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?
$\alpha$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.
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.
