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The classic example of a supercooled liquid is supercooled water. The metastable state is easily perturbed and the whole sample transitions to water ice.

The classic example of a superheated liquid is liquid hydrogen in bubble chambers for particle physics. When a particle passes through the chamber, it leaves behind a train of bubbles: locally, the sample transitions to hydrogen gas.

Why is the phase transition global in the first example, and local in the second? Why does a nucleation point trigger a global freezing in the first case, but only a local vaporization in the second?

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    $\begingroup$ While a slightly supercooled water sample may completely freeze, that is not true in general (see recalescence). $\endgroup$
    – Jon Custer
    Commented Aug 26 at 15:52
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    $\begingroup$ I think that this has something to do with the actual mechanism of phase separation you look at. There is a fair amount of possible mechanisms and depending on the actual thermodynamic potential surface geometry of the process you could observe complete transitions to different states with almost no energy barrier, like in spinodal decomposition while in other situations there might be a relatively small energy barrier to overcome in order to observe a phase transition thus leading to an activated mechanism (like in nucleation). The two systems you are describing are different in nature. $\endgroup$
    – Giuseppe Basile
    Commented Aug 28 at 9:15

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It is not true in general that a superheated fluid cannot completely vaporise. Interpreting your question broadly, a hot saturated liquid in a pressure vessel which is suddenly depressurised can completely vaporise. Fluids that are capable of fully vaporising under these conditions are called 'retrograde' and have peculiarly shaped saturation lines (an isentrope can in some places link the saturated liquid and saturated vapour lines - meaning that the fluid can isentropically expand from saturated liquid to saturated vapour). See an example here for MD4M in CoolProp. MD4M P-s diagram Retrograde fluids tend to have high molecular complexity, which means high molar specific heat capacities. For normal (non-retrograde) fluids, isentropes never link the saturated liquid and vapour lines (they don't 'overhang' one another on a P-s diagram).

The situation I described allows very high superheats, because of the pressure vessel. Your question suggests complete evaporation under a more restrictive case, with the fluid superheated, at most, to the spinodal (otherwise the fluid enters the region of thermodynamic instability and rapidly evaporates) and is then perturbed at constant pressure. I think it is impossible. But you could test the idea using eg. REFPROP or the open source CoolProp (noting that getting metastable liquid properties in the latter is more complicated).

Liquid hydrogen is probably quite an unusual example given its low molecular weight and how close to absolute zero it is (it may also only be slightly superheated). Superheated liquids can boil violently when perturbed (see eg. here), this is probably a matter of the relative size of the available superheat energy to the latent heat of vaporisation.

Freezing is exothermic (meaning you can't run out of energy) and also the latent heats tend to be much smaller for freezing than for boiling (in part because the volume change is comparatively small).

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  • $\begingroup$ To state it more succintly: a thermodynamic analysis (to determine the equilibrium end state) would suggest that the end state of the phase transition is governed by the ratio of superheat/supercool sensible enthalpy to the phase change latent enthalpy. It is likely large in the case of supercooled water and small in the case of superheated liquid hydrogen. $\endgroup$
    – Nick Mason-Smith
    Commented Sep 9 at 1:21

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