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CH4 SECOND LAW OF THERMODYNAMICS 82
CHAPTER 04 THE SECOND LAW OF
THERMODYNAMICS
4-1 EFFICIENCY OF A HEAT ENGINE
Example 4-1
A diesel engine performs J2200 of mechanical work and
discards J4300 of heat each cycle.
(a) How much heat must be supplied to the engine in each
cycle?
(b) What is the thermal efficiency of the engine?
Solution
(a) JWQQ OUTIN 650022004300 =+=+=
(b) %3434.06500
2200or
Q
W
IN
===η
Example 4-2
An engine with %20 efficiency does J100 of work in each
cycle.
(a) How much heat is absorbed in each cycle?
(b) How much heat is rejected in each cycle?
Solution (a) The efficiency of a heat engine is given by
INQ
W=η
JQ
Q ININ 500
20.0
100===
η
(b) JWQQ INOUT 400100500 =−=−=
CH4 SECOND LAW OF THERMODYNAMICS 83
Example 4-3
An engine absorbs J100 and rejects J60 in each cycle.
(a) What is its efficiency?
(b) If each cycle takes s5.0 , find the power output of this
engine in watts.
Solution (a) The efficiency of the heat engine is defined as
%4040.0100
60100or
Q
Q
W
IN
OUTIN
IN
=−
=−
==η
(b) Wt
t
WP OUTIN 80
5.0
60100=
−=
−==
Example 4-4
Find the efficiency of a Carnot engine working between the
temperatures C0225 and C
025 . B.U. B.Sc. (Hons.) 1983A
Solution The efficiency of a Carnot engine is given by
%1001 ×
−=
H
L
T
Tη
%40%100273225
273251 =×
+
+−=η
Example 4-5
What is the maximum efficiency of a steam engine that
utilizes steam from a boiler at K480 and exhausts at
K373 . K.U. B.Sc. 2003
Solution The efficiency of a Carnot engine is given by
%1001 ×
−=
H
L
T
Tη
%3.22%100480
3731 =×
−=η
CH4 SECOND LAW OF THERMODYNAMICS 84
Example 4-6
A heat engine performs 200 joules of work and at the same
time rejects 3000 joules of heat to the cold body. Calculate
the efficiency. B.U. B.Sc. (Hons.) 1989A
Solution The efficiency of a Carnot engine is given by
%1001 ×
−=
IN
OUT
Q
Qη
%25.6%1002003000
30001 =×
+−=η
Example 4-7
A Carnot engine is operated between two heat reservoirs at
temperatures of K400 and K300 . If the engine receives
1200 joules of energy from the reservoir at K400 in each
cycle, how many joules of energy does it reject to the
reservoir at K300 ? B.U. B.Sc. (Hons.) 1986A
Solution The efficiency of a Carnot engine is given by
−=
−=
IN
OUT
H
L
Q
Q
T
T11η
IN
OUT
H
L
Q
Q
T
T=
JQT
TQ OUT
H
LOUT 900)1200(
400
300=
=
=
Example 4-8
A steam engine, operating between C0115 and C
060 ,
develops hp2500 . How much heat must be supplied if its
overall efficiency is 10 percent? What is the maximum
efficiency that an ideal heat engine can have when it
operates between the stated temperatures? (1 hp = 746 W).
B.U. B.Sc. 1999A
CH4 SECOND LAW OF THERMODYNAMICS 85
Solution
Let INQ be the supplied heat, then
ININ QQofW 1.0%10 ==
INQ1.07462500 =×
JQIN
710865.11.0
7462500×=
×=
The efficiency of ideal heat engine is given by
%2.14142.0273115
2736011 or
T
T
H
L =+
+−=
−=η
Example 4-9
A heat engine absorbs kJ4.52 of heat and exhausts kJ2.36
of heat per cycle. Calculate the efficiency and the work done
by the engine per cycle. P.U. B.Sc. 2003, 2009
Solution The efficiency of a Carnot engine is given by
%1001 ×
−=
IN
OUT
Q
Qη
%9.30%1004.52
2.361 =×
−=η
The work done by the engine per cycle is
kJQQW OUTIN 2.162.364.52 =−=−=
Example 4-10
A Carnot’s engine takes 61036.3 × joules of heat from a
reservoir at C0527 and gives some heat to sink at C
027 .
What is the efficiency of engine? How much work does it
perform in joules? K.U. B.Sc.2006
Solution The efficiency of heat engine is given by
−=
H
L
T
T1η
CH4 SECOND LAW OF THERMODYNAMICS 86
%5.62625.0273527
273271 or=
+
+−=η
The work done by the engine is
JQW IN
66 101.2)1036.3)(625.0( ×=×== η
Example 4-11
A Carnot engine extracts J240 of heat from a high
temperature reservoir during each cycle. It rejects J100 of
heat at reservoir at C015 .
