Two uniform rods of the same cross-section 4×10^-4 m^2 are joined end to end and lagged on their sides. Rod 1 has length 0.5 m and thermal…

The rods are in series, so the same heat current flows through both. Thermal resistances are $R_1=\dfrac{L_1}{K_1A}=\dfrac{0.5}{200\times4\times10^{-4}}=6.25\ \mathrm{K\,W^{-1}}$ and $R_2=\dfrac{1.0}{100\times4\times10^{-4}}=25\ \mathrm{K\,W^{-1}}$. The total resistance is $31.25\ \mathrm{K\,W^{-1}}$, so the steady heat current is $H=\dfrac{100-0}{31.25}=3.2\ \mathrm{W}$ (the junction is at $80^\circ\mathrm{C}$, which is not needed). All of this heat reaches the ice at $0^\circ\mathrm{C}$ and only melts it. In $525\ \mathrm{s}$ the heat is $3.2\times525=1680\ \mathrm{J}$, so $m=\dfrac{1680}{336}=5\ \mathrm{g}$. The answer is independent of the interface temperature, but one can check it: the drop across rod 1 is $3.2\times6.25=20\ \mathrm{K}$, so the junction is at $80^\circ\mathrm{C}$, consistent with the larger drop across the poorer conductor, rod 2.