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The problem is simple, I suppose...

I have a beverage at room temperature (let's say 23°C) in an aluminum can (it's a Barq's root beer.) I want to put it in a freezer (let's say it's at -17°C). How long do I have to wait for it to reach a chilled temperature (let's say 8°C).

I need to know for a couple reasons. One, I don't want the can to explode. Two, I don't want it to get too cold for my teeth (perhaps I shouldn't be drinking pop? :-) ) Three, I want don't my coworkers to steal my nice, cold beverage on the first hot day of the year.

If I could find the right equations, I would just plug in the numbers, but I can't seem to find the right ones.

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This problem is not simple at all. 1st semantics: Do You ask about "keeping chilled" or chilling down? 2nd look at my comment to Benjamin. –  Georg Jun 21 '11 at 10:30
I'm confident in my ability to drink the beverage before it returns to a undesirable state, so I'd say "chilling down". –  corsiKa Jun 21 '11 at 16:15

1 Answer 1

I would probably use Newton's Law of Cooling;

$\frac{dT}{dt}=-k(T - T_0)$

where $T_0$ is the ambient temperature.

As far as what k is, you could just do a little experiment. Find the temperature of the beer, put it in for 5 minutes, take it out, take the temperature again.

If you aren't the experimental type, $k = A \cdot H$ where $A$ is the surface area (a nice cylinder) and $H$ is the heat transfer coefficient.

For water to air, through metal, you get about $11 \frac{W}{m^2K}$.

But I bet non-linear effects will cause enough issues that you will probably have a pretty wide error, (Specific type of metal of a can (aluminum), the shape of it, thickness, fact that beer is not just water, etc.). Probably an experimental approach of finding k will give you the most accurate way of going about it.

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That is the very simple sight of world of a beginner in physics. Look up "Grashoff Number" and its use. –  Georg Jun 21 '11 at 10:15
I don't quite understand what you are saying? –  Benjamin Horowitz Jun 26 '11 at 21:43
EDIT (silly 5 minute limit): Obviously there are tons of complicated effects that go on, including convection in the can (which I guess is what you are referring to). I assume the asker is not looking for a complicated computer model to understand all the factors of the cooling of the can, but a relatively simple approximation. I have a feeling that this method will get you pretty close to the actual time. –  Benjamin Horowitz Jun 26 '11 at 21:49

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