heat capacity and thermal equilibrium

A $52\ \mathrm{g}$ piece of iron at $250\ \mathrm{^\circ C}$ is added to a mixture of $100\ \mathrm{g}$ of ice and $100\ \mathrm{g}$ of liquid water at thermal equilibrium $(0\ \mathrm{^\circ C})$. Calculate the mass of ice that melts (not all of the ice will melt, so you know the final temp of everything mixed together).


I think that the temperature of the mixture should stay same, if is at thermal equilibrium. Because then it less willing to accept change in temperature. But I am not sure how to use heat of fusion and heat capacity of liquid water to solve this.


My attempt:


$Q$ lost from iron = $Q$ gained by water + ice mixture 


$(0.93\ \mathrm{mol}\ \ce{Fe})(25.1\ \mathrm{J\ mol^{-1}\ K^{-1}})(T\_\text{f}-250\ \mathrm{^\circ C}) = -(200\ \mathrm{g}\ \text{mixture})(\text{what is }C\text{?})(T\_\text{f}-0\ \mathrm{^\circ C})$


