present at equilibrium.
IDENTIFY and SET UP: Now we can calculate the entropy changes. Use ∆S = Q / T for phase changes and the method of Example
20.6 to calculate ∆S for temperature changes.
EXECUTE: ice: The process takes ice at –15°C and produces water at 34.8°C. Calculate ∆S for a reversible process between these
two states, in which heat is added very slowly. ∆S is path independent, so ∆S for a reversible process is the same as ∆S for the actual
(irreversible) process as long as the initial and final states are the same.
∆S =
2
1
dQ / T , where T must be in kelvins
For a temperature change dQ = mc dT so ∆S =
T2
T1
( mc / T ) dT = mc ln(T2 / T1 ).
For a phase change, since it occurs at constant T,
∆S =
2
1
dQ / T = Q / T = ± mL / T .
Therefore ∆Sice = mcice ln(273 K/258 K) + mLf / 273 K + mcwater ln(308 K/273 K)
∆Sice = (0.0500 kg)[(2100 J/kg ⋅ K) ln(273 K/258 K) + (334 × 103 J/kg)/273 K + (4190 J/kg ⋅ K)ln(308 K/273 K)]
∆Sice = 5.93 J/K + 61.17 J/K + 25.27 J/K = 92.4 J/K
water: ∆S water = mc ln(T2 / T1 ) = (0.600 kg)(4190 J/kg ⋅ K)ln(308 K/318 K) = −80.3 J/K
For the system, ∆S = ∆Sice + ∆S water = 92.4 J/K − 80.3 J/K = +12 J/K
EVALUATE: Our calculation gives ∆S > 0, as it must for an irreversible process of an isolated system.