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(C) Let $R$ be the thermal resistance of each rod. The rate of heat flow is given by $H = \frac{dQ}{dt} = \frac{\Delta T}{R_{eq}}$. The rate of meltin

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$4/1$

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(C) Let $R$ be the thermal resistance of each rod. The rate of heat flow is given by $H = \frac{dQ}{dt} = \frac{\Delta T}{R_{eq}}$. The rate of melting of ice is $q = \frac{1}{L} \frac{dQ}{dt}$,where $L$ is the latent heat of fusion of ice.Case $1$: The rods are connected in parallel. The equivalent thermal resistance is $R_p = \frac{R 	imes R}{R + R} = \frac{R}{2}$.The rate of heat flow is $H_1 = \frac{100 - 0}{R/2} = \frac{200}{R}$.Thus,$q_1 = \frac{H_1}{L} = \frac{200}{RL}$.Case $2$: The rods are connected in series. The equivalent thermal resistance is $R_s = R + R = 2R$.The rate of heat flow is $H_2 = \frac{100 - 0}{2R} = \frac{100}{2R} = \frac{50}{R}$.Thus,$q_2 = \frac{H_2}{L} = \frac{50}{RL}$.Taking the ratio,$\frac{q_1}{q_2} = \frac{200/RL}{50/RL} = \frac{200}{50} = \frac{4}{1}$.

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