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The melting of ice is a reversible process according to most sources I have seen. However, the melting of ice increases the total entropy of the universe which shouldn't be possible if the process was reversible.

Can someone help clear up this concept?

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    $\begingroup$ It doesn't increase the entropy of the universe if done reversibly, by contacting the ice with a surroundings heat source at slightly above 0 C. $\endgroup$ – Chet Miller Jun 18 at 14:01
  • $\begingroup$ When considering entropy one should also consider enthalpy. $\endgroup$ – Jon Custer Jun 18 at 14:02
  • $\begingroup$ Could you provide the source of this claim and the exact conditions under which it is reversible? $\endgroup$ – Roger Vadim Jun 18 at 14:02
  • $\begingroup$ @RogerVadim See Mike Stone's answer. It couldn't be simpler than that. $\endgroup$ – Chet Miller Jun 18 at 14:11
  • $\begingroup$ @ChetMiller what does it have to do with the formulation of the question? Are you claiming that melting of ice is always reversible? $\endgroup$ – Roger Vadim Jun 18 at 14:14
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The melting of ice, in the idealized case, is reversible. Here, we have a mass of ice at 0°C (at standard conditions) and an infinitely large heat reservoir (or environment idealized to have infinite heat capacity) also at 0°C. Thermal energy $Q$ flows from the reservoir to the ice. (Don't ask why. When we consider reversible idealizations in thermodynamics, there's often no reason for anything to occur.) The melting of the ice produces a notable entropy increase because of the increased mobility of the water molecules. The same entropy (Q/273 K) is removed from the heat reservoir. Entropy is transferred, and none is generated.

Melting of ice when you personally observe it is irreversible when you expand the scope to include the cause. Ultimately, something hotter than 0°C (at 1°C, say) heats the ice to melt it. The flow of energy down a temperature gradient generates entropy. (The surrounding environment loses entropy ~Q/274 K, and the ice gains entropy Q/273 K, which is a larger number.) Correspondingly, the process is thermodynamically spontaneous, as is everything you actually observe. Moreover, the entire real, observable process is irreversible. Does this answer your question?

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  • $\begingroup$ Yes! That's cleared a lot of things up. I've always had an issue with the idea of a spontaneous process. Is a spontaneous process one where there is no done during the process? $\endgroup$ – ibrahim kanber Jun 18 at 15:05
  • $\begingroup$ Sorry, no what done during the process? I think your comment is missing a key word. A spontaneous process increases global entropy. $\endgroup$ – Chemomechanics Jun 18 at 15:08
  • $\begingroup$ *work. Sorry missed that $\endgroup$ – ibrahim kanber Jun 18 at 15:41
  • $\begingroup$ Work is not a factor. Some spontaneous processes involve work, and some don't. $\endgroup$ – Chemomechanics Jun 18 at 15:53
  • $\begingroup$ I saw a derivation of an equilibrium condition from Helmholtz free energy and in the derivation the work transferred was said to be zero for a spontaneous process which resulted in an inequality dF≤0. This tells us that spontaneous processes occur in the direction of decreasing F and U. Is this correct? $\endgroup$ – ibrahim kanber Jun 18 at 18:29
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Why do you think the entropy increases? The change in entropy of the heat source is $−𝑄/𝑇_{\rm melt}$ and the change in entropy of the water/ice mix is $+𝑄/𝑇_{\rm melt}$ with the same $𝑄$, so the sum of the entropy changes is zero.

The water in contact with the ice is always at the melting temperture, so no entropy is generated by the melting itself --- but if the ice is in warm water then entropy is generated by the heat conduction from the warm water to the melting-point water.

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