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Imagine we have two thermodynamic systems, one a mass of ice and the other an equal mass of liquid water, with both at 273.16K. Each system is isolated, except that they can interact with each other across a boundary that permits the exchange of heat but not matter or work.

What will the two systems look like at equilibrium? Somehow I want to automatically imagine that each system will be identical, a combination of liquid water, ice, and water vapor at 273.16K. But if this is true then the two systems were initially at the same temperature but not in thermodynamic equilibrium, an apparent violation of the zeroth law of thermodynamics.

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You have not specified how the pressure is controlled in the two systems. If they are each at the triple point pressure of 611.73 Pa there is no reason for heat to exchange and all will stay constant. If the pressures are different from this (and not on the freezing curve) energy can be released if there is heat flow by transferring heat between the reservoirs. There is a ratchet effect that will cause a small amount of heat to flow from the water to the ice because if each is 100% flowing the other way will make a temperature difference.

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  • $\begingroup$ I should have specified that each system started off at the triple point pressure, so that each was independently in equilibrium before they started interacting. My intuition is that each system would wind up with half of the total available energy, and I guess that's just wrong. $\endgroup$ Aug 29, 2014 at 4:07
  • $\begingroup$ In classical physics, temperature differences are the only thing that drive heat flow, so in your model there is nothing to drive heat. You are correct that the entropy will rise if heat flows from the water to the ice. Over the long term quantum effects will force energy to tunnel and equalize the energy per mass content. $\endgroup$ Aug 29, 2014 at 4:18
  • $\begingroup$ I disagree that the total entropy will increase if heat flows from the water to the ice, since Q/T(ice) = -Q/T(water). This is where my intuition hits the shoals - it seems like the energy should drift around and end up equally shared, but the laws of thermodynamics say it doesn't. $\endgroup$ Aug 29, 2014 at 5:41
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Erik, nothing will happen (as you already know!).

Water and ice are in equilibrium at the triple point. No heat can flow at constant temperature. The water and ice have the same free energy per mole, so no spontaneous change can occur, and the total entropy cannot rise as a consequence of heat transfer.

It is much simpler to just consider putting ice and water into an empty container, removing the air, and sealing it. Initially the system is out of equilibrium, so some ice will melt, or some water will freeze, and some will evaporate to fill the space. Once all three phases have reached the triple point, their proportions will not change - they were determined by the initial conditions.

You might ask why the system cannot wander about on the flat surface of constant free energy. I think one answer is as follows. To stay at the triple point, the temperature and pressure must be fixed. Of course the energy is also fixed, so we have three equations, which are (probably) sufficient to determine the proportions of the three phases. You might ask what happens when there are four phases in equilibrium - but there is a theorem that this cannot happen.

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    $\begingroup$ That's not true! Even though there will always be the same mass of ice, it will change shape due to the exchange with water and vapor. It's the same with the equilibrium of a weak acid. Even though the pH is constant, there is always an exchange of protons with the anion. $\endgroup$
    – LDC3
    Aug 30, 2014 at 1:02
  • $\begingroup$ I agree - the equilibrium is dynamic, and the ice could change shape. $\endgroup$
    – akrasia
    Aug 30, 2014 at 17:24
  • $\begingroup$ I agree with that also. Without the barrier the ice, vapor, and water systems would be free to exchange matter, resulting in the continuous random breakdown and reformation of the ice. What's not clear to me is why we wouldn't see a similar effect across a barrier due to the random movement of energy. $\endgroup$ Aug 30, 2014 at 20:35
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The original state is not in thermodynamic equilibrium. The liquid side has a lot more heat than the solid side due to the heat of fusion.

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  • $\begingroup$ Remove the barrier, we have a liquid/solid mix that is (if the pressure is right) at the equilibrium temperature. What makes heat flow? You are correct that if heat flows from the liquid to the solid the temperature of each will not change, compensated by freezing/melting. $\endgroup$ Aug 29, 2014 at 4:03
  • $\begingroup$ @RossMillikan If you remove the barrier, there is still an imbalance of entropy. Now that the ice can float, there is more submerged than when it distributes itself across the whole surface. $\endgroup$
    – LDC3
    Aug 29, 2014 at 4:35
  • $\begingroup$ There is no specification of gravity in the problem, so ice floating is not important. You are correct that entropy will increase if heat flows. $\endgroup$ Aug 29, 2014 at 4:51
  • $\begingroup$ @RossMillikan In a gravitational field, with the ice and water interchanging, the ice will tend to form along the water-air surface. Eventually, the ice will spread across the whole surface. So you will need to employ Maxwell's demon to maintain the status quo. In a micro-gravity environment, I would assume that the liquid would form a separate sphere from the ice. Once separated, there would be no liquid-solid interchange. If the liquid and solid are still in contact, then the ice would reform so the liquid could form as close to a spherical shape as possible. $\endgroup$
    – LDC3
    Aug 29, 2014 at 5:36

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