7-1C Yes. Because we used the relation (QH/TH) = (QL/TL) in the proof, which is the defining relation of
absolute temperature.
7-4C No. A system may reject more (or less) heat than it receives during a cycle. The steam in a steam
power plant, for example, receives more heat than it rejects during a cycle.
7-8C Yes.
7-9C That integral should be performed along a reversible path to determine the entropy change.
7-11C The value of this integral is always larger for reversible processes.
7-14C Sometimes.
7-16C Always.
7-17C Increase.
7-19C Decreases.
7-22C They are heat transfer, irreversibilities, and entropy transport with mass.
7-25 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy
changes are negligible. 3 Air is an ideal gas. 4 The process involves no internal irreversibilities such as
friction, and it is an isothermal, internally reversible process.
Noting that h = h(T) for ideal gases, we have h1 = h2 since T1 = T2 = 25C.
We take the compressor as the system. Noting that the enthalpy of air remains constant, the energy balance
for this steady-flow system can be expressed in the rate form as
P2
Q  W H  0
W in  Q out
·
Q
AIR
T = const.
Therefore,
Q out  W in  12 kW
Noting that the process is assumed to be an isothermal and internally reversible process, the rate of entropy
change of air is determined to be
S air  
Q out,air
Tsys

P1
12 kW
 0.0403 kW/K
298 K
7-28E (a) This is a reversible isothermal process, and the entropy change during such a process is given by
Q
T
Noting that heat transferred from the working fluid is equal to the heat transferred to the sink, the heat
transfer become
S 
Q fluid  T fluid S fluid  555 R  0.7 Btu/R   388 .5Btu  Q fluid,out  388.5 Btu
(b) The entropy change of the sink is determined from
Q
388.5 Btu
Ssink  sink,in 
 0.7 Btu / R
Tsink
555 R
(c) the total entropy change of the process is
Sgen  Stotal  S fluid  Ssink  0.7  0.7  0
Heat
SINK
95F
95F
Carnot heat engine
This is expected since all processes of the Carnot cycle are reversible processes, and no entropy is
generated during a reversible process.
7-30C Yes, because an internally reversible, adiabatic process involves no irreversibilities or heat transfer.
12kW
7-32 1 The tank is stationary and the kinetic and potential energy changes are zero. 2 There are no work
interactions.
(a) From the refrigerant tables (Tables A-11 through A-13),
P1  200 kPa 

x1  0.4

u1  u f  x1u fg  36 .69  0.4 221 .43  36 .69   110.59 kJ/kg
s1  s f  x1 s fg  0.1481  0.4 0.9253  0.1481   0.4590 kJ/kg  K
v1  v f  x1v fg  0.0007532  0.4 0.0993  0.0007532   0.04017 m 3 /kg
v2  v f
0.04017  0.0007904
 0.7859
v fg
0.0509  0.0007904
P2  400 kPa 
 u 2  u f  x 2 u fg  61 .69  0.7859 231 .97  61 .69   195.51 kJ/kg
v 2  v1
 s  s  x s  0.2399  0.7859 0.9145  0.2399   0.7701 kJ/kg  K
2
f
2 fg
x2 

The mass of the refrigerant is
m
0.5 m 3
V

 12.45 kg
v1 0.04017 m 3 /kg
Then the entropy change of the refrigerant becomes
S system  ms 2  s1   12.45kg 0.7701  0.4590 kJ/kg  K  3.873 kJ/K
(b) We take the tank as the system. This is a closed system since no mass enters or leaves. Noting that the
volume of the system is constant and there is no boundary work, the energy balance for this stationary
closed system can be expressed as
Q  W  U  m(u 2  u1 )
R-134a
200 kPa
W0
Substituting,
Q
Source
35C
Qin  mu 2  u1   12.45 kg 195.51  110.59   1057 kJ
The heat transfer for the source is equal in magnitude but opposite in direction. Therefore,
Qsource, out = - Qtank, in = - 1057 kJ, and
S source  
Qsource,out
Tsource