(i) How much work does the engine do in one cycle?
(ii) What is its efficiency?
(iii) What is the temperature of the hot reservoir?
K.U. B.Sc. 2001
Solution (i) The work done by the engine is given by
JQQW OUTIN 140100240 =−=−=
(ii) The efficiency of the given Carnot engine is
%3.58%100240
1001%1001 =×
−=×
−=
IN
OUT
Q
Qη
(iii) The efficiency of Carnot engine in terms of
temperatures of reservoirs is given by
%1001 ×
−=
H
L
T
Tη
10027315
13.58 ×
+−=
HT
HT
2881583.0
100
3.58−==
418.0583.01288
=−=HT
CorKTH
0418691418.0
288==
CH4 SECOND LAW OF THERMODYNAMICS 87
Example 4-12
A heat engine operates between two reservoirs at K300
and K500 . During each cycle it absorbs cal200 of heat
from the hot reservoir.
(a) What is maximum efficiency of the engine?
(b) Determine the maximum work the engine performs
during each cycle. B.U. B.Sc. 2008A
Solution (a) The efficiency of heat engine is given by
−=
H
L
T
T1η
%4040.0500
3001 or=−=η
(b) The heat absorbed by the engine from hot reservoir in
each cycle is
JJcalQIN 2.837)186.4)(200(200 ===
The work done by the engine in each cycle is
JQW IN 88.334)2.837)(40.0( === η
Example 4-13
A Carnot engine absorbs heat from a reservoir at a
temperature of C01000 and rejects heat to a cold reservoir
at a temperature of C00 . If the engine absorbs 1000 joules
from the high temperature reservoir, find heat rejected and
efficiency of heat engine. B.U. B.Sc.(Hons.) 1988A, 1989A
Solution Now
H
L
IN
OUT
T
T
Q
Q=
JQT
TQ IN
H
LOUT 214
)2731000(
)1000)(2730(=
+
+=
=
The efficiency of heat engine is given by
CH4 SECOND LAW OF THERMODYNAMICS 88
%1001 ×
−=
H
L
T
Tη
%6.78%1002731000
27301 =×
+
+−=η
Example 4-14
In a Carnot cycle, the isothermal expansion of an ideal gas
takes at C0140 and isothermal compression at C
025 . During
the compression, J2090 of heat energy is transferred to the
gas. Calculate
(i) The work performed by the gas during isothermal
expansion.
(ii) Heat rejected from the gas during isothermal
compression.
(iii) Work done on the gas during isothermal compression.
K.U. B.Sc. 2002, 2005
Solution
(a) JQW IN 2090−=−= The negative sign indicates that the
work is done by the gas.
(b) The efficiency of Carnot engine is given by
%1001 ×
−=
H
L
T
Tη
%8.27%100273140
373251 =×
+
+−=η
Now %1001 ×
−=
IN
OUT
Q
Qη
)100(2090
18.27
−= OUTQ
−==
20901278.0
100
8.27 OUTQ
722.0278.012090
=−=OUTQ
CH4 SECOND LAW OF THERMODYNAMICS 89
JQOUT 1509)2090)(722.0( ==
(c) Work done on the gas during isothermal compression
JQW OUT 1509==
Example 4-15
A Carnot cycle has efficiency %75 working between
highest and lowest temperatures. What is the lowest
temperature of its high temperature is C0927 ?
B.U. B.Sc. 2007S
Solution The efficiency of a Carnot engine is given by
−=
H
L
T
T1η
273927
1100
75
+−= LT
1200
175.0 LT−=
25.075.011200
=−=LT
CKTL
027300)1200)(25.0( ===
Example 4-16
A Carnot engine has an efficiency of %22 . It operates
between heat reservoirs differing in temperature by C075 .
Find the temperature of the reservoirs.
K.U. B.Sc. 2004, 2008
Solution
Let LT and HT be the temperatures of cold and hot reservoirs
(i.e. sink and source) respectively, then the efficiency of Carnot
engine is given by
H
L
T
T−= 1η
Now 22.0%22 ==η
CH4 SECOND LAW OF THERMODYNAMICS 90
KTT LH )75( +=
Hence
75
75
75122.0
+
−+=
+−=
L
LL
L
L
T
TT
T
T
75
7522.0
+=
LT
755.1622.0 =+LT
5.585.167522.0 =−=LT
CorKTL
0726622.0
5.58−==
CorKTT LH
0683417526675 =+=+=
Example 4-17
A Carnot heat engine with an efficiency of 0.20 operates
between two temperatures that differ from each other
by K100 . What are the temperatures between which the
cycle operates?
