Physics 41 Chapter 22 HW
1. A heat engine performs 200 J of work in each cycle and has an efficiency of 30.0%. For each cycle, how much
energy is (a) taken in and (b) expelled as heat?
Weng = Qh − Qc = 200 J
Weng
Qc
e= =
=
1−
0.300 (2)
Qh
Qh
(1)
Qc = 0.700 Qh
From (2),
(3)
Solving (3) and (1) simultaneously,
(a)
Qh = 667 J
we have
and
(b)
Qc = 467 J .
2. A refrigerator has a coefficient of performance of 3.00. The ice tray compartment is at –20.0°C, and the room
temperature is 22.0°C. The refrigerator can convert 30.0 g of water at 22.0°C to 30.0 g of ice at –20.0°C each
minute. What input power is required? Give your answer in watts.
Q
Qc
. Therefore, W = c .
3.00
W
The heat removed each minute is
QC
=
( 0.030 0 kg )( 4 186 J kg °C ) ( 22.0°C ) + ( 0.030 0 kg ) ( 3.33 × 105 J kg )
t
+ ( 0.030 0 kg )( 2 090 J kg °C ) ( 20.0°C )= 1.40 × 104 J min
COP
= 3.00
=
Qc
= 233 J s
t
Thus, the work done per second is
Js
P 233
=
= 77.8 W .
3.00
or,
3. An electric power plant that would make use of the temperature gradient in the ocean has been proposed.
The system is to operate between 20.0°C (surface water temperature) and 5.00°C (water temperature at a depth
of about 1 km). (a) What is the maximum efficiency of such a system? (b) If the useful power output of the plant
is 75.0 MW, how much energy is taken in from the warm reservoir per hour? (c) In view of your answer to part
(a), do you think such a system is worthwhile? Note that the “fuel” is free.
a)
1
emax =−
(b)
Tc
278
1
5.12 × 10−2 =5.12%
=−
=
293
Th
=
P
Weng
= 75.0 × 106 J s
∆t
(
From e =
(c)
)
Weng =
75.0 × 106 J s ( 3 600 s h ) =
2.70 × 1011 J h
Therefore,
Weng
Qh
we find
Weng 2.70 × 1011 J h
=
5.27 × 1012 J h =
5.27 TJ h
Qh = =
e
5.12 × 10−2
As fossil-fuel prices rise, this way to use solar energy will become a good buy.
4. The temperature at the surface of the Sun is approximately 5 700 K, and the temperature at the surface of the
Earth is approximately 290 K. What entropy change occurs when 1 000 J of energy is transferred by radiation
from the Sun to the Earth?
∆S=
Q2 Q1  1 000 1 000 
− = 
−
 J K=
T2 T1  290 5 700 
3.27 J K
5. A 2.00-L container has a center partition that divides it into two equal parts,
as shown. The left side contains H2 gas, and the right side contains O2 gas. Both
gases are at room temperature and at atmospheric pressure. The partition is
removed and the gases are allowed to mix. What is the entropy increase of the
system?
 Vf 
=
∆S nR=
ln  
 Vi 
6.
R ln 2
∆S
( 0.044 0) ( 2) =
0.088
=
0 ( 8.314) ln 2
0.507 J K
At point A in a Carnot cycle, 2.34 mol of a monatomic ideal gas has a pressure of 1 400 kPa, a volume
of 10.0 L, and a temperature of 720 K. It expands isothermally to point B, and then expands
adiabatically to point C where its volume is 24.0 L. An isothermal compression brings it to point D,
where its volume is 15.0 L. An adiabatic process returns the gas to point A. (a) Draw the PV diagram.
(b) Determine all the unknown pressures, volumes and temperatures as you fill in the following table.
(c) Find the energy added by heat, the work done by the engine, and the change in internal energy for
each of the steps A→B, B→C, C→D, and D→A. But values in the table. (d) Calculate the efficiency
Wnet/Qh. Show that it is equal to 1 – TC/TA, the Carnot efficiency.
Results:
State
A
B
C
D
P(kPa)
1 400
875
445
712
V(L)
10.0
16.0
24.0
15.0
T(K)
720
720
549
549
Process
Q(kJ)
W(kJ)
A →B
B→C
C→D
D→A
ABCDA
+6.58
0
–5.02
0
+1.56
–6.58
–4.98
+5.02
+4.98
–1.56
∆Eint (kJ)
0
–4.98
0
+4.98
0
#6. Calculations:
(a)
First, consider the adiabatic process D → A :
γ
γ
γ
so
PD VD = PA VA
 nRTD
Also 
 VD
53
 VA 
 10.0 L 
=
=
1
400
kPa
PD P=




