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(B) In steady-state,the rate of heat flow through both layers is the same.Let the thermal conductivity of $A$ be $K_A = 2K$ and that of $B$ be $K_B =

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$12$

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(B) In steady-state,the rate of heat flow through both layers is the same.Let the thermal conductivity of $A$ be $K_A = 2K$ and that of $B$ be $K_B = K$.Let the thickness of each layer be $x$ and the cross-sectional area be $A$.The rate of heat flow is given by $H = \frac{KA \Delta T}{x}$.Since the layers are in series,the rate of heat flow $H$ is constant:$H = \frac{K_A A \Delta T_A}{x} = \frac{K_B A \Delta T_B}{x}$Substituting the values: $\frac{2K A \Delta T_A}{x} = \frac{K A \Delta T_B}{x} \Rightarrow 2 \Delta T_A = \Delta T_B$.Given the total temperature difference $\Delta T_A + \Delta T_B = 36 ^\circ C$.Substituting $\Delta T_B = 2 \Delta T_A$ into the equation: $\Delta T_A + 2 \Delta T_A = 36 ^\circ C$.$3 \Delta T_A = 36 ^\circ C \Rightarrow \Delta T_A = 12 ^\circ C$.

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