Answer
a.
Aluminum
b.
I is best
Work Step by Step
$\displaystyle \Delta \mathrm{T}=\frac{\mathrm{Q}}{\mathrm{m}\mathrm{c}}$
so with Q being costant, $\Delta \mathrm{T}$ is inversely proportional to $\mathrm{m}\mathrm{c}$, the heat capacity.
Calculate $\mathrm{m}\mathrm{c}$ for
aluminum: $\mathrm{m}\mathrm{c}=(2 \mathrm{k}\mathrm{g} )(900 \mathrm{J}/\mathrm{k}\mathrm{g}\cdot \mathrm{K})=1800\mathrm{J}/\mathrm{K}$
ice:$\quad \mathrm{m}\mathrm{c}=(1 \mathrm{k}\mathrm{g} )(2090 \mathrm{J}/\mathrm{k}\mathrm{g}\cdot \mathrm{K})=2090\mathrm{J}/\mathrm{K}$
a.
Aluminum has less heat capacity, so the temperature change is greater.
b.
I is correct,
II is wrong (the mass of aluminum is greater),
III would be right if the heat capacities were the same.
Thus, I is best.
