A wet object or a volume of water will decrease in temperature due to the effect of evaporation. We understand this to be because of the molecular kinetics, where the faster water molecules escape and become a part of the air. Since those that escape have more energy than the average, the process decreases the temperature of the water, but how does it affect the temperature of the air?

I can imagine two different factors:

  1. The escaped water molecules have a higher energy than the air molecules, since the air was previously at equilibrium with the water
  2. The binding energy that makes water liquid and gives it surface tension reduces the energy at which a molecule escapes

While #1 would seem to predict that the air becomes hotter at the expense of the water becoming colder, #2 should imply that the net thermal energy of the entire system decreases because some of the kinetic energy is converted to a form of potential energy (not sure to call it surface tension, chemical potential, or something else). These two factors are in conflict, so the answer is non-trivial to me.

Let's not consider long term factors like convection eventually cooling the air. Considering only the action of a molecule escaping the surface of water through evaporation in a closed system, how does that affect the temperature of the gas it escapes into?

  • $\begingroup$ The air has nothing to do with that process at first order approximation. Vaporisation of water will happen in a vacuum with the same thermal effects, only faster. The air will slow the vaporisation speed (diffusion in a gas is slow) and the layer in contact with the water will be cooled down. $\endgroup$ – Georg Dec 5 '11 at 14:51
  • $\begingroup$ Georg offers some meaningful insight. One thing I considered adding is that I've heard it claimed that a molecule that evaporates is itself above the boiling point. I think this idea has several problems, again point #2 makes me wonder if it would be at that T after it escapes at all. Also, once in the gas, it adopts the gas T (this question is about the energetics of that). If it had to be above vaporization T to get in the gas, how can it remain below vapor T while suspended in the gas? That probably relates to the vapor equilibrium conditions which I don't have the background for. $\endgroup$ – Alan Rominger Dec 5 '11 at 16:07

If one goes to the wiki article on evaporation one sees that

For molecules of a liquid to evaporate, they must be located near the surface, be moving in the proper direction, and have sufficient kinetic energy to overcome liquid-phase intermolecular forces. Only a small proportion of the molecules meet these criteria, so the rate of evaporation is limited. Since the kinetic energy of a molecule is proportional to its temperature, evaporation proceeds more quickly at higher temperatures. As the faster-moving molecules escape, the remaining molecules have lower average kinetic energy, and the temperature of the liquid, thus, decreases.

This is due to the statistical distribution of the kinetic energy of the molecules, the molecules in the tails have enough energy to escape the surface tension of the liquid.

In a closed container energy should be conserved and the higher kinetic energy molecules released in the gas should increase the temperature while the liquid surface and the contact part of the gas will be cooling. In open systems with convection the contact of the gas with the liquid is continually renewed and thus a cooling of the gas can be obtained as in the evaporative coolers. Convection by continually replacing the air keeps the humidity low, which also allows higher evaporation rates.


The temperature of the air should definitely increase,

As the water cools ,its means the air molecules carrying the extra energy transfers it into the atmosphere.

And this has little to do with surface tension, as surface tension is a result of the top most layer of the liquid which more or less remains the same.

So its only beacause of the 1# that the air warms.

  • $\begingroup$ I have a feeling, it shouldnt have been down voted $\endgroup$ – Tomarinator Dec 5 '11 at 15:01
  • $\begingroup$ The escaped molecules take energy from the surface of the liquid. They escape because of the tail of the statistical distribution of the kinetic energy: a perercentage has more energy than the binding of surface tension . This percentage depends on the temperature of the liquid, more when it is hotter,. In a closed container from conservation of kinetic energy, the liquid will cool as well as the air in contact with it, but the bulk of the air would get a higher temperature. Convection is used and the air with the evaporate is removed in the water cooling systems. $\endgroup$ – anna v Dec 6 '11 at 7:16
  • $\begingroup$ @annav It sounds like Stom's answer is falling a little bit short of full correctness, but you seem to have a good grasp to argue that the air still warms. Try writing an answer, I'd like to select something. $\endgroup$ – Alan Rominger Dec 6 '11 at 14:40

If my intuition is right:

It depends. Obviously it will still depend on the temperature of the liquid.

For evaporation to heat up the air, the newly escaped molecules from the liquid would have to have higher kinetic energy than the average air molecule.

While it's probably true that the higher kinetic energy molecules are the ones that escape from the liquid, it's not clear weather they will still have high energy after they break the surface tension of the liquid and run off with the gas.

It's possible that the particular particle has enough energy to break from the liquid, but afterwords is left flaccid with a small kinetic energy, therefore bringing down temp of gas.

To figure this out, we would need:

-Force/work/something required for a molecule to escape the liquid.

-Then we look at the boltzman curve for liquid and see how much energy particles have at different temperatures.

-For heating of gas, we need temp of liquid which gives us plenty of particles with enough energy to overcome surface tension and still have a kinetic energy higher than that of gas.

-It would be nice to have superimposed on the boltzman curve of our liquid, a similar curve which describes which of the molecules are the ones that escape the liquid, otherwise we kinda speculate.


protected by ACuriousMind Feb 14 '16 at 21:05

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