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(B) In steady state,the rate of heat flow through layers in series is the same.Let the temperature at the interface be $	heta$. The temperature diffe

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The temperature difference across $A = 5^\circ C$

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(B) In steady state,the rate of heat flow through layers in series is the same.Let the temperature at the interface be $	heta$. The temperature difference across layer $A$ is $(	heta_1 - 	heta)$ and across layer $B$ is $(	heta - 	heta_2)$.The rate of heat flow is given by $H = \frac{KA(\Delta T)}{l}$.Since $H_A = H_B$ and the area $A$ and thickness $l$ are the same for both layers:$\frac{K_A A (	heta_1 - 	heta)}{l} = \frac{K_B A (	heta - 	heta_2)}{l}$$K_A (	heta_1 - 	heta) = K_B (	heta - 	heta_2)$Given $K_A = 3K_B$,we have:$3K_B (	heta_1 - 	heta) = K_B (	heta - 	heta_2)$$3(	heta_1 - 	heta) = (	heta - 	heta_2)$Let $\Delta T_A = (	heta_1 - 	heta)$ and $\Delta T_B = (	heta - 	heta_2)$.Then $3 \Delta T_A = \Delta T_B$.The total temperature difference is $\Delta T_A + \Delta T_B = 20^\circ C$.Substituting $\Delta T_B = 3 \Delta T_A$:$\Delta T_A + 3 \Delta T_A = 20^\circ C$$4 \Delta T_A = 20^\circ C$$\Delta T_A = 5^\circ C$.

$A$ wall has two layers $A$ and $B$ made of different materials. The thickness of both the layers is the same. The thermal conductivities of $A$ and $B$ are $K_A$ and $K_B$ such that $K_A = 3K_B$. The temperature difference across the wall is $20^\circ C$. In thermal equilibrium:

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