5-61
5-82 Long cylindrical steel rods are heat-treated in an oven. The rate of heat transfer to the rods in the oven is to be
determined.
Assumptions 1 The thermal properties of the rods are constant. 2 The changes in kinetic and potential energies are
negligible.
Properties The density and specific heat of the steel rods are given to be = 7833 kg/m3 and cp = 0.465 kJ/kg.C.
Analysis Noting that the rods enter the oven at a velocity of 2 m/min and exit at the same velocity, we can say that a 3–m
long section of the rod is heated in the oven in 1 min. Then the mass of the rod heated in 1 minute is
kg 75.78]4/)m 08.0([m) 2)(kg/m 7833()4/( 232
DLLAm
V
We take the 2-m section of the rod in the oven as the system. The
energy balance for this closed system can be expressed as
)()( 1212rodin
energies etc. potential,
kinetic, internal,in Change
system
mass and work,heat,by
nsferenergy traNet
outin
TTmcuumUQ
EEE
Substituting,
kJ 530,24C)30700(C)kJ/kg. 465.0)(kg 75.78()( 12in TTmcQ
Noting that this much heat is transferred in 1 min, the rate of heat transfer to the rod becomes
kW 409=kJ/min 24,530=min) kJ)/(1 530,24(/
inin tQQ
5-83 An electronic device is on for 5 minutes, and off for several hours. The temperature of the device at the end of the 5–
min operating period is to be determined for the cases of operation with and without a heat sink.
Assumptions 1 The device and the heat sink are isothermal. 2 The thermal properties of the device and of the sink are
constant. 3 Heat loss from the device during on time is disregarded since the highest possible temperature is to be
determined.
Properties The specific heat of the device is given to be cp = 850 J/kg.C. The specific heat of aluminum at room
temperature of 300 K is 902 J/kg.C (Table A-3).
Analysis We take the device to be the system. Noting that electrical energy is
)(
12ine,
TTmctW
Substituting, the temperature of the device at the end of the process is determined to be
C466=
22 C)25)(CJ/kg. 850)(kg 020.0(=s) 60J/s)(5 25(TT
(without the heat sink)
Case 2 When a heat sink is attached, the energy balance can be expressed as
Substituting, the temperature of the device-heat sink combination is determined to be
sink)heat (with
C)25)(CJ/kg. 902)(kg 500.0(C)25)(CJ/kg. 850)(kg 020.0(=s) 60J/s)(5 25(
2
22
= C41.0
T
TT
Discussion These are the maximum temperatures. In reality, the temperatures will be lower because of the heat losses to the
surroundings.
Oven, 900C
Steel rod, 30C
Electronic
5-65
Review Problems
5-88 For a 10C temperature change of air, the final velocity and final elevation of air are to be determined so that the
internal, kinetic and potential energy changes are equal.
Properties The constant-volume specific heat of air at room temperature is 0.718
kJ/kg.C (Table A-2).
Air
0C 10C
5-67
5-93 A well-insulated rigid vessel contains saturated liquid water. Electrical work is done on water. The final temperature is
to be determined.
Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 The thermal energy
stored in the tank itself is negligible.
Analysis We take the contents of the tank as the system. This is a closed system since no mass enters or leaves. The energy
)( 12in,
energies etc. potential,
uumUW
e
The amount of electrical work added during 30 minutes period is
kJ 900J 900,000
VA 1
W1
s) 60A)(30 V)(10 50(
ine,
tIW V
The properties at the initial state are (Table A-4)
/kg..m 001008.0
kJ/kg 53.167
3
C@40 1
C@40 1
f
f
uu
vv
Substituting,
kJ/kg 53.467
kg 3
kJ 900
kJ/kg 53.167)( in,
1212in,m
W
uuuumW e
e
The final state is compressed liquid. Noting that the specific volume is constant during the process, the final temperature is
determined using EES to be
EES) (from
/kgm 001008.0
kJ/kg 53.467
2
3
12
2C118.9
T
u
vv
v
1
2
5-68
5-94 A 1- ton (1000 kg) of water is to be cooled in a tank by pouring ice into it. The final equilibrium temperature in the
tank is to be determined.
Assumptions 1 Thermal properties of the ice and water are constant. 2 Heat transfer to the water tank is negligible. 3 There
is no stirring by hand or a mechanical device (it will add energy).
Properties The specific heat of water at room temperature is c = 4.18 kJ/kg·°C, and the specific heat of ice at about 0C is
c = 2.11 kJ/kg·°C (Table A-3). The melting temperature and the heat of fusion of ice at 1 atm are 0C and 333.7 kJ/kg.
Analysis We take the ice and the water as our system, and disregard any heat
transfer between the system and the surroundings. Then the energy balance for
this process can be written as
E E E
U
U U
in out
 
 
Net energy transfer
by heat, work, and mass
system
Change in internal, kinetic,
potential, etc. energies
ice water
   