1057 kJ
 3.432 kJ/K
308 K
(c) The total entropy change for this process is
S total  S system  S source  3.873   3.432   0.441 kJ/K
7-37E A cylinder is initially filled with R-134a at a specified state. The refrigerant is cooled and condensed
at constant pressure. The entropy change of refrigerant during this process is to be determined
From the refrigerant tables,
P1  120 psia 
 s1  0.2301 Btu/lbm  R
T1  120  F 

 s 2  s f @90 F  0.0836 Btu/lbm  R
P2  120 psia 
T2  90  F
R-134a
120 psia
120F
Q
Then the entropy change of the refrigerant becomes
S  ms 2  s1   3 lbm 0.0836  0.2301Btu/lbm  R  0.4395 Btu/R
7-44 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulated and heat
transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The process is stated
to be reversible.
This is a reversible adiabatic (i.e., isentropic) process, and s2 = s1. From the steam tables,
v  0.6339 m 3 /kg
P1  300 kPa  1
 u1  2570.8 kJ/kg
T1  150  C 
s  7.0778 kJ/kg  K
1
P2  1 MPa 
 u 2  2774.2 kJ/kg
s 2  s1

Also,
m
0.05 m 3
V

 0.0789 kg
v1 0.6339 m 3 /kg
H2O
300 kPa
150C
We take the contents of the cylinder as the system. This is a closed system since no mass enters or
leaves. The energy balance for this adiabatic closed system can be expressed as
Q  Wb ,in  U  m(u 2  u1 )
W done to system  
Q0
Substituting, the work input during this adiabatic process is determined to be
Wb,in  mu 2  u1   0.0789 kg 2774.2  2570.8 kJ/kg  16.05 kJ
7-49 1 Both the water and the copper block are incompressible substances with constant specific heats at
room temperature. 2 The system is stationary and the kinetic and potential energies are negligible. 3 The
tank is well-insulated and there is no heat transfer.
The density and specific heat of water at 25C are  = 997 kg/m3 and Cp = 4.18 kJ/kg.C. The specific heat
of copper at 27C is Cp = 0.386 kJ/kg.C (Table A-3).
We take the entire contents of the tank, water + copper block, as the system. This is a closed system since
no mass crosses the system boundary during the process. The energy balance for this system can be
expressed as
Q  W  U  0
WATER
or,
U Cu  U water  0
[mC (T2  T1 )] Cu  [mC (T2  T1 )] water  0
Copper
50 kg
where
mwater  V  (997kg/m3 )(0.120m3 )  120kg
Using specific heat values for copper and liquid water at room temperature and substituting,
(50kg)(0.386kJ/kg C)(T2  80) C  (120kg)(4.18kJ/kg C)(T2  25) C  0
T2 = 27.0C
The entropy generated during this process is determined from
T 
 300.0 K 
  3.140 kJ/K
S copper  mC ave ln  2   50 kg 0.386 kJ/kg  K ln 
 353 K 
 T1 
T 
 300.0 K 
  3.355 kJ/K
S water  mC ave ln  2   120 kg 4.18 kJ/kg  K ln 
 298 K 
 T1 
,
S total  S copper  S water  3.140  3.355  0.215 kJ/K
120 L
7-54C For ideal gases, Cp = Cv + R and
P2V2 P1V1
V
T P


 2  2 1
T2
T1
V1 T1 P2
,
T
s 2  s1  C v ln  2
 T1
T
 C v ln  2
 T1

V 
  R ln  2 

 V1 

T P 
  R ln  2 1 

 T1 P2 
T 
T 
P
 C v ln  2   R ln  2   R ln  2
 T1 
 T1 
 P1
T 
P 
 C p ln  2   R ln  2 
T
 1
 P1 



