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(B) In steady state, the rate of heat flow $(H)$ through both layers $A$ and $B$ is the same.$H = \frac{K_A A \Delta T_A}{L} = \frac{K_B A \Delta T_B}

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) In steady state, the rate of heat flow $(H)$ through both layers $A$ and $B$ is the same.$H = \frac{K_A A \Delta T_A}{L} = \frac{K_B A \Delta T_B}{L}$Since the area $(A)$ and thickness $(L)$ are the same for both layers, we have:$K_A \Delta T_A = K_B \Delta T_B$Given that the thermal conductivity of $A$ is twice that of $B$, $K_A = 2K_B$:$2K_B \Delta T_A = K_B \Delta T_B$$\Delta T_B = 2 \Delta T_A$The total temperature difference across the wall is $\Delta T_A + \Delta T_B = 36^{\circ}C$.Substituting $\Delta T_B = 2 \Delta T_A$ into the equation:$\Delta T_A + 2 \Delta T_A = 36^{\circ}C$$3 \Delta T_A = 36^{\circ}C$$\Delta T_A = 12^{\circ}C$.

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