A large well-insulated container holds a mixture of $75\ \mathrm g$ of ice and $100\ \mathrm g$ of water at $0\ \mathrm{^\circ C}$. Using the data given below, calculate the mass in grams of solid iron, $\ce{Fe}$, at $325\ \mathrm{^\circ C}$ that you would have to add to this mixture in order to melt all of the ice and raise the temperature of the resulting $175\ \mathrm g$ of water to $22\ \mathrm{^\circ C}$.
Specific heat capacity of water is $c_p(\ce{H2O})=4.184\ \mathrm{J\ g^{-1}\ ^\circ C^{-1}}$
Specific heat capacity of iron, $\ce{Fe}$, is $c_p(\ce{Fe})=0.45\ \mathrm{J\ g^{-1}\ ^\circ C^{-1}}$
Molar enthalpy of fusion (melting) of water is $\Delta_\text{fus}H_\mathrm m(\ce{H2O})=6.01\ \mathrm{kJ\ mol^{-1}}$
$M(\ce{H2O})=18\ \mathrm{g\ mol^{-1}}$
My answer was that you needed $25041.67\ \mathrm J$ to melt the ice and $16108.4\ \mathrm J$ to raise the temperature to 22 degrees Celsius.
Therefore $mc \Delta T(\ce{Fe}) = -(25041.67\ \mathrm J + 16108.4\ \mathrm J)$ and dividing this by the mass of iron and the change in temperature results in a mass of $301.8\ \mathrm g$.
Is this correct?