5-41
5-59E Paddle Wheel work is applied to nitrogen in a rigid container. The final temperature is to be determined.
Assumptions 1 Nitrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values
of 126.2 K (227.1 R) and 3.39 MPa (492 psia). 2 The kinetic and potential energy changes are negligible,
0peke
. 3
Constant specific heats at room temperature can be used for nitrogen.
Properties For nitrogen, c
v
= 0.177 Btu/lbmR at room temperature and R =
R/lbmftpsia 0.3830 3
(Table A-1E and A-
2Ea).
Analysis We take the nitrogen as the system. This is a closed system since no mass crosses the boundaries of the system.
The energy balance for this system can be expressed as
kinetic, internal,in Change
system
nsferenergy traNet
outin
EEE
R) R)(560/lbmftpsia (0.3830
3
1
Substituting,
F489R 949
2
12inpw,
ftlbf 17.778
Btu 1
)(
T
TTmcW
v
5-60 One part of an insulated rigid tank contains an ideal gas while the other side is evacuated. The final temperature and
pressure in the tank are to be determined when the partition is removed.
Assumptions 1 The kinetic and potential energy changes are negligible,
 ke pe 0
. 2 The tank is insulated and thus
heat transfer is negligible.
Analysis We take the entire tank as the system. This is a closed system since no mass crosses the boundaries of the system.
5-43
5-62 A student living in a room turns her 150-W fan on in the morning. The temperature in the room when she comes back
10 h later is to be determined.
Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of –
141C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible,
 ke pe 0
. 3 Constant specific heats at
room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning
applications. 4 All the doors and windows are tightly closed, and heat transfer through the walls and the windows is
disregarded.
Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Also, c
v
= 0.718 kJ/kg.K for air at room
temperature (Table A-2).
Analysis We take the room as the system. This is a closed system since the doors and the windows are said to be tightly
closed, and thus no mass crosses the system boundary during the process. The energy balance for this system can be
expressed as
)()( 1212,
,
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
TTmcuumW
EEE
ine
ine
outin
v
The mass of air is
kg 08.57
K) K)(293/kgmkPa (0.287
m 48443
3
3
1
1
3
RT
m
V
The electrical work done by the fan is
Discussion Note that a fan actually causes the internal temperature of a confined space to rise. In fact, a 100-W fan supplies
a room with as much energy as a 100-W resistance heater.
ROOM
3 m 4 m 4 m
Fan
5-45
5-64 Argon is compressed in a polytropic process. The work done and the heat transfer are to be determined.
Assumptions 1 Argon is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of
151 K and 4.86 MPa. 2 The kinetic and potential energy changes are negligible,
0peke
.
Properties The properties of argon are R = 0.2081kJ/kgK and cv = 0.3122 kJ/kgK (Table A-2a).
Analysis We take argon as the system. This is a closed system since no mass crosses the
5-47
5-66 Carbon dioxide contained in a spring-loaded piston-cylinder device is heated. The work done and the heat transfer are
to be determined.
cv = 0.657 kJ/kgK (Table A-2a).
Analysis We take CO2 as the system. This is a closed system
)( 12out,in
energies etc. potential,
TTmcUWQ
b
v
The initial and final specific volumes are
3
3
1
K) K)(298/kgmkPa kg)(0.1889 (1
mRT
P
(kPa)
V
(m3)
100
1
2
5-49
5-69 A cylinder is initially filled with air at a specified state. Air is heated electrically at constant pressure, and some heat is
lost in the process. The amount of electrical energy supplied is to be determined.
Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 Air is an ideal gas
with variable specific heats. 3 The thermal energy stored in the cylinder itself and the resistance wires is negligible. 4 The
compression or expansion process is quasi-equilibrium.
Properties The initial and final enthalpies of air are (Table A-21)
h h
h h
1
2
 
 
@
@
298 K
350 K
298.18 kJ /kg
350.49 kJ /kg
Analysis 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 closed
system can be expressed as
kinetic, internal,in Change
sy stem
nsferenergy traNet
EEE outin
or,
kWh0.235
kJ 3600
kWh 1
kJ)(845
,
ine
W
Alternative solution The specific heat of air at the average temperature of Tavg = (25+ 77)/2 = 51C = 324 K is, from Table
A-2b, cp,avg = 1.0065 kJ/kg.C. Substituting,
kJ845kJ 60C2577C)kJ/kg. kg)(1.0065 (15)( out12ine, QTTmcWp
or,
kWh0.235
kJ 3600
kWh 1
)kJ 845(
ine,
W
Discussion Note that for small temperature differences, both approaches give the same result.
AIR
P = const.
We
Q
5-52
n
Qout
kJ]
Win
[kJ]
1
1.044
1.089
1.133
1.178
1.222
1.267
1.311
1.356
1.4
134.9
121.8
108.2
94.25
79.8
64.86
49.42
33.45
16.93
–
0.1588
134.9
137
139.1
141.3
143.5
145.8
148.1
150.5
152.9
155.4
0 0.2 0.4 0.6 0.8 1
0
200
400
600
800
1000
1200
1400
1600
1800
0
500
1000
1500
2000
2500
3000
3500
4000
4500
spvplot [m^3/kg]
Pplot [kPa]
n=1.0
Pplot
n=1.3
n=2
Pressure vs. specific volume as function of polytropic exponent
40
60
80
100
120
140
160
Q or W (kJ)
Qout (kJ)
Win (kJ)
5-53
5-72E A cylinder initially contains air at a specified state. Heat is transferred to the air, and air expands isothermally. The
boundary work done is to be determined.
Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 The air is an ideal gas
with constant specific heats. 3 The compression or expansion process is quasi-equilibrium.
5-55
5-74 Air at a specified state contained in a piston-cylinder device with a set of stops is heated until a final temperature. The
amount of heat transfer is to be determined.
Assumptions 1 Air is an ideal gas since it is at a high temperature relative to its critical temperature of 304.2 K. 2 The
kinetic and potential energy changes are negligible,
0peke
.
Properties The properties of air are R = 0.287 kJ/kgK and cv = 0.718 kJ/kgK (Table A-2a).
Analysis We take air as the system. This is a closed system since no mass
crosses the boundaries of the system. The energy balance for this system can
be expressed as
kinetic, internal,in Change
system
nsferenergy traNet
outin
EEE
5-56
5-75 Air at a specified state contained in a piston-cylinder device undergoes an isothermal and constant volume process
until a final temperature. The process is to be sketched on the P–
V
diagram and the amount of heat transfer is to be
determined.
Assumptions 1 Air is an ideal gas since it is at a high temperature relative to its critical temperature of 304.2 K. 2 The
kinetic and potential energy changes are negligible,
0peke
.
Properties The properties of air are R = 0.287 kJ/kgK and cv = 0.718 kJ/kgK (Table A-2a).
Analysis (a) The processes 1-2 (isothermal) and 2-3 (constant-volume) are
sketched on the P-
V
diagram as shown.
(b) We take air as the system. This is a closed system since no mass crosses
the boundaries of the system. The energy balance for this system fort he
process 1-3 can be expressed as
kinetic, internal,in Change
system
nsferenergy traNet
outin
EEE
Air
100 kPa
27°C
Q
5-58
5-78E A person shakes a canned of drink in a iced water to cool it. The mass of the ice that will melt by the time the canned
drink is cooled to a specified temperature is to be determined.
Assumptions 1 The thermal properties of the drink are constant, and are taken to be the same as those of water. 2 The effect
of agitation on the amount of ice melting is negligible. 3 The thermal energy capacity of the can itself is negligible, and thus
it does not need to be considered in the analysis.
Properties The density and specific heat of water at the average temperature of (85+37)/2 = 61F are = 62.3 lbm/ft3, and
cp = 1.0 Btu/lbm.F (Table A-3E). The heat of fusion of water is 143.5 Btu/lbm.
Analysis We take a canned drink as the system. The energy balance for this closed system can be expressed as
)()( 21out12drink cannedout
energies etc. p otential,
kinetic, internal,in Change
sy stem
mass and work,heat,by
nsferenergy traNet
outin
TTmcQuumUQ
EEE
Noting that 1 gal = 128 oz and 1 ft3 = 7.48 gal = 957.5 oz, the total amount of
heat transfer from a ball is
Btu/can 5.37F)3785(F)Btu/lbm. 0.1)(lbm/can 781.0()(
lbm/can 781.0
oz fluid 128
gal 1
gal 7.48
ft 1
oz/can) 12)(lbm/ft 3.62(
21out
3
3
TTmcQ
m
V
Noting that the heat of fusion of water is 143.5 Btu/lbm, the amount of ice that will melt to cool the drink is
drink) ofcan (per
Btu/lbm 143.5
Btu/can 5.37
out
ice lbm 0.261
if
h
Q
m
since heat transfer to the ice must be equal to heat transfer from the can.
Discussion The actual amount of ice melted will be greater since agitation will also cause some ice to melt.
Cola
85F
5-60
5-80 Stainless steel ball bearings leaving the oven at a specified uniform temperature at a specified rate are exposed to air
and are cooled before they are dropped into the water for quenching. The rate of heat transfer from the ball bearing to the air
is to be determined.
Assumptions 1 The thermal properties of the bearing balls are constant. 2 The kinetic and potential energy changes of the
balls are negligible. 3 The balls are at a uniform temperature at the end of the process
Properties The density and specific heat of the ball bearings are given to be = 8085 kg/m3 and cp = 0.480 kJ/kg.C.
Analysis We take a single bearing ball as the system.
The energy balance for this closed system can be
expressed as
)(
)(
21out
12ballout
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
TTmcQ
uumUQ
EEE
The total amount of heat transfer from a ball is
kg 007315.0
m) 012.0(
)kg/m 8085(
3
3
3
D
m
V
5-81E Large brass plates are heated in an oven at a rate of 300/min. The rate of heat transfer to the plates in the oven is to be
determined.
Assumptions 1 The thermal properties of the plates are constant. 2 The changes in kinetic and potential energies are
negligible.
Properties The density and specific heat of the brass are given to be
Steel balls, 900C
Water, 25C
Furnace