Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This isn't a duplicate to "Why did my liquid soda freeze once I pulled it out of the fridge?". My question is why soda froze after it was opened. Opening a can or bottle seems to have a larger effect than just jostling it.

Is it because of the disturbance noted in the previous question? Is it related to the pressure decrease? Is it because of the release of some CO2 when it was opened?

share|cite|improve this question
up vote 7 down vote accepted

It's not a disturbance, the liquid isn't supercooled in this case. It's right about at 0 degrees, though.

It isn't the pressure drop directly, because you can give an upper-bound estimate to how much cooling the pressure drop directly does based on the observation that the pressure is not more than a few atmospheres. That means that space the gas in the bottle gets multiplied by a factor of, say 10 (that's too big), and (kT times) the entropy gain per atom is

$kT\Delta S = kT log(10)$

and kT is 1/30 eV, so you get at best 1/10 eV per gas atom. The density of gas on top is 1/300 the density of water, and the maximum total cooling from expanding that gas is negligible per atom of liquid. There is no significant cooling and heating in compressing a liquid, since essentially no work is done in the process, so the pressure drop does nothing to the liquid.

But the pressure drop makes the soda supersaturated with respect to its dissolved CO2, and this CO2 is outgassing from a liquid state to a gas state, and this is a huge gain in entropy. By outgassing the dissolved CO2 into bubbles, each CO2 atom gets a 300-fold increase of available volume to roam, and this can cool the liquid by

$kT log(300) \approx {1\over 4} eV $

So you get $.25 eV$ per released CO2 molecule, which is comperable to the binding energy of a water molecule. So each outgassed CO2 freezes order 1 H2O, and this freezes a network of filaments of water around the outgassed bubbles, making a slush.

share|cite|improve this answer
"There is no significant cooling and heating in compressing a liquid, since essentially no work is done in the process" I don't understand this bit - surely it takes significant work to compress a liquid? – chris838 Feb 26 at 21:10
Part of the definition of a liquid is that it is (for all intents and purposes) incompressible. As a result, you can apply or release pressure to a liquid and will not affect it, so no work is done. – WhatRoughBeast Mar 3 at 18:09

Farenheit calibrated his thermometer to have 0 degrees be the freezing point of salt water. Fresh water freezes at 32F. Water with stuff dissolved in it freezes at a lower temperature than water with less stuff dissolved in it.

Another hypothesis for why soda freezes after opening is that when you open it some CO2 outgasses and less CO2 is dissolved in the water. This will raise the temperature at which the soda freezes.

It is possible that the soda was cooled to 27F, and that the freezing point of soda with lots of CO2 is 25F so it stays a liquid. When you open it, enough CO2 escapes to change the freezing point to 30F and some of the soda freezes.

A test would be to see if the soda gets colder or warmer when it is opened. In the example above, the soda would warm up from 27F to 30F when opened. My guess is that that a soda that freezes when opened will get slightly warmer, but that a soda that does not freeze when opened will get slightly cooler - so slightly that I doubt I could measure it.

share|cite|improve this answer
The zero of the Fahrenheit scale is a mixture of ice, water and ammonium chloride in a 1:1:1 ratio, not seawater. – John Rennie Mar 3 at 18:04
Thank you, I meant salt water. – Wade Satterfield Mar 3 at 20:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.