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(B) Let the thickness of each layer be $x$ and the thermal conductivity of layer $B$ be $K$. Then the thermal conductivity of layer $A$ is $2K$.In ste

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$12$

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(B) Let the thickness of each layer be $x$ and the thermal conductivity of layer $B$ be $K$. Then the thermal conductivity of layer $A$ is $2K$.In steady state,the rate of heat flow $H$ through both layers is the same.$H = \frac{Q}{t} = \frac{K_A A \Delta T_A}{x} = \frac{K_B A \Delta T_B}{x}$Since the cross-sectional area $A$ and thickness $x$ are the same for both layers,we have:$K_A \Delta T_A = K_B \Delta T_B$$(2K) \Delta T_A = (K) \Delta T_B$$\Delta T_B = 2 \Delta T_A$The total temperature difference is given as $\Delta T_A + \Delta T_B = 36\,^{\circ}C$.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$Therefore,the temperature difference across layer $A$ is $12\,^{\circ}C$.

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