7-57C Setting s = 0 gives
T 
P
C p ln  2   R ln  2
 T1 
 P1

T
  0 
 ln  2

 T1
 R  P2 
P 
T
 
ln   
 2   2 
T1  P1 
 C p  P1 
But
R Cp
k 1 k
C p  Cv
P 
T
R
1 k 1

 1 
since k  C p / C v . Thus, 2   2 
Cp
Cp
k
k
T1  P1 
7-59C The entropy of a gas can change during an isothermal process since entropy of an ideal gas depends
on the pressure as well as the temperature.
7-62 At specified conditions, CO2 can be treated as an ideal gas with constant specific heats at room
temperature.
The specific heat of CO2 is Cv = 0.657 kJ/kg.K (Table A-2).
Using the ideal gas relation, the entropy change is determined to be
P2V P1V
T
P
120 kPa


 2  2 
 1.2
T2
T1
T1
P1 100 kPa
,
0


T
V
T
S  ms 2  s1   m C v , ave ln 2  R ln 2   mC v ,ave ln 2

T1
V1 
T1

 2.7 kg 0.657 kJ/kg  K ln 1.2 
 0.323 kJ/K
CO2
1.5 m3
100 kPa
1.2 kg
7-64 1 At specified conditions, N2 can be treated as an ideal gas. 2 Nitrogen has constant specific heats at
room temperature.
The gas constant of nitrogen is R = 0.297 kJ/kg.K (Table A-1). The constant volume specific heat of
nitrogen at room temperature is Cv = 0.743 kJ/kg.K (Table A-2).
From the polytropic relation,
T2  v1 
 
T1  v 2 
n 1
v

 T2  T1  1
 v2



n 1
 300 K 21.31  369 .3K
Then the entropy change of nitrogen becomes
N2
PV1.3 = C

T
V 
S N 2  m C v,ave ln 2  R ln 2 
T1
V1 



369.3 K
 1.2 kg  0.743 kJ/kg  K ln
 0.297 kJ/kg  K ln 0.5 
300 K


 0.0617 kJ/K
7-70 1 At specified conditions, air can be treated as an ideal gas. 2 The process is given to be reversible and
adiabatic, and isentropic. Therefore, isentropic relations of ideal gases apply.
The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The specific heat ratio of air at low to
moderately high temperatures is k = 1.4 (Table A-2).
(a) Assuming constant specific heats, the ideal gas isentropic relations give
k 1 k
0.4 1.4
P 
 800 kPa 

T2  T1  2 
 290 K 
 525.3K
 100 kPa 
 P1 
Then,
Tave  290  525.3/2  407.7K 
 Cv,ave  0.727kJ/kg  K
We take the air in the cylinder as the system. The energy balance for this
stationary closed system can be expressed as
Q  Win  U  m(u 2  u1 )  mCv (T2  T1 )
AIR
Reversible
Q0
W done to system  
win  Cv,ave T2  T1   0.727kJ/kg K525.3 290K  171.1kJ/kg
(b) Assuming variable specific heats, the final temperature can be determined using
the relative pressure data (Table A-17),
Pr1  1.2311
T1  290 K 

u1  206.91 kJ/kg
and
T  522.4K
P
800 kPa
1.2311   9.849  2
Pr2  2 Pr1 
u 2  376.16 kJ/kg
P1
100 kPa
Then the work input becomes
win  u 2  u1  376.16  206.91 kJ/kg  169.25kJ/kg
7-72 1 Helium is an ideal gas with constant specific heats. 2 The process is given to be reversible and
adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply.
The specific heats and the specific heat ratio of helium are Cv = 3.1156 kJ/kg.K, Cp = 5.1926 kJ/kg.K, and
k = 1.667 (Table A-2).
(a) From the ideal gas isentropic relations,
P
T2  T1  2
 P1