A
 15.0 L 
 VD 
 γ  nRTA
 VD = 

 VA
 γ
 VA

γ −1
or
23
 VA 
 10.0 
=
=
720
K
TD T=



A 
 15.0 
 VD 
Now, consider the isothermal process C → D : T=
T=
C
D
445 kPa
V 
PA VAγ
from above. Also considering the isothermal process, PB = PA  A 
γ −1
VCVD
 VB 
V
Hence, PA  A
 VB
 γ  PA VAγ  γ
VA VC 10.0 L ( 24.0 L )
V which reduces to
=
VB =
=
 VB = 
γ −1  C
VD
15.0 L

 VCVD 
 VA 
 10.0 L 
=
PB P=
Finally,
 1 400 kPa  =

A 
V
 16.0 L 
 B
(b)
549 K
549 K
γ
53
1 400 kPa ( 10.0 L )
 VD    VA    VD  PA VAγ
=
P
=
PC P=
P
=
=




 
C
D 
A 
23
γ −1
24.0 L ( 15.0 L )
 VC    VD    VC  VCVD
Next, consider the adiabatic process B → C : PBVBγ = PCVCγ
But, PC =
712 kPa
For the isothermal process A → B :
16.0 L
875 kPa
∆Eint= nCV ∆T=
0
V 
 16.0 
−W =
2.34 mol ( 8.314 J mol ⋅ K ) ( 720 K ) ln 
nRT ln  B  =
so Q =
 =+6.58 kJ
 10.0 
 VA 
For the adiabatic process B → C :
Q= 0
3

2.34 mol  ( 8.314 J mol ⋅ K )  ( 549 − 720) K =−4.98 kJ
∆Eint =
nCV ( TC − TB ) =
2

and W = −Q + ∆Eint = 0 + ( −4.98 kJ) = −4.98 kJ
For the isothermal process C → D :
∆Eint= nCV ∆T=
0
V 
 15.0 
−W =
2.34 mol ( 8.314 J mol ⋅ K ) ( 549 K ) ln 
nRT ln  D  =
and Q =
 =−5.02 kJ
 24.0 
 VC 
Finally, for the adiabatic process D → A :
Q= 0
3

2.34 mol  ( 8.314 J mol ⋅ K )  ( 720 − 549) K =+4.98 kJ
∆Eint =
nCV ( TA − TD ) =
2

