About the mechanism behind regelation, I read that "increasing the pressure causes a decrease in the melting point of ice and hence the ice at the point of application melts and the water thus formed goes to the rear side of the wire where it refreezes and the released energy is what is needed for melting in the first place".

This was all good but when I thought about it , I got stuck.

Suppose we have a block of ice at 0°C with which we try to do the experiment of regelation. Now applying some pressure results in decrease in the melting point of the compressed part which means that we have an ice of melting point some -x°C at 0°C (btw how is it even possible ?) .

Now even if the ice melts , it has to first lose its energy to lower it's temperature from 0°C to -x°C and then gain some energy to account for the latent heat of fusion .

But in order to lose energy there should be a temperature difference but I don't think there is any difference at all since the surrounding ice is also at 0°C.

So how does the ice reach its new melting point ? Where does it deposit it's energy ?

  • $\begingroup$ The ice under pressure is already above its melting point...? $\endgroup$
    – HTNW
    Jul 9, 2022 at 3:34
  • $\begingroup$ @HTNW Because initially the ice was at 0°C and the pressure just changed its melting point and not it's temperature (I guess).. $\endgroup$
    – Ankit
    Jul 9, 2022 at 3:55
  • $\begingroup$ yes, and ice above its melting temperature melts. Substances don't have to be at the melting temperature to melt! $\endgroup$
    – HTNW
    Jul 9, 2022 at 3:56
  • $\begingroup$ @HTNW but doesn't that mean that the melted water will still be at 0°C ? And if that's the case how will it regelate ? $\endgroup$
    – Ankit
    Jul 9, 2022 at 3:58
  • 1
    $\begingroup$ Ice melting does not require temperature difference. It requires only addition of energy, and in this case it is supplied by the gravity force, pulling the wire down. $\endgroup$ Jul 9, 2022 at 21:38

1 Answer 1


First, let's address some possible misconceptions: (1) that the ice can't melt because it's at the same temperature as the surroundings or because we're not actively heating it, or (2) that heat transfer doesn't occur toward the melting ice (regarding your statements "how is it even possible ?", "there could be no energy flow since there is no temperature difference", and "How will it absorb energy from the surrounding ice which is also at the same temperature ?").

In regelation, we force melting by making the ice unstable in a way that's distinct from heating it. (More on that below.) The latent heat associated with the phase change (specifically, the enthalpy of fusion) must be paid for by sensible heating from elsewhere. In this way, the melting absorbs energy from the immediate surroundings, cooling them, creating a temperature gradient, and thus driving heat transfer from the broader surroundings through Newton's law of cooling.

Now, let's look at the thermodynamic reason for melting.

A phase transition temperatures is simply the intersection in the Gibbs free energy (or the chemical potential, which is just the molar Gibbs free energy) of two phases:

The Gibbs free energy $G$ incorporates the enthalpy $H$, temperature $T$, and entropy $S$ as $G\equiv H-TS$, so we find that the temperature-dependent slope of every curve is that phase's entropy. (Observe that high-entropy gas always has a steeper curve than low-entropy condensed matter.) The lower $G$ curve is the more stable one. Accordingly, we see solids melt into liquids and then boil into gases as the temperature increases.

Regelation occurs because applying a pressure to a region of solid ice increases its Gibbs free energy as $G=F+PV$, where $F$ is the Helmholtz free energy, $P$ is pressure, and $V$ is volume. Pushing the blue curve up in the diagram clearly lowers the equilibrium temperature with liquid water. (The green curve doesn't move because the liquid, being a fluid, can simply move away from a region of local compression; thus, the Gibbs free energy of the liquid remains unchanged.)

Then, let's look at the kinetics and heat transfer during the process.

Assume that compression (from a weighted wire stretched over the top of the ice, as you reference) lowers the melting temperature from $T_\text{m}=0^\circ \text{C}$ to $T^\prime_\text{m}<0^\circ \text C$. We have compressed solid ice at 0°C (or somewhat below, but assumed to be above $T^\prime_\text{m}$). The compressed ice accordingly melts and cools below its original temperature, which draws thermal energy from the surrounding region (uncompressed ice that started at the original temperature and is now cooling) to supply that latent heat of fusion. The temperature of the adjacent surroundings is lowered correspondingly by that sensible heat transfer.

Depending on the geometry, the melted water may flow away (probably to refreeze elsewhere). This exposes fresh ice that consequently melts under the applied pressure. Now, if the surrounding ice has cooled to the new compression melting temperature $T^\prime_\text{m}$ because of all this latent heat being absorbed, melting—even under compression—is no longer thermodynamically favorable. This is a temporary situation, though: heat transfer from the surrounding ice ($>T^\prime_\text{m}$) warms the compressed ice and allows continued progression. In this thought experiment, therefore, the wire passes downward entirely through the solid ice, melting the ice below it and allowing refreezing above it.

  • $\begingroup$ thanks for your answer but I didn't get your last papa . How will it absorb energy from the surrounding ice which is also at the same temperature ? $\endgroup$
    – Ankit
    Jul 9, 2022 at 4:01
  • $\begingroup$ @Ankit The part of the ice that melts locally drops in temperature. It is similar to blowing dry air over liquid water to cool it. $\endgroup$
    – HTNW
    Jul 9, 2022 at 4:34
  • $\begingroup$ @Ankit Please see my edit describing why a temperature difference arises that drives heat transfer. $\endgroup$ Jul 9, 2022 at 5:51
  • $\begingroup$ @Chemomechanics the water thus formed is also at 0°C (right ? ) . So what did you mean by saying that it cools ? $\endgroup$
    – Ankit
    Jul 9, 2022 at 9:51
  • $\begingroup$ The water and ice in that area cool down because sensible heat is being drawn away to pay for latent heat. I wonder if the confusion comes from the fact that we normally actively heat a solid to melt it. Here, we don’t actively heat anything; we put a solid in a situation where it’s not stable (but the liquid is). This forces melting, which necessarily draws heat from the surroundings. $\endgroup$ Jul 9, 2022 at 15:17

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