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

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

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    $\begingroup$ i attempted it, can you tell me if it's right please $\endgroup$ – xia Apr 1 '15 at 12:58
  • $\begingroup$ Is the first paragraph part of the homework question? Is it given that all ice won't melt? And if all ice doesn't melt, what should be the final temperature of the mixture? $\endgroup$ – Del Pate Apr 1 '15 at 17:21
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    $\begingroup$ Seems like you need the heat capacities of iron and ice $\endgroup$ – Gimelist Apr 3 '15 at 11:19
  • $\begingroup$ You'll also need heat of fusion of water $\endgroup$ – dark32 May 4 '17 at 1:57

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