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(D) In the steady state,the rate of heat flow through cube $A$ must be equal to the rate of heat flow through cube $B$.Let $T$ be the temperature of t

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$60$

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(D) In the steady state,the rate of heat flow through cube $A$ must be equal to the rate of heat flow through cube $B$.Let $T$ be the temperature of the interface.The rate of heat flow $H$ is given by $H = \frac{KA(T_1 - T_2)}{L}$.Since the cubes have the same size,their cross-sectional area $A$ and length $L$ are equal.For cube $A$: $H_A = \frac{K_A A (100 - T)}{L}$For cube $B$: $H_B = \frac{K_B A (T - 0)}{L}$Equating $H_A = H_B$:$\frac{K_A A (100 - T)}{L} = \frac{K_B A (T - 0)}{L}$$K_A (100 - T) = K_B T$$300(100 - T) = 200T$$3(100 - T) = 2T$$300 - 3T = 2T$$5T = 300$$T = 60^{\circ}C$Thus,the temperature of the interface is $60^{\circ}C$.

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