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Chapter 5
The First Law of Thermodynamics
Closed System Energy Analysis: Solids and Liquids
545
A number of brass balls are to be quenched in a water bath at a specified rate. The rate at which heat
needs to be removed from the water in order to keep its temperature constant is to be determined.
Assumptions
1
The thermal properties of the balls are constant.
2
The balls are at a uniform temperature
before and after quenching.
3
The changes in kinetic and potential energies are negligible.
Properties
The density and specific heat of the brass balls are given to be
ρ
= 8522 kg/m
3
and
C
p
= 0.385
kJ/kg.
°
C.
Analysis
We take a single ball as the system. The energy balance for this closed
system can be expressed as
EE
E
QU
m
u
u
Qm
C
T
T
in
out
out
ball
out
−=
=
−
=−
Net energy transfer
by heat, work, and mass
system
Change in internal, kinetic,
potential, etc. energies
12
43
4
4
Δ
Δ
()
21
Brass balls, 120
°
C
Water bath, 5
°
C
The total amount of heat transfer from a ball is
mV
D
C
T
T
out
==
=
=
=
°−
°
=
ρρ
ππ
3
6
8522
005
0558
0 558
0 385
120
74
9 88
(.
.
(
.)
(
.
(
).
kg /m
m)
6
kg
kg
kJ /kg. C)
C
kJ /ball
3
3
Then the rate of heat transfer from the balls to the water becomes
&
&
((
.
)
Qn
Q
total
×
=
ball
ball
balls/min)
kJ /ball
100
9 88
988 kJ / min
Therefore, heat must be removed from the water at a rate of 988 kJ/min in order to keep its temperature
constant at 50
°
C since energy input must be equal to energy output for a system whose energy level
remains constant. That is,
E
in
out
=
=
when
system
Δ
0 .
542
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View Full DocumentChapter 5
The First Law of Thermodynamics
546
A number of aluminum balls are to be quenched in a water bath at a specified rate. The rate at which
heat needs to be removed from the water in order to keep its temperature constant is to be determined.
Assumptions
1
The thermal properties of the balls are constant.
2
The balls are at a uniform temperature
before and after quenching.
3
The changes in kinetic and potential energies are negligible.
Properties
The density and specific heat of aluminum at the average temperature of (120+74)/2 = 97
°
C =
370 K are
ρ
= 2700 kg/m
3
and
C
p
= 0.937 kJ/kg.
°
C (Table A3).
Analysis
We take a single ball as the system. The energy balance for this closed
system can be expressed as
EE
E
QU
m
u
u
Qm
C
T
T
in
out
out
ball
out
−=
=
−
=−
Net energy transfer
by heat, work, and mass
system
Change in internal, kinetic,
potential, etc. energies
12
43
4
4
Δ
Δ
()
21
Aluminum balls, 120
°
C
Water bath, 50
°
C
The total amount of heat transfer from a ball is
kJ/ball
62
.
7
C
)
74
120
(
C)
kJ/kg.
937
.
0
)(
kg
1767
.
0
(
)
(
kg
1767
.
0
6
m)
05
.
0
(
)
kg/m
2700
(
6
2
1
3
3
3
=
°
−
°
=
−
=
=
=
=
=
T
T
mC
Q
D
V
m
out
π
ρρ
Then the rate of heat transfer from the balls to the water becomes
kJ/min
762
=
×
=
=
)
kJ/ball
62
.
7
(
balls/min)
100
(
ball
ball
Q
n
Q
total
&
&
Therefore, heat must be removed from the water at a rate of 762 kJ/min in order to keep its temperature
constant at 50
°
C since energy input must be equal to energy output for a system whose energy level
remains constant. That is,
E
in
out
=
=
when
system
Δ
0 .
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 Spring '07
 Dr.Kinne

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