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(C) In a steady state, the rate of heat flow through materials $A$ and $B$ must be equal.Let $T$ be the temperature of the common surface.The rate of

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$55$

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(C) In a steady state, the rate of heat flow through materials $A$ and $B$ must be equal.Let $T$ be the temperature of the common surface.The rate of heat flow $Q$ is given by $Q = \frac{KA \Delta T}{X}$. Assuming the cross-sectional area $A$ is the same for both slabs, we have:$Q_A = Q_B$$\frac{K_A (75^{\circ} C - T)}{X_A} = \frac{K_B (T - 50^{\circ} C)}{X_B}$Given $K_A = \frac{K_B}{2}$ and $X_A = 2 X_B$, we substitute these values into the equation:$\frac{(K_B / 2) (75^{\circ} C - T)}{2 X_B} = \frac{K_B (T - 50^{\circ} C)}{X_B}$$\frac{75^{\circ} C - T}{4} = T - 50^{\circ} C$$75^{\circ} C - T = 4T - 200^{\circ} C$$5T = 275^{\circ} C$$T = 55^{\circ} C$Thus, the temperature of the common surface is $55^{\circ} C$.

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