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If water evaporates at room temperature because a small percentage of the molecules have enough energy to escape into the air, then why does a kitchen counter with a small amount of water eventually evaporate completely when at room temperature?

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  • $\begingroup$ In the system (glass of water etc.) the water molecules have a distribution of different kinetic energies. Some of the molecules with a higher kinetic energy are therefore able to break free from the bulk and evaporate. $\endgroup$ – Bdrs Oct 15 '17 at 11:55
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    $\begingroup$ This is because of diffusion. It not "evaporates", it "diffuses" :-) $\endgroup$ – ParaH2 Oct 15 '17 at 11:57
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    $\begingroup$ It doesn't. There's still a microscopic amount of water present. $\endgroup$ – Hot Licks Oct 15 '17 at 18:37
  • $\begingroup$ @Bdrs: your explanation doesn't describe how all the water can eventually evaporate, by your explanation all of the molecules have enough energy since the beginning. $\endgroup$ – whatsisname Oct 16 '17 at 3:40
  • $\begingroup$ While I understand the underlying question(Why does water evaporate at room temperature?), I'm really confused by the wording of this question, to me it essentially equates to "If water evaporates at room temperature, then why does more water evaporate at room temperature, too?" $\endgroup$ – Taegost Oct 16 '17 at 12:43
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As your small percentage of molecules with high enough kinetic energy evaporates, the remaining liquid water cools down. But in doing so, it drains heat from its surroundings and thus stays at room temperature (or close to it), so there is still some fraction of molecules that can evaporate, and they do so, and more heat is transferred from the surroundings, and so it continues until all water is gone.

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    $\begingroup$ And, in particular, a small amount of water evaporating won't lower the temperature or raise the humidity of the room in any measurable way. $\endgroup$ – David Richerby Oct 15 '17 at 18:25
  • $\begingroup$ Thanks for the answer ! Still got a question, because it seems that everything relie on The system had energy at start. How the layer of molecules get this starter energy from ? Or is it unrealistic to imagine not this case ? Edit Might be the answer ? $\endgroup$ – Grégoire Fruleux Oct 16 '17 at 14:16
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    $\begingroup$ Everything has thermal energy, unless it is cooled down to absolute zero. If you do that with your water (and prevent thermal exchange), then of course it won't evaporate. $\endgroup$ – Ivan Neretin Oct 16 '17 at 14:20
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This happens because the rate of evaporation is higher than the rate of condensation.

$$\ce{ H2O (l) <=>> H2O (g) }$$

This is also due to the fact that you have an open system: matter and energy can be exchanged with its surroundings. The evaporated water can evaporate from the glass and condense somewhere else.

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  • $\begingroup$ States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different. $\endgroup$ – Martin - マーチン Oct 16 '17 at 11:21
  • $\begingroup$ Then when is the rate of evaporation equal to the rate of condensation? $\endgroup$ – BadAtChemistry Sep 8 at 3:43
  • $\begingroup$ This depends on a number of factors (see van.physics.illinois.edu/qa/…), e.g. the temperature of the water surface, the temperature of the air, the humidity, and of course the area of the surface. Also, you have to take into account the airflow past the surface as well as the water currents convecting heat away from the surface. $\endgroup$ – basseur Sep 9 at 15:25
  • $\begingroup$ Also a good read: e-education.psu.edu/meteo3/l4_p4.html $\endgroup$ – basseur Sep 9 at 15:27
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The water on the surface does not exist in isolation it is in contact with the air and with the surface. Random higher energy molecules in the surface and in the air will add energy by collision to the water molecules leading some of them to escape the liquid (evaporate).

This is why evaporating water leads to cooling of the air and surfaces around it.

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    $\begingroup$ This answer almost hits the point. It is not a "percentage of the total water" which escaped the liquid state, but a percentage of the water molecules close enough to the surface. As the surface retains a constant size (assuming a standard-issue cylindrical pot), we observe a more constant rate of evaporation (in the "percentage of total water" case we would observe some sort of half-life - an evaporation rate going down to zero asymptotically as the amount of water decreases). $\endgroup$ – Klaws Oct 16 '17 at 14:58
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Let's say, $q\in {]}0,100{]}$ is the minimal net percentage of the volume (or of the mass) that evaporates at any second $t$ (for every $t>0$). Saying "net", we assume that more water molecules leave the kitchen counter than return, and that the fraction of the molecules leaving the surface in relation to the number of molecules getting back has a constant, positive lower bound. (Other answers explain why it is likely to be so in kitchen conditions.)

Then, at most $100-q$ percent are left per time unit. So, after $t$ time units, the amount of water left will be at most $\mathrm{a_0}\Bigl(\frac{100-q}{100}\Bigr)^t$, where $a_0$ is the initial amount. Since $100-q<100$, we obtain $$\lim_{t\to+\infty}\,\mathrm{a_0}\Bigl(\frac{100-q}{100}\Bigr)^t \ = \ 0\,.$$ In particular, after a certain point of time, the amount of water will be lower than the minimal possible amount (the volume of one $\mathrm{H}_2\mathrm{O}$ molecule or its mass, simplified, of course).

If the assumption made is not valid (say, due to great humidity somewhere in Asia), the result would be wrong: the water will NOT fully evaporate.

(An aside has to be made. Note that the above mathematical treatment is a gross simplification. To get a more realistic evaporation model, we should take into account that the evaporation happens only from the surface, and not from the whole volume, and that both the surface and the volume change with time. Also, bear in mind that even within one second, the evaporation rate changes.)

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    $\begingroup$ I think this ignores the physics aspect of the question. The question seems to me to be asking, "Wehn water evaporates, it's because the high-energy molecules leave the body of the liquid. OK, but once they've all gone, why doesn't evaporation stop?" $\endgroup$ – David Richerby Oct 16 '17 at 7:21
  • $\begingroup$ @DavidRicherby The question doesn't say it that way. What you wrote is your purely subjective interpretation of the question. $\endgroup$ – user53443 Oct 16 '17 at 10:44
  • $\begingroup$ The question talks about the small percentage of molecules that have enough energy to escape and asks how water is able to evaporate completely. Having said that they understand how the energetic molecules are able to escape, it's certainly a natural interpretation that the question is asking how the unenergetic ones can escape, too. $\endgroup$ – David Richerby Oct 16 '17 at 11:29
  • $\begingroup$ @DavidRicherby It seems "natural" only subjectively, to you, but not necessarily to the others. If you wish to go with your own interpretation, post your own answer. $\endgroup$ – user53443 Oct 16 '17 at 13:03
  • $\begingroup$ I don't need to post an answer because it's already been posted and received a lot of upvotes. $\endgroup$ – David Richerby Oct 16 '17 at 13:30
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Water always evaporates when above 0 degrees Celsius at normal atmospheric pressure.
Which means when above 0 deg there always are molecules with high enough energies to leave the liquid phase.

See this Quora answer (note the Quora answer points out evaporation at 0 degrees C is possible with the right pressure), this Physics.SE answer

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    $\begingroup$ Just a comment on temperature and evaporation: there is nothing magical about the temperature of 0C when we're talking about the evaporation of water. So long as the relative humidity is less than 100%, water at any temperature, above or below 0C, will evaporate. $\endgroup$ – airhuff Oct 17 '17 at 6:24

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