I understand that evaporative cooling takes place thanks to small pores contained in the pot and that allow some water to go through and evaporate. However I couldn't understand clearly whether water inside the pot stays at its original temperature or would it cool further? If it will become cooler then how?
Evaporative cooling works by removing the high-velocity tail of the kinetic energy distribution. That is, only the fastest molecules escape the liquid, leaving the rest to thermalize at a lower temperature. If there is capillary action taking water to the outside of the pot and that is evaporating, then the pot cools down as it is losing heat to the leaving molecules' kinetic energy. This then cools down the water inside by conduction.
One can then ask why, if the air is hotter than the water, can heat flow from the water into the hot air? The answer to this is that there is also a reverse process which is also possible: on a wet day, water molecules in the air can rejoin the water on the pot walls, and if the air is hot then in the mean this process will heat the pot and the water.
In an equilibrium situation, both of these processes happen at the same rate and there is no heat flow. For evaporative cooling to work, the air needs to be dry so that more molecules leave the water than condense into it. In a closed environment, though, evaporation will raise the air's humidity until both processes are equally likely and everything thermalizes. On a windy day, though, the pot is trying to raise the humidity of the whole atmosphere, which is not going to happen soon. This is an open system, in contact with an infinitely dry reservoir of dryness.
I heartily recommend Feynman's lectures on thermodynamics, such as The distinction of Past and Future, for getting a gut understanding of this aspect of the arrow of time.
Latent heat of evaporation cools the pot in the same way it cools your skin. If the pot is then at a lower temperature than the water in it, heat energy will be transfered to the pot from the water in it by conduction.
Imho, this process is driven not by energy considerations but by kinetic considerations. That should be why it naively seems weird that water absorbs heat from a cooler object and evaporates. Note: This is an explanation I came up with on-the-fly and have no references to back up with.
Since the earthen pot has small pores, water "flows" through those (probably by capillary action) to the outside surface of the pot. There, it encounters a relatively dry environment and it is kinetically encouraged to leave the surface of the pot and "evaporate". Presumably, while it is on the surface of the pot, it doesn't/can't interact much with the atmosphere and so it needs to take the necessary latent heat from the pot1,2. If that happens, it would help keep the earthen pot cool. Since the water evaporates, by Le Chatelier's principle, the reaction equilibrium gets a strong forward push and the water keeps evaporating.
 This seems like the weakest part of the argument. It would be interesting if someone could comment on this aspect.
 Since this is a surface effect, the cooling rate should now be propotional to the surface area of the pot. This would mean that smaller pots cool better than larger pots due to a better surface/volume ratio. This claims seems quite easily testable.
protected by Qmechanic♦ Feb 24 at 17:28
Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.
Would you like to answer one of these unanswered questions instead?