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Question:An ice box made of 1.75 cm thick wood has on the inner side a 2.0 cm thick layer of cork.If the temperature of the inner surface of the cork 0 degree C and that of the outer surface of wood is 12 degree C ,then find the temperature of the interface.thermal conductivity of wood is 5 times that of cork .

parmanand tiwari , 7 Years ago
Grade 12th pass
anser 1 Answers
Rituraj Tiwari

Last Activity: 5 Years ago

Thickness of wood = d = 1.75 cm = 0.0175 m

Thickness of cork = 2 cm = 0.02 m

During steady state,

( Q/t ) w = (Q/t) c ---here w = wood, c = cork

[KA(Θ1 - Θ2)/d] wood = [KA(Θ1 - Θ2)/d ] cork

5KA(12-Θ) / 0.0175 = KAΘ /0.02

Solving, above we get Θ = 10.2° C
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