Or should it be the other way around, because hotter water evaporates easier than cold water?

I know this sounds stupid, so just hear me out.

I use an evaporative cooler and add normal room temperature water. Someone told me to add cold water to get better results. I intuitively thought that indeed cold water will produce colder air.

But as we all know evaporative coolers cool air by using the air to evaporate water which takes away heat from the air. So, we need the water to evaporate to cool the air. But we know that hotter water evaporates easier than cold water.

So, should it not be the case that colder water will actually not evaporate as efficiently and hence not produce cooler air than in case of hotter water?

For what its worth, i tried both and could not feel any noticeable difference, but of course it was nowhere close to a controlled experiment. So, I would appreciate an answer about what does the physics predict about what would happen?

  • $\begingroup$ The energy change from the evaporation is much larger than from the warming up of the water. The temperature will make barely any difference so it's not worth cooling the water as you will be wasting energy to get this cold water. $\endgroup$ – Jonathan R Swift May 1 at 22:23
  • $\begingroup$ But you are saying that the colder water will in theory produce more cooling, even if it is not considerable enough to be worth it, right ? But my point is, will it actually even produce less cooling, since it wont evaporate as much and hence will not be able to make use of energy change from the evaporation? $\endgroup$ – silverrahul May 1 at 22:27
  • $\begingroup$ It will produce more cooling but increase the humidity in the room causing other problems. As in, it does not work. $\endgroup$ – StainlessSteelRat May 2 at 0:08
  • $\begingroup$ What does the air actually give up heat to? The evaporating water cools the remaining water. The water takes heat from the air only if the air is warmer than the water. The added heat raises the vapor pressure of the water until the heat of vaporization equals the heat given up by the air, and steady state is reached. Hot water won't cool the air until it has cooled down cooler than the air. $\endgroup$ – Phil Sweet May 2 at 1:54
  • $\begingroup$ @StainlessSteelRat "It will produce more cooling but increase the humidity in the room causing other problems" . I know it will increase humidity in the long run. My question is only about , when i put colder water in, will the exit air have lower temperature or not $\endgroup$ – silverrahul May 2 at 5:21

An evaporative cooler works by removing the suspended layer of saturated vapor from around the filter, encouraging it to use the ambient heat from the stream of the air thus making it colder, to evaporate to replace the washed-out moisture.

The latent heat for water is 2460 J/4.18 J per gram = 588.5 times more than specific heat. (Thanks To Phil Sweet for correcting me)

So the heat removed from the air is mostly done by water vaporization not by being near cold water.

Also, saturated water pressure increases with water temperature.

Bottom line just room temperature water is the most practical.


there is new technolgy called two stage evaporative cooling. this technology is more efficient and can be used in humid climates as well. it doesn't add moisture to the conditioned air.

it is sold in US. i don't want to plug them, but you can search indirect evaporative cooler.

  • $\begingroup$ @Phil sweet, thanks Phil. i will correct my answer. $\endgroup$ – kamran May 2 at 2:11
  • $\begingroup$ " so the heat removed from the air is mostly done by water vaporization not by being near cold water. " It is not about which process absorbs MOST heat, it is about which process absorbs heat QUICKEST . The latent heat vs specific heat addresses which would absorb most heat. Is there any similar figures which help figure out which absorbs heat most QUICKLY. $\endgroup$ – silverrahul May 2 at 5:28
  • $\begingroup$ @silverrahul, of course, vaporization is faster. think about the contact area of a small body of water being in contact with the airflow absorbing its heat. Now compare it to the contact area of the same water atomized into vapor and peeled off the surface.? $\endgroup$ – kamran May 2 at 18:27
  • $\begingroup$ I might make a tiny change: 'Bottom line is whatever temperature water is most easily available' $\endgroup$ – Carl Witthoft May 3 at 14:16
  • $\begingroup$ @CarlWitthoft Availability of water is not an issue. Any temperature water from freezing to boiling is easily available $\endgroup$ – silverrahul May 3 at 14:44

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