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Suppose we have an experiment with two containers, one with ice at 0°C and the other with water also at 0°C (equal masses), and a thermally conductive barrier (also at 0°C) in contact with both the water and the ice, and in between them. The whole experiment itself is insulated from the environment.

What is the final state of the system? Does it remain unchanged, because with no temperature difference there is no heat flow? Or does it random walk itself somehow to a final state with 50/50 water/ice slush on both sides?

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  • $\begingroup$ What do you think and why? Please show your attempt to answer your own question. $\endgroup$ – sammy gerbil Nov 30 '16 at 15:26
  • $\begingroup$ I think the system stays the same, that even though the final state has a greater entropy, that without a temperature difference, there isn't a way for heat to flow through the conducting barrier. $\endgroup$ – David Elm Nov 30 '16 at 15:37
  • $\begingroup$ What about random fluctuations causing a temperature difference? There is (in theory) no limit to how small the temperature difference needs to be for heat to flow. $\endgroup$ – sammy gerbil Nov 30 '16 at 15:44
  • $\begingroup$ Do water and ice respectively fill completely the two containers? $\endgroup$ – valerio Dec 2 '16 at 10:11
  • $\begingroup$ No, they each have an air gap on top of them that is at 0°C. $\endgroup$ – David Elm Dec 2 '16 at 12:30
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When the system has evolved towards thermal equilibrium it will have maximized its entropy. Having a lot of energy concentrated in one of the vessels corresponds to a low entropy state. The system will therefore move to a state where the energy is spread over the two vessels.

Note that no temperature difference is necessary to achieve this. As in any thermodynamical system on a microscopic level there will be continuous exchanges of heat back and forth between the two vessels.

edit: As Pieter pointed out, my answer ignores the effects of surface tension. When surface tension is taken into account, it becomes energetically favorable to reduce the area of the boundary between water and ice and thus to keep all the ice on one side.

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    $\begingroup$ This would increase the surface of the ice-water interface which costs more energy than what can be offset by an entropy increase. It is not going to happen. $\endgroup$ – Pieter Nov 25 '17 at 21:23
  • $\begingroup$ i agree, see edit $\endgroup$ – Crimson Nov 29 '17 at 8:37
  • $\begingroup$ Regarding your edit, do you mean that if the system is initially in a slush state, ice will spontaneously migrate toward a container and water toward the other, separating the phases this way? $\endgroup$ – thermomagnetic condensed boson Nov 29 '17 at 19:17
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There is a surface tension associated with the ice-water interface (about 29 mJ/m$^2$ according to Hardy, Phil. Mag. 35 (1977) 471--484). This makes a minimal surface the lowest energy state. There is an entropy gain for point defects, but the entropy of slush cannot compete with the cost of 2-dimensional structures (not even with line faults).

So the slush created after sudden partial crystallization of undercooled water should slowly change into more clearly separated regions of ice and water. I have not been able to find data about this, but there are some simulations in https://arxiv.org/abs/1612.00363

Searches with "Ostwald ripening" or "migratory recrystallization" of undercooled ice then mostly lead to ice cream research.

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  • $\begingroup$ I don't think this is right. Surface tension is (I guess), not the only factor involved in the total energy. With your conclusion, a vertical planar interface between ice/water would have the lowest energy state, hence be the final state of the system. Don't you think that a horizontal interface is much more probable, assuming that separated water and ice are involved in the final equilibrium state? I assume that the crosss section area in the horizontal plane is greater than in the vertical plane. $\endgroup$ – thermomagnetic condensed boson Nov 25 '17 at 17:13
  • $\begingroup$ @no_choice99 Horizontal/vertical is only relevant if there is a gravitational field. The problem did not mention this, does not mention the relative position of the two containers. $\endgroup$ – Pieter Nov 25 '17 at 20:39
  • $\begingroup$ While I agree with you here, it's stated in the comment section of the original post that he considers air above the ice and water (this only complicates the problem, but this suggests there is gravitation). The point is that we could safely assume gravitation coming into play. Anyway, for the sake of it, let's assume there could or not, be gravitation included. You are still left with an infinity of possible planar interphases because they have the same area. Why would the system pick one and stick to it? If it switches between those planar interphases, then there would be no "final state". $\endgroup$ – thermomagnetic condensed boson Nov 25 '17 at 23:38
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    $\begingroup$ @no_choice99 The initial situation (one container ice, the second water, no interface) could not go to the infinite number of situations with two interfaces without an increase in energy. This would be a process that could only happen the other way around: theoretically, the horizontal interfaces would move randomly up and down, until one of them disappeared. I think it would take forever... $\endgroup$ – Pieter Nov 26 '17 at 1:48
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They will stay unchanged. The entropy in each container will not change if you look at the equation where there is no heat exchange. $$dS=\frac{\delta Q}{T}$$

If they change to 50%-50%, then one container needs to give heat out or to take heat in to or from the other one container, which we don't see.

If there is no barrier, then they mixes to 50%-50%. The entropy increase. The increase is not due to heat exchange but because of the increase of the number of micro state .

