A wall has two layers A and B, each made of different material. Both the layers have the same thickness. The thermal conductivity for A is twice that of B. Under steady state, the temperature difference across the whole wall is 36°C. Then the temperature difference across the layer A is

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$6 ° C$

$12 ° C$

$24 ° C$

$18 ° C$

answer is B.

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Detailed Solution

$θ_{A} - θ_{B} = 36^{\circ} C (G i v e n)$
$K_{A} = 2 K_{B} (g i v e n)$
$θ_{C} = \frac{\frac{K_{A}}{l} θ_{A} + \frac{K_{B}}{l} θ_{B}}{\frac{K_{A}}{l} + \frac{K_{B}}{l}} \Rightarrow θ_{C} = \frac{K_{A} θ_{A} + K_{B} θ_{B}}{K_{A} + K_{B}}$

$∴ θ_{C} = \frac{2 θ_{A} + θ_{A} - 36}{3} = \frac{3 (θ_{A} - 12)}{3}$
∴θ_A− θ_C= 12 ⁰C

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