and W = −Q + ∆Eint = 0 + 4.98 kJ = +4.98 kJ
The work done by the engine is the negative of the work input. The output work Weng is given by the
work column in the table with all signs reversed.
(c) =
e
Weng −WABCD 1.56 kJ
=
=
= 0.237 or 23.7%
6.58 kJ
Qh
QA → B
ec =1 −
Tc
549
=1 −
=0.237 or 23.7%
720
Th
7. A heat engine with a diatomic gas as the working
substance uses the closed cycle shown. How much work does
this engine do per cycle and what is its thermal efficiency?
But all your values in the tables as usual.
P(atm)
3
3
1
1
1
2
3
4
V(m3)
1
2
2
1
T(K)
900
2700
900
300
W (kJ)
Q (kJ)
∆Eint (kJ)
1→2
-608
2127
1519
2→3
0
-1519
-1519
3→4
+203
-709
-506
4→ 1
0
5.06
506
-405
405
0
Net
8 Otto Cycle: Approximates the process occurring in an internal combustion engine. The Otto Cycle represents
a four-stroke cycle consisting of two upstrokes and two downstrokes.
The compression ratio (V1/V2)=8 and gas is diatomic:
Otto Cycle: e =−
1
1
T
T
1 A =−
1 D,
=−
γ −1
(V1 / V2 )
TB
TC
The tables look like:
State
T(K)
P (kPa)
A
B
C
D
A
293
673
1 023
445
293
100
1 840
2 790
152
100
Process
Q(J)
output W(J)
AB
BC
CD
DA
ABCDA
0
149
0
–65.0
84.3
(a), (b)
V ( cm 3 )
500
62.5
62.5
500
500
–162
0
246
0
84.3
Eint (J)
125
287
436
190
125
∆Eint (J)
162
149
–246
–65.0
0
The quantity of gas is
=
n
PA VA
=
RTA
(100 × 10
Pa)( 500 × 10−6 m 3 )
= 0.020 5 mol
( 8.314 J mol ⋅ K ) ( 293 K )
3
5
5
5
Eint, A =nRTA =PA VA =( 100 × 103 Pa)( 500 × 10−6 m 3 ) =125 J
2
2
2
γ
V 
In process AB, PB =
PA  A  =
1.84 × 106 Pa
(100 × 103 Pa) ( 8.00)1.40 =
V
 B
=
TB
PBVB
=
nR
Pa)( 500 × 10−6 m 3 8.00)
=
( 0.020 5 mol )( 8.314 J mol ⋅ K )
(1.84 × 10
6
673 K
5
5
287 J
Eint, B =nRTB =( 0.020 5 mol )( 8.314 J mol ⋅ K ) ( 673 K ) =
2
2
so
∆Eint, AB =
287 J− 125 J =
162 J =
Q − Wout =
0 − Wout
WAB =
−162 J
Process BC takes us to:
nRTC ( 0.020 5 mol )( 8.314 J mol ⋅ K )( 1 023 K )
=
= 2.79 × 106 Pa
62.5 × 10−6 m 3
VC
5
5
436 J
Eint, C =
nRTC =
( 0.020 5 mol )( 8.314 J mol ⋅ K )(1 023 K ) =
2
2
436 J− 287 J =149 J =−
Eint, BC =
Q Wout =−
Q 0
=
PC
QBC = 149 J
In process CD:
γ
1.40
V 
 1 
PD =
PC  C  =
2.79 × 106 Pa) 
1.52 × 105 Pa
(
 =
 8.00 
 VD 
=
TD
PD VD
=
nR
(1.52 × 10
Pa)( 500 × 10−6 m 3 )
=
( 0.020 5 mol )( 8.314 J mol ⋅ K )
5
445 K
5
5
Eint, D =nRTD =( 0.020 5 mol )( 8.314 J mol ⋅ K ) ( 445 K ) =
190 J
2
2
Q − Wout = 0 − Wout
∆Eint, CD =
−246 J =
190 J− 436 J =
WCD = 246 J
and
∆Eint, DA =
Eint, A − Eint, D =
125 J− 190 J =−65.0 J =−
Q Wout =−
Q 0
QDA =
−65.0 J
For the entire cycle, ∆Eint, net = 162 J+ 149 − 246 − 65.0 =
0 . The net work is
Weng =−162 J+ 0 + 246 J+ 0 = 84.3 J
Qnet = 0 + 149 J+ 0 − 65.0 J = 84.3 J
(c)
The input energy is Qh = 149 J , the waste is Qc = 65.0 J , and Weng = 84.3 J .
Weng 84.3 J
=
=
149 J
Qh
(d)
The efficiency is:
=
e
(e)
Let f represent the angular speed of the crankshaft. Then
0.565
obtain work in the amount of 84.3 J/cycle:
f
1 000 J s =   ( 84.3 J cycle)
 2
2 000 J s
f
s
=
= 23.7 rev
=
84.3 J cycle
f
is the frequency at which we
2
1.42 × 103 rev min