A copper rod (K=400 W m^-1K^-1, length 0.50 m) is joined end to end with a brass rod (K=100 W m^-1K^-1, length 0.50 m); both have cross-sectional…

In steady state the same heat current flows through both rods, which behave as thermal resistances in series, $R_{th}=\dfrac{L}{KA}$. The total is $\dfrac{1}{A}\left(\dfrac{0.5}{400}+\dfrac{0.5}{100}\right)=\dfrac{1}{1.4\times10^{-4}}\times6.25\times10^{-3}=44.64\ \mathrm{K\,W^{-1}}$. The heat current is $H=\dfrac{\Delta T}{R_{th}}=\dfrac{100}{44.64}=2.24\ \mathrm{W}$. The temperature of the junction is $100-H\times\dfrac{0.5}{400\times1.4\times10^{-4}}=100-2.24\times8.93=80^\circ\mathrm{C}$, showing that most of the drop occurs across the brass rod. In $20\ \mathrm{min}=1200\ \mathrm{s}$ the heat conducted is $2.24\times1200=2688\ \mathrm{J}$, which melts $2688/336=8\ \mathrm{g}$ of ice. The conversion uses $336\ \mathrm{J\,g^{-1}}$ for the latent heat, so each gram absorbs $336$ J, and $2688/336$ is exactly $8$, the required integer.