Writing up the evaporative cooling post on cold atom techniques, I used the standard analogy that people in the field use for describing the process: cooling an atomic vapor to BEC is like the cooling of a cup of coffee, where the hottest component particles manage to escape the system of interest, and what’s left behind is colder. The departing atoms or coffee molecules carry off more than the average energy per particle, leaving a lower average energy (and thus a lower temperature) for the remainder.
A question that sometimes comes up when I talk about this is how you can possibly use this to get a substantial reduction in temperature, given that coffee doesn’t really cool down by all that much– you don’t see people’s drinks Bose condensing at Starbucks when they fall down the YouTube rabbit hole and leave them sitting untouched for too long, after all. The answer (at least part of it) is that when AMO physicists do evaporative cooling, they can continue to lower the energy threshold at which “hot” atoms are removed from the sample. In the coffee analogy, the energy threshold is fixed– it takes some amount of energy to get a molecule of water out of the liquid phase. The re-thermalization of the remaining liquid will bump a few of the leftover molecules up to sufficient energy to get them out, but as long as you’re dealing with a finite sample, you’ll eventually hit a point where there just aren’t any molecules at that energy to remove, so the evaporative cooling process cuts off.
Which got me to idly wondering whether you could actually do this with hot liquid. That is, would it be possible to substantially cool a container of hot liquid by evaporative cooling through somehow lowering the energy at which molecules can leave the system. For purposes of idle speculation, we’ll say that “substantially cool” means “reduce to a temperature below room temperature, assuming excellent thermal isolation of the container from the environment, in a time less than would be required to reach that temperature by ordinary cooling processes.”
Of course, there is a well-known way to do this kind of thing: you reduce the pressure. The boiling point of water depends on pressure, dropping to a bit above room temperature at a pressure of 20-30 torr, which is not a particularly impressive vacuum by experimental physics standards. So, in principle, it seems like you ought to be able to do it– stick a beaker of hot water with a thermometer in it into a bell jar, and start pumping air out. If you could do it right, maybe you could get a substantial reduction in temperature at the cost of throwing a lot of your liquid away, just the way that AMO physicists do in making BEC.
And, of course, you can find videos demonstrating this sort of thing, such as this one, where Edwards Vacuum freezes a bunch of water by pumping on it:
This is the same basic physics, but cast in different language than we usually use when talking about evaporative cooling of BEC, where you try to do the reduction somewhat more slowly. It’s cast as making a change in the boiling and freezing points of water– at low pressure, the beaker of water bubbles dramatically because the whole thing is now above the boiling point, which is dropped below the temperature of the sample.
The question I was wondering about is– and I don’t know if this is a sensible question– whether you can lower the temperature of the liquid by evaporation without the dramatic boiling, in a manner analogous to evaporative cooling in ultra-cold atom experiments. That is, rather than lowering the boiling point for the entire sample quickly, making the bulk boil, lower it slowly, so that only the “hot” molecules escape, but those with lower energy happily remain in the liquid phase. Which might involve losing less of the liquid to boiling (or maybe more; I’m not sure how you would calculate that).
The physics of the phase change complicates this whole business, of course– thermodynamics was not one of my better subjects– and there’s also a lot more than just evaporation going on. A real cup of coffee cooling will lose heat not just through evaporation, but also direct conduction through the container and convection of the air directly about the liquid (this is, after all, why my insulated Starbucks cup works well to keep my tea hot). I don’t know how big a factor the evaporation actually is relative to those. And, of course, when they boil/freeze water in vacuum, the container is in contact with a large vacuum apparatus at room temperature, which will tend to make it hard to change the temperature of the liquid.
This makes me want to go in the lab and start screwing around with vacuum pumps and thermocouple gauges, though. Happily, I let the Mechanical Engineering depertment borrow our big bell jar, so I can’t easily add yet another project to the loooooong list of interesting physics things I don’t have time to actually do.
(The “featured image” at the top is taken from this discussion at the Physics Stack Exchange, which takes on a vaguely related question in a fun way, including a simple experiment.)