Calculate the specific heat of a metal from the following data. A container made of the metal weighs 8.0 lb and contains in addition 30 lb of water. A 4.0-lb piece of metal initially at a temperature of 350°F is dropped into the water. The water and container initially have a temperature of 60°F and the final temperature of the entire system is 65°F.
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Explanation:
Let m
w
=14 kg,m
c
=3.6 kg,m
m
=1.8 kg,T
Ω
=180
o
C,T
Ω
=16.0
o
C, and T
f
=18.0
o
C
The specific heat c
m
of the metal then satisfies
(m
w
c
w
+m
c
c
m
)(T
f
−T
Ω
)+m
m
c
m
(T
f
−T
ω
)=0
which we solve for c
m
:
c
m
=
m
c
(T
f
−T
Ω
)+m
m
(T
f
−T
Ω
m
w
c
w
(T
Ω
−T
f
)
=
(3.6 kg)(18.0
o
C−16.0
o
C)+(1.8 kg)(18.0
o
C−180
o
C)
(14 kg)(4.18 kJ/kg.K)(16.0
o
C−18.0
o
C)
0.41 kJ/kg.C
o
=0.41 kJ/kg.K.
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