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?
Best Answer
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 $T$.