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(C) In a steady state, the rate of heat flow $(H)$ through both slabs connected in series is the same.$H = \frac{k_1 A (\Delta T_A)}{L} = \frac{k_2 A

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$8$

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(C) In a steady state, the rate of heat flow $(H)$ through both slabs connected in series is the same.$H = \frac{k_1 A (\Delta T_A)}{L} = \frac{k_2 A (\Delta T_B)}{L}$Since the thickness $(L)$ and cross-sectional area $(A)$ are the same for both slabs, we have:$k_1 (\Delta T_A) = k_2 (\Delta T_B)$Given $k_1 = \frac{k_2}{2}$, we can write $k_2 = 2k_1$.Substituting this into the equation:$k_1 (\Delta T_A) = 2k_1 (\Delta T_B)$$\Delta T_A = 2 \Delta T_B$We know the total temperature difference is $\Delta T_A + \Delta T_B = 12^{\circ} C$.Substituting $\Delta T_A = 2 \Delta T_B$ into the total temperature equation:$2 \Delta T_B + \Delta T_B = 12^{\circ} C$$3 \Delta T_B = 12^{\circ} C \implies \Delta T_B = 4^{\circ} C$Therefore, the temperature difference across slab $A$ is:$\Delta T_A = 12^{\circ} C - 4^{\circ} C = 8^{\circ} C$.

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