Why did they provide the value for the enthalpy change of fusion?

Question

The metal gallium melts when held in the hand; its melting point is $\pu{29.76 ^\circ C}$. How much energy as heat is removed from the hand when $\pu{5.00 grams}$ of gallium initially at $\pu{20.0^\circ C}$ melts? The value of $\Delta H_{fusion}$ is $5.576~\mathrm{kJ~mol^{-1}}$ and the specific heat of gallium is $0.374~\mathrm{J~g^{-1}K^{-1}}$. Take the final temperature to be $\pu{29.76 ^\circ C}$.

I got:

$$q = (\pu{5.00 g})(0.374~\mathrm{J~g^{-1}K^{-1}}) (\pu{9.76^\circ C})$$

$$q=\pu{18.3 J}$$

So I reason that since $\pu{18.3 J}$ of energy in the form of heat is required to melt gallium from at an initial temperature of $\pu{20.0 ^\circ C}$ to a final of $\pu{29.76 ^\circ C}$ then this is all the energy required to do this but why did they give $\Delta H_\text{fusion}$?

Answer 1

You calculated the energy required to change solid gallium at 20°C to solid gallium at 29.76°C. Once solid gallium reaches 29.76°C, additional energy is required to melt it. The energy required is given by the $\Delta H_{\text{fusion}}$. So to complete the question, transform the $\Delta H$ units to a grams basis instead of a mole basis, multiply it by the weight of gallium in the problem, and then add that result to the one you already calculated. That will be the energy required to change solid gallium at 20°C to liquid gallium at 29.76°C.