Let's say a solid has a stable phase (A) that exists below some temperature, T, and A transitions to another stable phase B above that temperature. If you take a solid sample of B and rapidly quench it such that it forms some metastable phase (A'), then upon heating past T will you see the following sequence:

A' -> A -> B


A' -> B ?

In other words, does the stable phase, A, have to appear between the metastable phase A' and B?

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    $\begingroup$ It might, it might not. The kinetics of the phase transition are different than the thermodynamics. $\endgroup$ – Jon Custer Jul 23 at 23:10
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    $\begingroup$ So if you heat very slowly to the transition temperature, you should see A'-> A -> B, but if you go quickly it's possible to jsut see A' -> B? $\endgroup$ – Drew Lilley Jul 23 at 23:18
  • $\begingroup$ Recall that ‘diamonds are forever’ yet metastable at room temperature. You can even heat them quite high in oxygen-free environments. All depends on nucleation rates. $\endgroup$ – Jon Custer Jul 24 at 0:38
  • $\begingroup$ I don’t quite understand: if you quickly cool the substance in the high-temperature phase B below the phase transition point, then this phase (B) will become metastable with decreasing temperature, where does any phase A' have to do with it? $\endgroup$ – Aleksey Druggist Sep 8 at 10:55

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