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(A) Let the length of each block be $L$,cross-sectional area be $A$,and thermal resistance of the block with conductivity $k$ be $R = \frac{L}{kA}$. T

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$2.0$

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(A) Let the length of each block be $L$,cross-sectional area be $A$,and thermal resistance of the block with conductivity $k$ be $R = \frac{L}{kA}$. Then the resistance of the block with conductivity $2k$ is $R' = \frac{L}{2kA} = \frac{R}{2}$.In configuration $I$ (series): The blocks are in series. The equivalent thermal resistance is $R_{eq,I} = R + \frac{R}{2} = \frac{3R}{2}$.The heat current $H_I = \frac{\Delta T}{R_{eq,I}} = \frac{2\Delta T}{3R}$.In configuration $II$ (parallel): The blocks are in parallel. The equivalent thermal resistance is given by $\frac{1}{R_{eq,II}} = \frac{1}{R} + \frac{1}{R/2} = \frac{1}{R} + \frac{2}{R} = \frac{3}{R}$. So,$R_{eq,II} = \frac{R}{3}$.The heat current $H_{II} = \frac{\Delta T}{R_{eq,II}} = \frac{3\Delta T}{R}$.Since the amount of heat $Q$ transported is the same,$Q = H_I t_I = H_{II} t_II$.$t_{II} = t_I 	imes \frac{H_I}{H_{II}} = 9 	imes \frac{2\Delta T / 3R}{3\Delta T / R} = 9 	imes \frac{2}{9} = 2.0 \ s$.

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