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(D) Let $K$ be the thermal conductivity,$A$ be the cross-sectional area,and $\ell$ be the length of each rod.In the first case,the two rods are connec

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$1/4$

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(D) Let $K$ be the thermal conductivity,$A$ be the cross-sectional area,and $\ell$ be the length of each rod.In the first case,the two rods are connected in parallel between the two reservoirs at $T_1 = 100^{\circ}C$ and $T_2 = 0^{\circ}C$.The rate of heat flow is $\frac{dH}{dt_1} = \frac{KA(T_1-T_2)}{\ell} + \frac{KA(T_1-T_2)}{\ell} = \frac{2KA(T_1-T_2)}{\ell}$.Since $q_1 \propto \frac{dH}{dt_1}$,we have $q_1 = \frac{2KA(T_1-T_2)}{L_f \ell}$,where $L_f$ is the latent heat of fusion.In the second case,the rods are connected in series. The equivalent thermal resistance is $R_{eq} = \frac{\ell}{KA} + \frac{\ell}{KA} = \frac{2\ell}{KA}$.The rate of heat flow is $\frac{dH}{dt_2} = \frac{T_1-T_2}{R_{eq}} = \frac{KA(T_1-T_2)}{2\ell}$.Thus,$q_2 = \frac{KA(T_1-T_2)}{2 L_f \ell}$.The ratio $q_2/q_1 = \frac{KA(T_1-T_2) / (2 L_f \ell)}{2KA(T_1-T_2) / (L_f \ell)} = \frac{1/2}{2} = \frac{1}{4}$.

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