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I have a hot tub which I keep at 100 degrees F, and the water has "a lot" of dissolved salts in it. If I leave it open, will it evaporate faster when it is hot and humid outside, or when it is cold and dry? What are the relevant equations governing this? I would imagine the evaporation rate may depend upon the temperature of the water and the air and the humidity of the air and the amount of dissolved solids in the water.

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    $\begingroup$ What's wrong with the option of hot and dry outside? $\endgroup$ – Chet Miller Jun 18 '16 at 15:54
  • $\begingroup$ This question depends on what specific values you mean by "hot" "humid", "cold" and "dry". It could be either depending on how cold is cold, how dry is dry, etc... $\endgroup$ – JMac Jul 30 '17 at 14:01
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You could bound the rate of evaporation by the product of (1) the mean velocity of impingement of molecules upon the surface, (2) a Boltzmann factor representing the fraction with enough kinetic energy to surmount the energy barrier, and (3) one minus the relative humidity of the air above the water. The mean impingement speed is roughly 400 m/s, and the heat of vaporization from 40°C is about 600 cal/g, so the Boltzmann factor is $\exp (-H/kT)=\exp (-18)$. But there are two alligators in your tub, and they make evaporation very hard to analyze:

  • Water molecules in the liquid state form aggregates, which move more slowly than free molecules.
  • Unless there is a very strong breeze, or maybe you are soaking in a hot tub on the moon, there will be a thin boundary layer of moist (almost saturated) air above the surface, and the rate of evaporation will be limited by diffusion through this boundary layer.
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Evaporation depends on partial vapor pressure. So the control parameters are, - temperature - partial pressure - vapor pressure at the temperature

For the hot tub that you are studying, humidity will slow down evaporation and dry will help. Cold is not good for evaporation. In reality, you may not be able to make all happens at the same time.

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  • $\begingroup$ If it's an outdoor hot tub, wind speed is a major factor. $\endgroup$ – Ben51 Jan 19 '18 at 5:00
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Liquid water evaporates when the ambient partial pressure of water vapor is smaller than a value set by the "vapor pressure". Humidity is defined as the ratio between the actual value of the partial pressure, and this reference.

$$ h = \frac{p_{H_2O}}{p_{ref}(T_{air})} $$

The relationship between vapor pressure and temperature is given by the Clausius–Clapeyron relation.

From a kinetic perspective, a balance can only be reached when the flow of water molecules escaping the liquid due to thermal agitation is exactly compensated by the flow of water molecules in the air colliding with the surface. If the humidity is low, not enough molecules are getting in the liquid, and a net flux is leaving the bath (that's evaporation). If the humidity is such that the partial pressure of water vapor matches exactly the vapor pressure of water at the temperature of the bath, evaporation stops.

Up to the first order, the evaporation rate will depend on the difference between the partial pressure of water vapor in the air, and the vapor pressure at the bath temperature.

So your question is pretty much : is the partial pressure of water larger in a hot and humid air, or in a cold and dry one ? To which the answer is of course: it depends, tell me the exact conditions you are considering, and we will calculate the partial pressure in each case.

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