This question already has an answer here:

When I leave wet clothes in the open air, they will get dry over time by themselves even at room temperature. I know that somehow the water becomes vaporized; it's not "disappearing". For that, it needs to get energy from its environment and probably it's getting this energy from heat. But since it is at the same temperature with the clothes or its environment, I don't understand how an energy transfer can occur. Doesn't this violate the definition of temperature, which is only defined to determine where heat will flow?

So the question can also be expressed as: How can water get heat (if not heat, what?) from its environment when there is no temperature difference so that it can evaporate?


marked as duplicate by ACuriousMind, Daniel Griscom, user36790, Gert, Ali Jan 12 '16 at 7:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ The temperature of the cloth is lowered. That is the basis for evaporative cooling. $\endgroup$ – Lewis Miller Jan 11 '16 at 19:31
  • $\begingroup$ Yes it has to be lowered since it gives heat to water. But what drives this heat transfer in the first place, that is my question. Why should it give heat to water when they had the same initial temperature? $\endgroup$ – ozgeneral Jan 11 '16 at 19:33
  • $\begingroup$ The heat transfer is driven by the temperature difference. When the water evaporates the temperature of the cloth sinks and this temperature difference drives the heat transfer. The temperature difference will settle into an equilibrium with the heat transfer until all the water that can evaporate is gone. The cloth will, of course, not be completely dry since the atmosphere is not completely dry, either. $\endgroup$ – CuriousOne Jan 11 '16 at 19:36
  • $\begingroup$ I don't think this is a duplicate because this question is principally interested in how heat can low from cold(er) air to warm(er) water, which isn't addressed by the proposed duplicate. $\endgroup$ – John Rennie Jan 12 '16 at 7:25

Microscopically, both the water molecules in the air and the water molecules on the clothing are rapidly moving around due to their thermal energy. Every once in a while, a molecule on the clothing will have enough energy to break free; every once in a while, a molecule in the air will stick to your clothes. Because the humidity in your room is less than 100%, the first process will happen more often, so water will go into the air. Conversely, if you put your clothes in a sauna, where the humidity is more than 100%, your clothes will get more wet over time.

Now let's look macroscopically. Why should evaporation happen at all, if it costs energy to do it? The reason is that there's "more room" for the water molecules in the air than on your clothes, so it's more likely for things to bounce into the air (which is big) than land on your clothes (which are small). Formally, we say that the process is entropy-driven. (Since entropy counts available microstates, this is just the same thing.)

Increasing entropy and decreasing energy are separate goals. In this case, the energy and entropy effects oppose each other, but entropy wins; in general, you can tell which one wins by looking at the change in Helmholtz free energy, $F = U - TS$.

  • $\begingroup$ Very insightful, thank you. I just had follow up question when I read your answer which might be unrelated if I am wrong. The small particles in air, like dust, exist almost all the time. Is it because of the same reason? For example do everything in my home contribute the dust(Idk if there is a better word for it) particles in air with the break free process you just described? $\endgroup$ – ozgeneral Jan 11 '16 at 19:55
  • 1
    $\begingroup$ Not quite. Dust is much heavier than water molecules, so the energy factor wins here. If you don't disturb anything, it'll almost all fall onto the floor. But if you walk around in the room, it'll get picked up by tiny air currents. $\endgroup$ – knzhou Jan 11 '16 at 20:02

The air molecules are like little balls constantly hitting the surface of water. The air molecules strike at different velocities and the one with higher velocities knock out, lets say, 2 water molecules. Other molecules rush to the site. The knocked out water molecule is still bombarded by other air molecules which drift it away.

Now, the density of knocked out water molecules are higher near to the surface and are always trying to get back and to form aggregates. For water there is a high tenancy to get aggregated, and these aggregates will get back to the surface if they are not transferred from the surface by collective high velocity bombardment of air(wind!). If there are higher concentration of water molecules in air(high humidity), they could form aggregates easier and thus take more time to dry the clothes. Converse happen in less humidity.


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