k 1 k
 450 kPa 

 303 K 
 90 kPa 
0.667 1.667
 576.9 K
(a) We take the air in the cylinder as the system. The energy balance for this
stationary closed system can be expressed as
Q  W  U
Q0
W done to system  
Win  U  m(u 2  u1 )  mCv (T2  T1 )
Thus,
win  C v T2  T1   (3.1156 kJ/kg  K )(576.9  303)K  853.4kJ/kg
(b) If the process takes place in a steady-flow device, the final temperature will remain the same but the
work done should be determined from an energy balance on this steady-flow device,
Q  W in  m (h2  h1 )
Q0
W done to system  
W in  m (h2  h1 ) m C p (T2  T1 )
Thus,
win  C p T2  T1   (5.1926 kJ/kg  K)(576.9  303)K  1422.3 kJ/kg
7-78C The work associated with steady-flow devices is proportional to the specific volume of the gas.
Cooling a gas during compression will reduce its specific volume, and thus the power consumed by the
compressor.
7-82 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3
The process is assumed to be reversible since we will determine the limiting case.
The specific volume of saturated liquid water at 20 kPa is v1 = vf @ 20 kPa = 0.001017 m3/kg (Table A-5).
The highest pressure the liquid can have at the pump exit can be determined from the reversible steady-flow
work relation for a liquid,
W in  m 


2
1
P2
vdP  k e0  pe0   m v1 P2  P1 

10 kW
Thus,
 1 kJ
10 kJ/s  (5 kg/s )(0.001017 m 3 /kg )( P2  100 )k Pa
3
 1 kPa  m




PUMP
It yields
100 kPa
P2  2100 kPa
7-84 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3
The process is reversible. 4 The pump and the turbine are adiabatic.
The specific volume of saturated liquid water at 20 kPa is v1 = vf @ 20 kPa = 0.001017 m3/kg (Table A-5).
Both the compression and expansion processes are reversible and adiabatic, and isentropic,
s1 = s2 and s3 = s4. Then the of the steam are
P4  20 kPa  h4  h g @ 20kPa  2609.7 kJ/kg

sat.vapor  s 4  s g @ 20kPa  7.9085 kJ/kg  K
P3  10 MPa 
 h3  4712.8 kJ/kg
s3  s 4

Also, v1 = vf @ 20 kPa = 0.001017 m3/kg.
2
The work output to this isentropic turbine is determined from the steady-flow energy balance to3 be
Q  W  m (h4  h3 )
Work done by  
W out  m (h3  h4 )
H2O
H2O
Substituting,
1
4
wturb, out  h3  h4  4712.8  2609.7  2103.1 kJ / kg
The pump work input is determined from the steady-flow
work relation to be
w pump,in 
2
 vdP  ke
0
1
 pe0  v1 P2  P1 
 1kJ 
  10.15 kJ/kg
 (0.001017 m 3 /kg )(10,000  20)kPa
3 
 1kPa  m 
,
wturb,out
w pump,in

2103.1
 207.2
10.15
7-112 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the
surroundings is negligible and heat transfer from the hot fluid is equal to the heat transfer to the cold fluid.
3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid are constant.
The specific heats of water and ethylene glycol are given to be 4.18 and 2.56 kJ/kg.C, respectively.
(a) We take the ethylene glycol tubes as the system, which is a control volume. The energy
balance for this steady-flow system can be expressed in the rate form as
Cold Water
20C
Q  W i  m (h2  h1 )
Q  m (h2  h1 ) m C p (T2  T1 )
Hot Glycol
80C
2 kg/s
Then the rate of heat transfer becomes
40C
 p (Tin  Tout )]glycol  (2 kg / s)(2.56 kJ / kg.  C)(80 C  40 C) = 204.8 kW
Q out  [ mC
The rate of heat transfer from water must be equal to the rate of heat transfer to the glycol. Then,
 p (Tout  Tin )]water 
Q in  [ mC
 m water 

Q in
C p (Tout  Tin )
204.8 kJ / s
= 1.4 kg / s
(4.18 kJ / kg.  C)(55 C  20 C)
(b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the
entropy balance on the entire heat exchanger:
S in  S out


Rate of net entropy transfer
by heat and mass

S gen

 S system0 (steady)

