If an ice cube at $0$ °C is in a thermally isolated system on its own, will any of it melt?
The chemistry teacher says it will reach a state of equilibrium of half ice and half water due to variation in kinetic energy of the particles but I don't understand where the latent heat energy would come from, unless half of the ice cube ends up as several degrees below zero to supply the energy to melt the other half, which goes against thermal equilibrium principles.
Best Answer
I agree with your assessment.
I'm also not sure what your teacher is trying to get at here.
If it's pure ice, and the system is thermally isolated, there's no reason for it to gain additional energy required to melt some of the ice. Like you also said, if the temperature of the ice dropped to provide that energy, it would violate the equilibrium of the system; so the system would never settle in that state.
If the ice were in thermal equilibrium with its surroundings at $0°C$, it would not make a difference; because there would be no net heat transfer. It would still be the same as if it were isolated.
I could see some merit in what your teacher said. That said, it's still a bit misleading.At the phase transition temperature, the mixture would be quasi-stable in a solid-liquid mixture. For such a mixture, the descriptions we've been using for equilibrium and quasi-equilibrium mechanics cannot be applied to the same level of accuracy. Basically, you get into the realm of statistics and microscopic effects. Because you aren't in equilibrium at all times; the path you take to get the system to $0°C$ will also matter.
See this question for some more insights on what happens with the mixture at $0°C$.Thanks to John Bollinger in the comments for letting me see that my second part of the answer wasn't really relevant (and also makes it less clear to me what the teacher was talking about).