Chemistry ? regarding jockey box

[Tweek]

So how does carbonation work in regards to a jockeybox? My guess is that it will be slightly less (read more co2 in solution) than the level in the keg, assuming the keg is actually warmer than the beer coming out after the jockey box.

So if I have a keg that is 70 degrees (f) and I run it through a jockeybox, the beer comes out at 50 degrees (f) will teh beer be fizzy like it probabally would be at 70 or would it regulate itself back down in the short amount of time it goes through the coils?

I think pehpas if I wasnt a few pints o homebrew and a bottle o wine (shared with the wife) into the night I might be able to figure this out rather easily, but in my current state this is proving to be a difficult problem.

any thoughts?

[Fast_Eddy]

I thought thermal shock causes CO2 to come out solution. So it should have more head and more bead, IMO.

[Tweek]

that may be so. Looks like I am going to have to actually dig out the chem book on this one.

[kgaugler]

Good question Tweek, I'm thinking that because the beer in the keg is still under pressure (not exposed to ambient pressure) as it gets cold, it should maintain equilibrium of dissolved CO2 for the new temperature. So it should not be too foamy IMO. Think you better pull out the P-chem books for this one though, I sucked at P-chem.:confused:

[Tweek]

Ok. so I did some reading and a little asking around and the best I can figure is it wont work. The reason I am thinking this is because the beer is served from the bottom of the keg and the co2 that has come out of suspension is at the top, so when you push it through the jockey box the original co2 that was in suspension when the beer was cold will no longer be with it thus not able to dissolve back into solution.


Cheers!

-Tweek