Rate of entropy
generation
Rate of change
of entropy
m 1 s1  m 3 s 3  m 2 s 2  m 3 s 4  S gen  0 (since Q  0)
m glycols1  m water s 3  m glycols 2  m water s 4  S gen  0
S gen  m glycol ( s 2  s1 )  m water ( s 4  s 3 )
Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is
determined to be
T
T
Sgen  m glycol C p ln 2  m water C p ln 4
T1
T3
 (2 kg / s)(2.56 kJ / kg.K)ln
40 + 273
55 + 273
 (1.4 kg / s)(4.18 kJ / kg.K)ln
 0.0446 kW / K
80 + 273
20 + 273
7-114 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the
surroundings is negligible and heat transfer from the hot fluid is equal to the heat transfer to the cold fluid.
3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid are constant.
The specific heats of cold and hot water are given to be 4.18 and 4.19 kJ/kg.C, respectively.
We take the cold water tubes as the system, which is a control volume. The energy balance for this
steady-flow system can be expressed in the rate form as
Cold Water
15C
0.25 kg/s
Q  W i  m (h2  h1 )
Q  m (h2  h1 ) m C p (T2  T1 )
Hot water
100C
Then the rate of heat transfer to the cold water in this heat exchanger becomes
3 kg/s
 p (Tout  Tin )]cold water  (0.25 kg / s)(4.18 kJ / kg.  C)(45 C  15 C) = 31.35 kW
Q in  [ mC
Noting that heat gain by the cold water is equal to the heat loss by the hot water, the outlet temperature of
the hot water is determined to be
Q
Q  [m C p (Tin  Tout )] hot water 
 Tout  Tin 
m C p
 100 C 
31.35 kW
 97.5C
(3 kg/s)(4.19 kJ/kg. C)
(b) The rate of entropy generation within the heat exchanger is determined by applying the rate form of the
entropy balance on the entire heat exchanger:
S in  S out



Rate of net entropy transfer
by heat and mass
S gen

 S system0 (steady)

Rate of entropy
generation
Rate of change
of entropy
m 1 s1  m 3 s 3  m 2 s 2  m 3 s 4  S gen  0 (since Q  0)
m cold s1  m hot s 3  m cold s 2  m hot s 4  S gen  0
S gen  m cold ( s 2  s1 )  m hot ( s 4  s 3 )
Noting that both fluid streams are liquids (incompressible substances), the rate of entropy generation is
determined to be
T
T
Sgen  m cold C p ln 2  m hot C p ln 4
T1
T3
45 + 273
97.5 + 273
 (3 kg / s)(4.19 kJ / kg.K)ln
 0.0190 kW / K
15 + 273
100 + 273
7-121 1 The thermal of the balls are constant. 2 There are no changes in kinetic and potential energies. 3
The balls are at a uniform temperature at the end of the process
The density and specific heat of the balls are given to be  = 7833 kg/m3 and Cp = 0.465 kJ/kg.C.
 (0.25 kg / s)(4.18 kJ / kg.K)ln
(a) We take a single ball as the system. The energy balance for this closed
system can be expressed as
Q  W i  m (h2  h1 )
Q  m (h2  h1 ) m C p (T2  T1 )
The amount of heat transfer from a single ball is
m  V  
Qout
D 3
 (7833 kg / m 3 )
 (0.008 m) 3
 0.00210 kg
6
6
 mC p (T1  T2 )  (0.0021 kg)(0.465 kJ / kg.  C)(900  100) C  781 J = 0.781 kJ (per ball)
Then the total rate of heat transfer from the balls to the ambient air becomes
Q  n Q  (2500 balls / h)  (0.781 kJ / ball)  1953
,
kJ / h  542 W
out
ball
out
(b) We again take a single ball as the system The entropy generated during this process can be determined
by applying an entropy balance on an extended system that includes the ball and its immediate surroundings
so that the boundary temperature of the extended system is at 35C at all times:
S in  S out