Solution The efficiency of a Carnot engine is given by
H
L
T
T−= 1η
Now 20.0=η
KTT LH )100( +=
Hence
100
100
100120.0
+
−+=
+−=
L
LL
L
L
T
TT
T
T
100
10020.0
+=
LT
1002020.0 =+LT
802010020.0 =−=LT
KTL 40020.0
80==
CH4 SECOND LAW OF THERMODYNAMICS 91
KTT LH 500100400100 =+=+=
The given engine operates between 400 K and 500 K.
Example 4-18
The exhaust temperature of a heat engine is C0220 . What
must be the high temperature if the Carnot efficiency is to
be %36 ?
Solution
Now H
L
T
T−= 1η
HT
2732201
100
36 +−=
HT
493136.0 −=
64.036.01493
=−=HT
CorKTH
049777064.0
493==
Example 4-19
In one cycle of operation a heat engine takes in J2200 of
heat and performs J620 of work.
(a) What is the efficiency of this engine?
(b) How much heat is exhausted in each cycle?
(c) If the engine completes a cycle each s033.0 , then at
what rate is heat added, heat exhausted, work done?
Solution (a) The efficiency of heat engine is defined as
%2828.02200
620or
Q
W
IN
===η
(b) The heat exhausted in each cycle is
JWQQ INOUT 15806202200 =−=−=
(c) kWWt
QP IN
IN 7.661067.6033.0
2200 4 =×===
CH4 SECOND LAW OF THERMODYNAMICS 92
kWWt
QP OUT
OUT 9.471079.4033.0
1580 4 =×===
kWWt
WP 8.181088.1
033.0
620 4 =×===
Example 4-20
The low temperature reservoir of a Carnot engine is at
C010 and has an efficiency of %50 . How much the
temperature of the high temperature reservoir is increased
to measure the efficiency to %50 ? K.U. B.Sc. 2007
Solution The efficiency of a Carnot engine is given by
−=
H
L
T
T1η
HT
27310150.0
+−=
50.050.01283
=−=HT
KTH 56650.0
283==
Let 1η be the new efficiency of heat engine with new source
temperature as *
HT , then *1 1
H
L
T
T−=η
*
283160.0
HT−=
40.060.01283
*=−=
HT
KTH 5.70740.0
283* ==
The desired increase in temperature of hot reservoir is
KTTT HH 5.1415665.707* =−=−=∆
CH4 SECOND LAW OF THERMODYNAMICS 93
Example 4-21
A heat engine utilizes a heat source at C0580 and has a
Carnot efficiency of 29 percent. To increase the efficiency to
35 percent, what must be the temperature of the heat
source?
Solution The temperature of the cold reservoir i.e. sink is calculated as
under
H
L
T
T−= 1η
273580
129.0+
−= LT
71.029.01853
=−=LT
KTL 606)853)(71.0( ==
Let 1η be the new efficiency of heat engine with new source
temperature as *
HT , then
*1 1
H
L
T
T−=η
*
606135.0
HT−=
65.035.01606
*=−=
HT
CorKTH
0* 65993265.0
606==
CH4 SECOND LAW OF THERMODYNAMICS 94
Example 4-22
A Carnot heat engine operates between a reservoir at
K950 and has heat transfer of J400 exhausted to another
reservoir at K300 during each cycle. What is the amount of
work done by the heat engine during each cycle?
Solution
Now IN
OUT
H
L
Q
Q
T
T=
JQT
TQ OUT
L
HIN 1267)400(
300
950=
=
=
JQQW OUTIN 8674001267 =−=−=
The given heat engine performs 867 J of work in each cycle.
CH4 SECOND LAW OF THERMODYNAMICS 95
4-2 COEFFICIENT OF PERFORMANCE OF A REFRIGERATOR ENGINE
Example 4-23
The low temperature of a freezer cooling coil is C015− and
the discharge temperature is C030 . What is the maximum
theoretical coefficient of performance?
Solution The coefficient of performance is defined as
LH
L
TT
TK
−=
733.5)27315()27330(
)27315(=
+−−+
+−=K
Example 4-24
A heat pump acting as a refrigerator is used to heat a house.
