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Question

A slab of stone of area 0.34 m2 and thickness 10 cm is exposed on the lower face to steam at 100 C. A block of ice at 0 C rests on the upper face of the slab. In one hour, 3.6 kg of ice is melted. Assume that the heat loss from the sides is negligible. The latent heat of fusion of ice is 3.4×105 J kg1. What is the thermal conductivity of the stone in units of Js1m1 C1?


A

1.0

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B

1.5

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C

2.0

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D

2.5

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Solution

The correct option is A

1.0


Amount of heat flowing through the stone in 1 hour = mL, where m is the mass of ice melted and L its latent heat.
Now, time (t) = 1 hour = 3600 s and thickness d = 10 cm = 0.1 m.
The rate of flow of heat is Q=mLt
Thermal conductivity k=QdA(θ2θ1)=mLdtA(θ2θ1)
=3.6×3.4×105×0.13600×0.34×(1000)=1 Js1m1 C1
Hence, the correct choice is (a).


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