Net entropy transfer
by heat and mass

 S gen  S system




Entropy
generation
Change
in entropy
Qout
Q
 S gen  S system  S gen  out  S system
Tb
Tb
where
Ssystem  m(s2  s1 )  mCav ln
T2
100 + 273
 (0.00210 kg)(0.465 kJ / kg.K)ln
 0.00112 kJ / K
T1
900 + 273
Substituting,
Q
0.781 kJ
Sgen  out  Ssystem 
 0.00112 kJ / K  0.00142 kJ / K (per ball)
Tb
308 K
Then the rate of entropy generation becomes
S gen  S gen n ball  (0.00142 kJ/K  ball)(2500 balls/h) = 3.55 kJ/h.K = 0.986 W/K
7-122 1 The egg is spherical in shape with a radius of r0 = 2.75 cm. 2 The thermal of the egg are constant.
3 Energy absorption or release associated with any chemical and/or phase changes within the egg is
negligible. 4 There are no changes in kinetic and potential energies.
The density and specific heat of the egg are given to be  = 1020 kg/m3 and Cp = 3.32 kJ/kg.C.
We take the egg as the system. This is a closes system since no mass enters or leaves the egg. The energy
balance for this closed system can be expressed as
E  Eout
in



Net energy transfer
by heat, work, and mass
Esystem

 

Boiling
Water
Change in internal, kinetic,
potential, etc. energies
Qin  Uegg  m(u2  u1 )  mC(T2  T1 )
Then the mass of the egg and the amount of heat transfer become
m  V  
D 3
 (1020 kg/m 3 )
Egg
8C
 (0.055 m) 3
 0.0889 kg
6
6
Qin  mC p (T2  T1 )  (0.0889 kg )(3.32 kJ/kg. C)(70  8)C  18.3 kJ
We again take a single egg as the system The entropy generated during this process can be determined by
applying an entropy balance on an extended system that includes the egg and its immediate surroundings so
that the boundary temperature of the extended system is at 97C at all times:
S in  S out


Net entropy transfer
by heat and mass
 S gen  S system




Entropy
generation
Change
in entropy
Qin
Q
 S gen  S system  S gen   in  S system
Tb
Tb
where
Ssystem  m(s2  s1 )  mCav ln
T2
70 + 273
 (0.0889 kg)(3.32 kJ / kg.K)ln
 0.0588 kJ / K
T1
8 + 273
Substituting,
S gen  
Qin
18.3 kJ
 S system  
 0.0588 kJ/K  0.00961 kJ/K (per egg)
Tb
370 K
7-140 1 The room is well insulated and well sealed. 2 The thermal of water and air are constant at room
temperature. 3 The system is stationary and the kinetic and potential energy changes are zero. 4 There are
no work interactions involved.
The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). The specific heat of water at room
temperature is C = 4.18 kJ/kgC (Table A-3). For air is Cv = 0.718 kJ/kgC at room temperature.
The volume and the mass of the air in the room are
4m  5m  6m
V = 4x5x6 = 120 m³
PV
100 kPa  120 m 3
m air  1 1 
 141.7 kg
RT1
0.2870 kPa  m 3 /kg  K 295 K 




ROOM
22C
100 kPa
Taking the contents of the room, including the water, as our system,
the energy balance can be written as
E  E out
in

Net energy transfer
by heat, work, and mass
or

Heat
 0  U  U water  U air
E system



Water
80C
Change in internal,kinetic,
potential,etc. energies
mC T2  T1 water  mC v T2  T1 air  0
Substituting,
1000 kg 4.18 kJ/kg CT f




 80  C  141.7 kg  0.718 kJ/kg C T f  22  C  0
It gives the final equilibrium temperature in the room to be
Tf = 78.6C
(b) Considering that the system is well-insulated and no mass is entering and leaving, the total entropy
change during this process is the sum of the entropy changes of water and the room air,
S total  S gen  S air  S water
where
S air  mC v ln
S water  mC ln
T2
V
 mR ln 2
T1
V1
0
 141.7 kg 0.718 kJ/kg  K ln
351.6 K
 17.86 kJ/K
295 K
T2
351.6 K
 1000 kg 4.18 kJ/kg  K ln
 16.61 kJ/K
T1
353 K
Substituting, the total entropy change is determined to be
Stotal = 17.86 - 16.61 = 1.25 kJ/K