One end of a 0.25m long metal bar is in steam and the other is in contact with ice .If 12g of ice melts per minute, then the thermal conductivity of the metal is (Given cross section of the bar=5×10^−4 m^2 and latent heat of ice is 80calg
−1 )
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Explanation:
Given: The length of the metal bar is x=0.25m
The change temperature across the ends of the rod is T
1
−T
2
=100−0=100
o
C
The time taken to melt the ice is t=1min=60s
A=5×10
−4
m
2
,L=80calg
−1
The amount of heat required by the ice is:
Q=mL=12×80=960cal
But,Q=
x
KA(T
1
−T
2
)t
or
960=
0.25
K×5×10
−4
×100×60
∴K= 5×10 −2 ×60960×0.25
⇒30cals −1m −oC−1
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