A copper rod (length 40 cm, cross-section 2.0 cm^2, K=400 W m^-1K^-1) is joined end to end to an iron rod (length 20 cm, same cross-section, K=80 W…

In steady state the heat current is the same through both rods, so the thermal resistances add in series. For copper, $R_1=\dfrac{l}{KA}=\dfrac{0.40}{400\times2\times10^{-4}}=5.0\ \mathrm{K\,W^{-1}}$. For iron, $R_2=\dfrac{0.20}{80\times2\times10^{-4}}=12.5\ \mathrm{K\,W^{-1}}$. The total is $17.5\ \mathrm{K\,W^{-1}}$, so the heat current is $H=\dfrac{100}{17.5}=5.714\ \mathrm{W}$ (the junction is at $100-5.714\times5=71.4^\circ\mathrm{C}$). The heat delivered to the ice in $600\ \mathrm{s}$ is $Q=5.714\times600=3428.6\ \mathrm{J}$, and the mass melted is $m=Q/L_f=3428.6/334000=0.010265\ \mathrm{kg}=10.27\ \mathrm{g}$. A common error is to treat the rods as parallel paths or to use only the copper rod, which would give $100/5=20\ \mathrm{W}$ and about $36\ \mathrm{g}$. Since both rods share the same area and the ice stays at $0^\circ\mathrm{C}$, the whole $100\ \mathrm{K}$ drop is shared in proportion to the resistances.