 
0
0
0)]([])C0()C0([ water12iceliquid2solid1 TTmcTmcmhTmc if
Substituting,
(
()( )( )
80 2
2
kg){(2.11 kJ /kg C)[0 (-5)] C + 333.7 kJ /kg + (4.18 kJ /kg C)( 0) C}
1000 kg 4.18 kJ /kg C20 C 0
 
  
 
 
T
T
It gives
T2 = 12.4C
which is the final equilibrium temperature in the tank.
WATER
1 ton
ice
-5C
80 kg
5-70
5-96 A cylinder is initially filled with saturated R-134a vapor at a specified pressure. The refrigerant is heated both
electrically and by heat transfer at constant pressure for 6 min. The electric current is to be determined, and the process is to
be shown on a T-
v
diagram.
Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are negligible. 2 The thermal
energy stored in the cylinder itself and the wires is negligible. 3 The compression or expansion process is quasi-equilibrium.
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 stationary closed system can be expressed as
energies etc. potential,
kinetic, internal,in Change
sy stem
mass and work,heat,by
nsferenergy traNet
outin
EEE
R-134a
5-71
5-97 Saturated water vapor contained in a spring-loaded piston-cylinder device is condensed until it is saturated liquid at a
specified temperature. The heat transfer is to be determined.
Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 There are no work
interactions involved other than the boundary work. 3 The thermal energy stored in the cylinder itself is negligible. 4 The
compression or expansion process is quasi-equilibrium.
Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The
5-73
5-99 A cylinder is initially filled with helium gas at a specified state. Helium is compressed polytropically to a specified
temperature and pressure. The heat transfer during the process is to be determined.
Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The cylinder is stationary and thus the kinetic and
potential energy changes are negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression
or expansion process is quasi-equilibrium.
5-75
5-101 Nitrogen gas is expanded in a polytropic process. The work done and the heat transfer are 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 and 3.39 MPa. 2 The kinetic and potential energy changes are negligible,
0peke
. 3 Constant specific
heats can be used.
Properties The properties of nitrogen are R = 0.2968 kJ/kgK and cv = 0.743 kJ/kgK (Table A-2a).
Analysis We take nitrogen in the cylinder 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
)( 12out,in
energies etc. potential,
kinetic, internal,in Change
system
nsferenergy traNet
outin
TTmcUWQ
EEE
b
v
Using the boundary work relation for the polytropic process of an ideal gas gives
kJ/kg 526
1
2000
200
1.251
K) 1200)(KkJ/kg 2968.0(
1
1
25.1/25.0
/)1(
1
21
out,
nn
bP
P
n
RT
w
The temperature at the final state is
K 1.757
kPa 2000
kPa 200
K) 1200(
25.1/25.0
1(
1
2
12
)/nn-
P
P
TT
Substituting into the energy balance equation,
kJ/kg 197K)1200K)(757.1kJ/kg 743.0(kJ/kg 526)( 12out,in TTcwq b
v
Q
N2
1200 K
5-77
5-103 One ton of liquid water at 50°C is brought into a room. The final equilibrium temperature in the room is to be
determined.
Assumptions 1 The room is well insulated and well sealed. 2 The thermal properties of water and air are constant.
Properties 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).
Analysis The volume and the mass of the air in the room are
 
 
K 288K/kgmkPa 0.2870
3
1
Taking the contents of the room, including the water, as our system, the
energy balance can be written as
kinetic, internal,in Change
system
nsferenergy traNet
outin
EEE
ROOM
15C
95 kPa
Water
Heat
5-78
5-104 Water is boiled at sea level (1 atm pressure) in a coffee maker, and half of the water evaporates in 25 min. The power
rating of the electric heating element and the time it takes to heat the cold water to the boiling temperature are to be
determined.
Assumptions 1 The electric power consumption by the heater is constant. 2 Heat losses from the coffee maker are
negligible.
Properties The enthalpy of vaporization of water at the saturation temperature
of 100C is hfg = 2256.4 kJ/kg (Table A-4). At an average temperature of
(100+18)/2 = 59C, the specific heat of water is c = 4.18 kJ/kg.C, and the
density is about 1 kg/L (Table A-3).
Analysis The density of water at room temperature is very nearly 1 kg/L, and
thus the mass of 1 L water at 18C is nearly 1 kg. Noting that the enthalpy of
vaporization represents the amount of energy needed to vaporize a liquid at a
specified temperature, the amount of electrical energy needed to vaporize 0.5 kg
Therefore, the electric heater consumes (and transfers to water) 0.752 kW of electric power.
Noting that the specific heat of water at the average temperature of (18+100)/2 = 59C is c = 4.18 kJ/kgC, the
time it takes for the entire water to be heated from 18C to 100C is determined to be
C18)C)(100kJ/kg kg)(4.18 1(
Tmc
Water
100C
Heater