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1eyedjack

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So you want to reduce the temperature of 15 kg of liquid by some 25 kelvin. That is 375 kilojoule.

 

I thought the cooling power of the freeze is about half the excess power it uses when freezing in something relative to what it uses at steady state. That may be something like 20 watt. 20000 secs is about 6 hours.

 

You may notice that this formula means that 1 bottle would take 50 times less, i.e. 7.5 minutes, which is obviously absurd. For small quantities one has to take the limited heat conductance of the air, glass and beer into account, so the time will be longer than what my formula says. But for 15 kg I think it doesn't matter much.

 

This is a very rough calculation, but I would say it must be of that order of magnitude i.e. between 1 hour and 48 hours unless my theoretical understanding of the problem is way off, which it probably is :)

 

Hopefully Nuno can give you more accurate estimates, he is the official Water Cooler thermodynamics expert.

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  • 2 weeks later...

A very good method for getting this right is, insert all beers in the freezer, go back in 10 minutes and open 1 beer and drink it, (decide how cold it is), go back, 10 minutes later. open and drink another beer (decide how cold it is), keep repeating this exercise until you are satisfied the beer is cold enough.

 

with methods like this, there is no reason for a scientific explanation. (Helene, you really overcomplicate such a simple thing as beer temperature) try my method, it is not only more accurate it is a lot more fun.

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