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(A) Let the length of each block be $\ell$ and the cross-sectional area be $A$. The thermal resistance of a block is given by $R = \frac{\ell}{K A}$.F

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$2.0$

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(A) Let the length of each block be $\ell$ and the cross-sectional area be $A$. The thermal resistance of a block is given by $R = \frac{\ell}{K A}$.For the block with conductivity $k$,$R_k = \frac{\ell}{k A}$. Let $R_k = 2R_0$,then the resistance of the block with conductivity $2k$ is $R_{2k} = \frac{\ell}{2k A} = R_0$.In configuration $I$ (series): The total thermal resistance is $R_I = R_k + R_{2k} = 2R_0 + R_0 = 3R_0$. The heat flow rate is $H_I = \frac{\Delta T}{R_I} = \frac{\Delta T}{3R_0}$. The heat transported in time $t_I = 9 \ s$ is $Q = H_I \cdot t_I = \frac{\Delta T}{3R_0} \cdot 9 = \frac{3 \Delta T}{R_0}$.In configuration $II$ (parallel): The total thermal resistance is $\frac{1}{R_{II}} = \frac{1}{R_k} + \frac{1}{R_{2k}} = \frac{1}{2R_0} + \frac{1}{R_0} = \frac{3}{2R_0}$,so $R_{II} = \frac{2R_0}{3}$. The heat flow rate is $H_{II} = \frac{\Delta T}{R_{II}} = \frac{3 \Delta T}{2R_0}$.To transport the same amount of heat $Q$,the time $t_{II}$ is given by $Q = H_{II} \cdot t_{II} \Rightarrow \frac{3 \Delta T}{R_0} = \frac{3 \Delta T}{2R_0} \cdot t_{II}$.Solving for $t_{II}$,we get $t_{II} = 2 \ s$.

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