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Let us say I put ice starting at a temperature $T \lt 0^\circ \textrm{C}$, and held at constant pressure of $1atm$ in contact with an infinite heat reservoir at temperature exactly $0^\circ \textrm{C}$. When we have reached an equilibrium state will we have just ice, water and ice, or just water?

I doubt the answer is going to be just water, but between the other two I do not know.

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    $\begingroup$ @lemon Can you give your specific reasons? $\endgroup$ Commented Mar 31, 2016 at 11:22

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TL;DR: If the ice has surfaces then it will completely melt to form water. If the ice has no surfaces (it's infinitely large) then it will remain completely ice.

Here's the reasoning:

If we imagine an idealised system that is infinitely large in extent (i.e. no surfaces and therefore no heterogeneous nucleation) then the ice will remain in an ice state, with small water nuclei that spontaneously form and then disappear again. To see why, consider how the free energy changes when a water nucleus of radius $r$ forms:

$$ \Delta G=-\Delta G_V (4\pi r^3/3) + \gamma(4\pi r^2) $$

where $\Delta G_V$ is the difference in free energy between the bulk phases, and $\gamma$ is the surface free energy. At the coexistence point, $\Delta G_V=0$ by definition, and so the the growth of a water phase will only ever cost energy (i.e. there's no critical size). The size distribution of these random water droplets will be given by the Boltzmann distribution:

$$ P\propto \exp(-4\pi r^2 \gamma/k_BT) $$

In the real world, the ice will have surfaces. And the surface energy of ice is greater than that of water, so the outer layers will melt, resulting in a block of ice surrounded by a layer of water. As before, since neither bulk phase is preferred, then the system will move towards the state that minimises the interfacial free energy, and so the ice in the core will shrink in size until it disappears.

The result is pure water with tiny ice nuclei spontaneously forming and melting with the same size distribution as before ($P\propto \exp(-4\pi r^2 \gamma/k_BT)$).

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  • $\begingroup$ One could also find the same result by considering entropy. Since liquid water has higher entropy at 0C than ice, random fluctuations will eventually melt the ice. $\endgroup$
    – Echows
    Commented Apr 1, 2016 at 11:58
  • $\begingroup$ @Echows Not true. Entropy may favour water but enthalpy favours ice by an equal amount. And when you consider the free energy, neither the bulk water nor the bulk ice is favoured. It's all down to the free energy of the interface. $\endgroup$
    – lemon
    Commented Apr 1, 2016 at 12:02
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The ice would rise to the same temperature as the reservoir (0 C). Once it got to that temperature, there would be no more driving force for heat transfer, and the system would be in equilibrium. So it would be the same as if you took any solid block of any material at T < 0 C and put it in contact with an ideal infinite reservoir at 0 C. There would be transient conduction within the block until the system equilibrated at 0 C.

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  • $\begingroup$ My answer disagrees with this; let me know if you see any fault in my reasoning. $\endgroup$
    – lemon
    Commented Mar 31, 2016 at 12:04
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    $\begingroup$ I am not able to follow your analysis. Are you saying that, if I put ice at 0 C into an adiabatic container with a huge amount of water at 0 C and allow the system to equilibrate, basically all the ice will melt? $\endgroup$ Commented Mar 31, 2016 at 13:02
  • $\begingroup$ If the volume of water is large enough, yes. Would you agree that a decrease in the size of the ice (in particular, the surface area) will correspond to a decrease in free energy? $\endgroup$
    – lemon
    Commented Mar 31, 2016 at 14:12
  • $\begingroup$ What if the original ice were sealed in a plastic bag so that it did not directly contact the surrounding water in the reservoir? (The OP didn't say anything about the reservoir being water at 0 C. It could have been some entirely different reservoir liquid, not even in direct contact with the ice.) $\endgroup$ Commented Mar 31, 2016 at 14:30
  • $\begingroup$ The surface of the ice will melt to form a thin layer of water (which is why ice is slippery) and you're left with the same situation; ice surrounded by water. So yes, it will still melt. $\endgroup$
    – lemon
    Commented Mar 31, 2016 at 14:35
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The answer of Lemon is Practically correct (-> Water), but I want to give another approach; QED.

First we need just to think what is HEAT? It's just kinetic "jiggling" of atoms. The carousel is rotating faster/with bigger radius when there is more heat.

What is melting of ice? (Or what is Solid-liquid change on state in matter?) It's the process where the atoms starts to flow freely compared to each other.

The basic question here is; Does a single molecule "know" if it's solid or if it's liquid. The Answer is of course; it doesn't, there is no difference to single atom if it's part of solid / liquid / gas / plasma. It's merely an issue of enough electromagnetic forces able to hold the molecule in it's position compared to others.

This thinking can be brought even below the molecule-level. Water molecule $H_2O$ is 2x Hydrogen + 1 x Oxygen. And with normal flowing water, the hydrogen atom doesn't even know which oxygen atom it's supposed to form the molecule with. And thus in fluid the hydrogen bonds are continually breaking and reforming at timescales faster than 200 femtoseconds.

This is the major difference of Solid ice and liquid water; on solid-ice the atoms are locked on the lattice structure.

Answer; The whole ice will turn in to water over time, because at melting temperature, the Heat of hydrogen is too great; there is nothing which would maintain the lattice structure; not even in idealized infinitely large system. The Melting will happen spontaneously throughout the whole matter, according to QED probabilities if there just is the needed Heat provided to keep up this process.

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Since the heat reservoir has exactly 0 deg. the ice would heat to reach the same temperature, but ideally it would not change since it's directly the freezing/melting point. Melting ice would require additional energy.

In case of being submerged in water at freezing point:

It depends on the initial temperature of ice. Since the thermal equilibrium is 0 deg. C. the ice would heat up to 0 deg. The energy lost by ice would be consumed by water for phase transition, thus creating additional ice layer.

Simple response: the amount of ice would increase depending on initial ice temperature.

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    $\begingroup$ The ice content would increase? Are you imagining the ice to be physically submerged in an ocean? Because that's not what is meant by an idealised heat reservoir... $\endgroup$
    – lemon
    Commented Mar 31, 2016 at 11:14
  • $\begingroup$ Right, for some reason I took heat reservoir for a water container. I'll update the answer. $\endgroup$
    – Rev
    Commented Mar 31, 2016 at 11:32
  • $\begingroup$ Melting the ice would not require additional work because the two bulk phases have the same free energy. The melting of the ice corresponds to the interface shrinking in size (and therefore a reduction in surface free energy). $\endgroup$
    – lemon
    Commented Mar 31, 2016 at 12:03

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