The temperature outside the house is C010− and the
interior to be kept at C022 . Find the maximum coefficient of
performance of heat pump. P.U. B.Sc. 2005, 2006
Solution The coefficient of performance of heat pump acting as a
refrigerator is given by
LH
L
TT
TK
−=
Now KKCTL 263)27310(100 =+−=−=
KKCTH 295)27322(220 =+==
KTT LH 32263295 =−=−
Hence 218.832
263==K
Example 4-25
What is the coefficient of performance of a refrigerator that
operates with Carnot efficiency between temperatures
C03− and C
027 ?
CH4 SECOND LAW OF THERMODYNAMICS 96
The coefficient of performance of heat pump acting as a
refrigerator is given by
LH
L
TT
TK
−=
Now KKCTL 270)2733(30 =+−=−=
KKCTH 300)27327(270 =+==
KTT LH 30270300 =−=−
Hence 930
270==K
Example 4-26
A household refrigerator, whose coefficient of performance is 4.7, extracts heat from the cooling chamber at rate of 250
J per second. How much work per cycle is required to
operate the refrigerator? P.U. B.Sc. 2002
Solution The coefficient of performance K is defined as
W
Q
QK
L
LH
L=
−=
cycleJK
QW
L/2.53
7.4
250===
Example 4-27
A refrigerator does 153 J of work to transfer 568 J of heat
from its cold compartment.
(a) Calculate the refrigerator’s coefficient of performance.
(b) How much heat is exhausted to the kitchen?
Solution (a) The coefficient of performance K is defined as
71.3153
568===
W
QK
L
(b) Heat exhausted to kitchen will be
JWQL 721153568 =+=+
CH4 SECOND LAW OF THERMODYNAMICS 97
Example 4-28
Apparatus that liquefies helium is in a laboratory at K296 .
The helium in the apparatus is at K4 . If mJ150 of heat is
transferred from helium, find the minimum amount of heat
delivered to the laboratory.
Solution The coefficient of performance is defined as
LH
L
TT
TK
−=
21037.14296
4 −×=+
=K
Now
K
QQWQQ
L
LLH +=+=
JQL 1.111037.1
11)10150(
2
3 =
×+×=
−
−
CH4 SECOND LAW OF THERMODYNAMICS 98
ADDITIONAL PROBLEMS
(1) A Carnot engine is operated between two reservoirs
at temperatures C0227 and C
0127 . Evaluate the
efficiency of the engine. B.U. B.Sc.(Hons.) 1988S
(2) A heat engine performs 1000 joules of work and at
the same time rejects 4000 joules of heat energy to
the cold reservoir. What is the efficiency of the
engine? B.U. B.Sc.(Hons.) 1991A
(3) A heat engine receives heat 120 joules and the work
done by it is 90 joules. Calculate the efficiency in
percentage. B.U. B.Sc. 1997A
(4) A heat engine performs 2000 joules of work and at
the same time rejects 6000 joules of heat energy to
the cold reservoir. What is the efficiency of the
engine? B.U. B.Sc. 1998A
(5) A Carnot engine whose heat source is at C0127 takes
100 calories of heat at this temperature in each cycle
and gives up 80 calories to the heat sink. Find the
temperature of heat sink. B.U. B.Sc. 1988S
(6) A Carnot engine absorbs heat from a reservoir at a
temperature of C0100 and rejects heat to a cold
reservoir at a temperature of C00 . If the engine
absorbs 1000 joules from the high temperature
reservoir, find heat rejected and efficiency of heat
engine. B.U. B.Sc. 1986S
(7) What is the maximum efficiency of a steam engine
that utilizes steam from a boiler at K480 and
exhausts at K373 ? K.U. B.Sc. 2003
(8) A heat engine performs 200 joules of work and at the
same time rejects 300 joules of heat to the cold body.
Calculate its efficiency. B.U. B.Sc.(Hons.) 1989A
(9) Calculate the efficiency of a Carnot engine working
between the temperatures C0927 and C
027 .
P.U. B.Sc. 1989
CH4 SECOND LAW OF THERMODYNAMICS 99
(10) A Carnot engine’s working substance is water. It
uses steam at C0300 and condenses to water at C
040 .
What is the maximum theoretical efficiency of this
engine? B.P.S.C. 1995
(11) A Carnot engine operates between the temperatures
K850 and K300 . The engine performs J1200 of work
each cycle, which takes s25.0 . Calculate its efficiency
and its average power. What are the rates of heat input
and heat exhaust per cycle? F.P.S.C. 2009
Answers
(1) 20 % (2) 20 % (3) 75 % (4) 25 %
(5) 320 K (6) JQOUT 732= , %8.26=η (7) 22.3 %
(8) 6.25 % (9) 75 % (10) 37 %
(11) 65 %, 4800 W, 13600 W and 8800 W