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Q.

A doubled layered wall has layer A 10 cm thick and B 20 cm thick. The thermal conductivity of A is thrice that of B. In the steady state, the temperature difference across the wall is $35^{0} C$ . The temperature difference across the layer A is

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a

$28^{0} C$

b

$14^{0} C$

c

$5^{0} C$

d

$7^{0} C$

answer is D.

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Detailed Solution

In the steady state, rate of heat across each layer of the wall is the same.

$\frac{K_{A} A (∆ T)}{10} = \frac{K_{B} A (35 - ∆ T)}{20}$

Using $K_{A} = 3 K_{B}, Solving, ∆ T = 5^{0} C$

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