# Equilibrium temperature and phase changes

Suppose we have $$M$$ kg of ice at temperature $$t_{ice}<0°C$$ and $$m$$ kg of water at temperature $$t_w>0°C$$ that are put together in a calorimeter (so we can ignore all dispersions of heat to the environment).

How do I determine the equilibrium temperature and how much ice/water there will be at equilibrium temperature?

Now, since the temperature of the water is higher than that of the ice then heat will flow from the water to the ice and it seems to me that to determine if (at least part) of the water will freeze it should be enough to see if $$t_{*}=\frac{m_{ice}c_{ice}t_{ice}+m_{w}c_{w}t_{w}}{m_{ice}c_{ice}+m_{w}c_{w}}<0.$$ If it is, and we are not even considering the energy that the water will lose when it solidifies, then it surely will freeze (at least part of it).

My question now is: how much of the water will freeze?

By setting up the equation $$\sum Q=0$$ for the water$$+$$ice system I have that: $$m_wc_{w}(0-t_w)+m_w \lambda_c+\delta m_w c_{ice}(t_e-0)+m_{ice} c_{ice} (t_e-t_{ice})=0$$ but I don't see how to get the part of the mass of water that will freeze $$\delta m_w$$. Thanks.

1. Pick a temperature and fraction solid/liquid for the final composition. This choice will almost certainly be wrong, but will bring us closer to the correct final solution. For example, we could choose all liquid at $$25^o \text{ C}$$ Clearly, this trial final state should be somewhere in between the two initial temperatures, but any guess will do.
2. Next, calculate the individual heat flows in/out necessary to bring all the individual components to this trial final state. In this example, you would need to warm the ice, melt the ice, and warm the now-liquid water to $$25^o \text{ C}$$ All of these work with the same fixed amount of water, and require heat to be added. In addition, you would need to remove some heat to cool the original liquid water from its original temperature to the trial temperature.