# Calculating the change in entropy in a melting process

I have a homework question that I'm completely stumped on and need help solving it.

I have a $50\, \mathrm{g}$ ice cube at $-15\, \mathrm{C}$ that is in a container of $200\, \mathrm{g}$ of water at $25\, \mathrm{C}$. No heat is lost to the surroundings and there is negligible heat loss to the container. I have to find the total change in entropy as a result of the mixing process.

I would assume that I can use $\Delta S = \int \frac{dQ}{T}$, but I'm unsure how to set up the $Q$ portion of the integral. Can I just use $Q=mc\Delta T+mL_f$ for the ice melting process?

For the melting process You should use $Q=mc\Delta T_1+mL_f + mc\Delta T_2$, assuming that after the ice cube melts, there is still heat exchange between the warm water and the cold water (former ice cube) and $\Delta T_2$ is the temperature difference between the final temperature of the mixture and the melting temperature.
For the total enthropy change I would simply use $\Delta S =\frac{ Q}{T}$, since $Q$ is the heat acquired by the ice cube and it is equal to the heat removed from the water. Integration is necessary only if You want to be accurate, if the result may be rounded, than You can come up with a meaningfull approximation for the average temperature of heat exchange $T$.
The actual way hinges on the fact there is infinitesimal change in temperatures if they are not undergoing phase changes. Hence there will be $dT$ term inside the integral if they are not undergoing phase change. Otherwise the entropy chage is simply $Q/T = mL_{f}/T$.
So Hint: Hence you would end up seeing 2 terms (assuming there is a change in water temperature as well) containing $ln(T2/T1)$ where one of the temperatures is $273 K$.