# Ice cube in a thermally isolated system … Will any of it melt?

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.

• Related (maybe duplicate?): Water and Ice with a Barrier – valerio Feb 5 '18 at 18:09
• A related topic in thermodynamics (but with steam/liquid mixture instead of liquid/ice mixture) is the idea of vapor quality en.wikipedia.org/wiki/Vapor_quality Changing the ratio between two phases in a mixture is one of many places that energy entering/leaving a system can go. – poompt Feb 5 '18 at 23:23
• Your teacher is wrong...but don't tell her/him. Very few teachers can cope with the loss of authority when a pupil displays superiority. Yes, even if you don't mean to be disrespectful - the teacher may subjectively perceive the loss of authority, even if this is not the case. – Klaws Feb 6 '18 at 10:00
• It may be partially melted, but asserting that half of it would be solid and half liquid is completely unfounded. Most likely, different parts of the ice cube would be solid or molten at different times, with the proportion determined by the variations in energy in the system. Thermodynamics is inherently statistical, and liquid vs. solid aren't distinct phases on microscopic levels; that doesn't allow the whole (or even half) of the cube to spontaneously turn into a liquid, but it's not going to be 100% solid either. The average energy stays the same, but it shifts around a bit randomly. – Luaan Feb 6 '18 at 13:44

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).

• You're soft-pedaling this, I think. If the system is thermally isolated, then not only is there no reason for the ice to gain the energy needed for any of it to melt, there is no way for it to do so. Moreover, if the ice were in thermal equilibrium with its surroundings at exactly its freezing point then it still would not melt, because that would require a net heat transfer, contradicting the stipulation of thermal equilibrium. And if it did melt, reaching equilibrium would not necessarily require exactly half of the ice to melt. – John Bollinger Feb 5 '18 at 21:45
• @JohnBollinger I don't really understand what you mean by "soft-pedaling" here. There would in theory be ways for the ice to gain the needed energy; and that was actually addressed in the question. In theory; the system could exist outside equilibrium; and have some part of the mixture at lower temperature while another part melts. This state would just be short lived and meaningless when considering the bulk. Good point about equilibrium though. I was thinking equilibrium mixture at 0°C could be in each phase; but yeah, thermal equilibrium would be the same as no heat transfer here. – JMac Feb 5 '18 at 21:59
• I mostly meant that you seemed to be going out of your way to avoid a firm and unequivocal denunciation of the claim attributed to the teacher. The claim is rubbish as relayed by the question, and I don't see any plausible interpretation of it that would make it correct. – John Bollinger Feb 5 '18 at 22:20
• @JohnBollinger Okay, that would be what I just changed based on your first comment then. All irrelevant now, and I took away the lenient wording towards the teacher because it no longer applies. – JMac Feb 5 '18 at 22:24

To go from water(solid) to water (liquid) you need to increase the internal energy of the water (break bonds).
As there is no heat input to the water or work done on the water, the internal energy of the water will not change and thus the water will stay in the solid state.

Your teacher seems to be mislead, it would follow that if the system is in isolation at 0 degrees, and the water was already in a solid state, then there would have to be some energy input to provide latent heat for the phase change. While it is true that the temperatures of the individual particles will follow the Maxwell-Boltzman Distribution, the total energy of the system is still not enough to trigger a phase change.