This is something like you have two containers with red balls in one container and black balls in the other. If you don't mix them but put them side by side, the entropy will not change. So mixing is the key with distracting heat element in the problem.

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  • $\begingroup$ Why do you say we don't see any exchange of heat? The barrier is stated to be thermally conducting. $\endgroup$ – sammy gerbil Dec 1 '16 at 3:00
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    $\begingroup$ I think there is no heat flux through a perfect thermally conductive material if the temperatures gradient is zero based on Fourier law. I don't understand micro level heat transfer. Do you have any reference? Regards, $\endgroup$ – user115350 Dec 1 '16 at 6:38
  • $\begingroup$ On a side thinking, if we remove the barrier and let the ice and the water blend. When it reaches equilibrium, it will not be 50%-50%. The entropy increase from $S_1+S_2$ to $S_{equilibrium}$ needs accompany heat change. $\endgroup$ – user115350 Dec 1 '16 at 15:52
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My answer is that the state of the system remains stable: that is, the ice remains ice and the water remains water.

Let's try to bring a few concepts together: since the whole system is at T=0 and isolated from the environment, a phase transition (in either direction) can't complete. Hence, everything moving away from the original state is driven by stochastic fluctuations. We want to prove that the original state is an equilibrium point: every fluctuation can produce a local phase transition, but this can't extend any further; actually it will disappear in a short time and the system will go back to the original state.

Let's assume that a stochastic fluctuation brings a localised infinitesimal volume of the water below T=0, lets say that the local temperature is $T_l=-\epsilon$. (Everything of course will work the same way around in the ice, melting locally inside the water at $T_l=+\epsilon$).

Normally a metastable phase is present and the fluctuation will average out to zero before anything the intermolecular bonds can be destroyed or created and the phase transition doesn't even start. But let's suppose that this is not the case and that a local phase transition actually starts happening. Then we will have a locally $T_l=-\epsilon$ and a ice sphere of ice of radius $R=\delta$ with both $\epsilon$ and $\delta$ much smaller than the typical dimensions of the system.

The phase transition follows the theory of nucleation (see Wikipedia). The results of this theory tell us that exists a critical radius $R_c$ which has the following property:

  • if $R<R_c$ the ice sphere will disappear and its radius shining exponentially in time
  • il$R>R_c$ the ice sphere will instead grow exponentially and the phase transition will that place

again the second case can't happen: because if the radius starts to grow, it will soon encounter the border $\Sigma$ of the infinitesimal volume of the fluctuation and the phase transition will stop.

A different (but connected) example

The following is not directly connected to the answer, but should remark that the described situation is stable. Suppose that we have the entire water at a state $T_c=-\epsilon$. Let's try to calculate the mean time to have a phase transition - that is, a fluctuation which creates an ice sphere of radius $R>R_c$.

The difference in free energy is of the form: $$\Delta F = T_s R^2 -\Delta f R^3$$ where $T_s$ is the surface tension coefficient (the partial ordering of the molecules of ice against the water phase cost a certain amount of free energy which scales with the surface, hence the $R^2$ dependence) and $\Delta f$ is the difference between free energy per unit volume of the ice and the water (see this image).

The $R_c$ is defined the maximum point of the function $\Delta F(R)$ because for $R>R_c$ the derivative is negative and an increase of the radius of the ice sphere diminish the free energy of the system and for $R<R_c$ the opposite holds. From this definition ve get that the critical radius is: $$R_c= \frac{2 T_s}{3\Delta f}$$ Hence we get $$\Delta F_c=\Delta F(R_c)\sim\frac{1}{\Delta f^2}$$ which is the variation of the free energy given by the fluctuation needed to generate the phase transition. Near $T=0$ we can assume $\Delta f \sim \Delta T=\epsilon$ (see the same picture for clarification). Arrhenius law tells us that the average waiting time for a fluctuation is: $$\tau= \tau_0 e^{\beta\Delta F}$$ hence: $$\tau \sim e^{\frac{\beta}{\epsilon^2}}$$ This tells us that is the temperature variation is small we should wait an insanely long time to see the transition happening.

I want to remark that this is not a proof, but we can convince ourselves that if a system with the entire water undergoing the fluctuation has a characteristic waiting time that long, the system described in the question is stable.

Hope that helps!

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  • $\begingroup$ If I'm not wrong, you're showing that homogeneous nucleation has very low chances to occur. But what about heterogeneous nucleation? I.e. the most common one, and the one we can expect near the interface water/ice in the system? $\endgroup$ – thermomagnetic condensed boson Nov 29 '17 at 19:41
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The ice and water phases are in equilibrium with each other at $0°C$ (provided your water is 100% pure. However, for your ice to melt you need to supply energy (latent heat of fusion). Given that the compound system water/ice is isolated from the environment, the fraction ice will not diminish neither increase. What you cannot say however, what part (molecules) will remain solid and what fluid, you only know the quantity of water that will be solid and the quantity that will be fluid.

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