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When white tin ($\beta$-tin) is cooled to a temperature below $13.2\ \mathrm{^\circ C}$, it creates the allotrope of gray ($\alpha$-tin), a gray, amorphous powder.

My question is that once you have the powdered gray tin, doe just raising the temperature above the point of stability turn it back to white tin, but in a powdered form?

Is there a change in appearance of the gray powder when that happens, or is it mostly unnoticeable?

Does the conversion to gray tin change the melting point?

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    $\begingroup$ I'd love to know; this seems like it would make an amazing demonstration. All I can find is that reheating will convert the tin back into its beta form. I imagine that would happen below its melting point at 231 C, but I don't know for sure. As long as it did, you'd wind up with tin powder. Really cool 20 second video of this, for anyone interested. $\endgroup$ – Jason Patterson Feb 16 '15 at 2:25
  • $\begingroup$ Yes, powdered beta tin can be made from powdered alpha tin, down to micrometer size. See, for example, this paper on exactly this topic: Powder Technology 1995, 84 (1), 35–38. $\endgroup$ – N. Böwering Aug 27 '17 at 12:05
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$\alpha$Sn and $\beta$Sn are the two solid allotropes of Sn. As you not, below 13C the stable phase is $\alpha$Sn, which has a diamond cubic crystal structure (like diamond, Si, and Ge) and is a semi-metal. Above 13C, the thermodynamically stable phase is $\beta$Sn, a body-centered tetragonal crystal. So, cycling back and forth around 13C varies the thermodynamically stable phase back and forth, with the only question being the kinetics of the transformation. The video in the comment by @JasonPatterson shows that the transformation indeed takes place reasonably quickly (unlike diamond to graphite for carbon).

As for melting, if you rapidly heated $\alpha$Sn and avoided the phase transformation, you would find that the melting temperature would be lower. Using the SGTE thermodynamic data (A.T. Dinsdale, CALPHAD 15(4) 317-425 (1991)), one finds that the $\alpha$Sn -> liquid phase transition would occur at about 430.7K, ~75K lower than the standard $\beta$Sn -> liquid melting point at 505.06K.

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  • $\begingroup$ Do you mean that an undercooled melt of tin would actually be produced by rapidly heating αSn > 430 K? I find that a bit hard to swallow. It's clear that the theoretical melting point of the unstable phase would be lower, but do you think this could be observable? $\endgroup$ – Karl Apr 10 '16 at 20:20
  • $\begingroup$ Why would it not be observable? For example, the melting point of amorphous silicon (a distinct thermodynamic phase, not a glass) was measured through pulsed laser irradiation (Phys Rev Lett 52 2360 (1984). So, yes, you would observe a lower melting point, provided you heat rapidly enough to bypass the (relatively slow) solid-solid transition. $\endgroup$ – Jon Custer Apr 11 '16 at 14:52
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Phase changes are fickle. Much like supersaturated solutions.

Add a crystal to a supersaturated solution and it will precipitate rapidly.

Bring a crystal of the thermodynamically favorable phase into contact with a "supersaturated" phase and it may just do this: https://www.youtube.com/watch?v=sXB83Heh3_c I am not sure how exactly the proceedure in the video was done, so do not take it as evidence that my statement is true. Also, Diamond is stable well below its phase transition temperature. Surely there are other compounds like that.

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