A wall has two layers A and B,each made of a | vedclass

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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 $B$,i.e.,$K_A = 2K_B$:$2K_B \Delta T_A = K_B \Delta T_B \implies \Delta T_B = 2 \Delta T_A$The total temperature difference across the wall is given as $36\,^{\circ}C$:$\Delta T_A + \Delta T_B = 36$Substituting $\Delta T_B = 2 \Delta T_A$ into the equation:$\Delta T_A + 2 \Delta T_A = 36$$3 \Delta T_A = 36$$\Delta T_A = 12\,^{\circ}C$

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