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Name: Oliver
Status: other
Grade: 12+
Country: Andorra
Date: Spring 2014

Does anybody know a formula for the cooling rate of Prince Rupert's Drop? If you have 200 grams of glass, how long would it take the core to cool to room temperature? Would the structure of a Prince Rupert's Drop cool slower?

Hi Oliver,

Two hundred grams (1/5th of a kilo!) of glass is far to large to compare with the fraction of a gram that makes up a Rupert Drop! You asked how long it would take for the core of a 200-gram "chunk" of glass to cool. The answer depends on the actual outside shape (and therefore the surface area) of the glass, and the method of cooling. The short answer is "a very long time"!

A Rupert Drop works because its small size allows its outer surface to cool extremely fast, and this skin becomes hard while the center is still liquid. The key here, is the large surface-area-to-volume ratio of the drop. Then after the surface skin is cool and hard, the center more slowly cools and shrinks. This shrinkage puts the already-cooled outer skin in compression. As you may know, glass is extremely strong in compression, so the compressive stresses in the outer skin protect the drop from damage even when the drop is hit with a hammer. It is the rapid cooling of the outer skin of the drop that matters here; the speed with which the inner core cools is not too important.

A 200g chunk of glass is simply far to large to achieve the ultra-rapid cooling rate that is necessary to get a sufficiently thick, compressively stressed, outer skin for the large chunk of glass to act like a Rupert Drop.

This rapid cooling effect is used to make "tempered glass", which is used in all car windows (except the front windshield). This very strong, but thin glass is made by exposing a nearly-molten sheet of glass to strong blasts of cold air. As with the Rupert Drop, the outer skin hardens first, then is subjected to strong compressive stress as the center more slowly cools.

Regards, Bob Wilson

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