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(B) Let the thickness of each layer be $x$ and the area of cross-section be $A$. Let the thermal conductivity of $B$ be $K$,then the thermal conductiv

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$12$

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(B) Let the thickness of each layer be $x$ and the area of cross-section be $A$. Let the thermal conductivity of $B$ be $K$,then the thermal conductivity of $A$ is $2K$.In steady state,the rate of heat flow through both layers is the same: $\frac{Q}{t} = \frac{K_A A (\Delta T_A)}{x} = \frac{K_B A (\Delta T_B)}{x}$.Since the areas and thicknesses are equal,we have $K_A (\Delta T_A) = K_B (\Delta T_B)$.Substituting the values: $(2K) (\Delta T_A) = K (\Delta T_B) \Rightarrow \Delta T_B = 2 \Delta T_A$.The total temperature difference is $\Delta T_A + \Delta T_B = 36^{\circ}C$.Substituting $\Delta T_B$: $\Delta T_A + 2 \Delta T_A = 36^{\circ}C \Rightarrow 3 \Delta T_A = 36^{\circ}C$.Therefore,$\Delta T_A = 12^{\circ}C$.

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