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 from the refrigerator 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 refrigerated water will actually not evaporate as efficiently and hence not produce cooler air than in case of room temperature water ?

For what it's 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 physics predict what would happen?

  • 1
    $\begingroup$ FWIW I can tell you from experience, growing up in Texas hot summers in the 1960s, all we had for cooling our house were evaporative coolers. When it was really hot we would run a lot of cold well water into them and they would definitely blow cooler for a while. $\endgroup$ Commented May 2, 2021 at 2:06
  • $\begingroup$ @AdrianHoward Okay, so we have your uncontrolled experiment saying colder water works and mine saying it does not. Hence, we need to see what other answers say about the physics involved. $\endgroup$ Commented May 2, 2021 at 5:30
  • 1
    $\begingroup$ If you use water from the refrigerator in your house, say goodbye to the nominal cooling - the refrigerator cools its contents by heating the kitchen! $\endgroup$
    – user121330
    Commented May 6, 2021 at 16:33
  • $\begingroup$ @user121330 that is irrelevant, because The fridge is not a variable. It stays on , whether or not i use refrigerated water in the cooler or room temperature water in the cooler. Hence, that is not an issue. You can just assume, the refrigerator is in another room or another house . $\endgroup$ Commented May 6, 2021 at 18:55
  • 1
    $\begingroup$ The radio uses more energy when the volume is high, the computer heats the house commensurate with processor use and your refrigerator has to put the heat from that room-temperature water somewhere. It's those coils on the back - touch them while it's running if you don't believe me. I get that you want to ignore this but part of physics is learning how to think about things. People leave fans on in closed rooms they don't occupy because they ignore part of a problem, but adding energy to a system always heats it up. $\endgroup$
    – user121330
    Commented May 6, 2021 at 19:24

2 Answers 2


It depends on whether or not you are calculating cooling per unit time or cooling per unit water.

Per unit water, you get (a little bit) more cooling. Besides the enthalpy of vaporization, you get a little bit more cooling as the water absorbs heat to come up to room temperature.

Per unit time, the speed of evaporation will be (slightly) less. As you mention, the warmer water will vaporize more rapidly.

Neither of these effects are large. The total budget is dominated by the vaporization. A 15 degree C difference in the water will be hard to notice.

  • $\begingroup$ I was concerned merely with temperature of the output air, and in which case will it be lower But i guess out of those 2 options, I am concerned with cooling per unit time i.e. which will provide me the most comfort from the heat. I dont care if more water is used. Water is cheap $\endgroup$ Commented May 1, 2021 at 20:46
  • $\begingroup$ " A 15 degree C difference in the water will be hard to notice. " The kind of colder water i was thinking was refrigerated water. So, the temperature difference will be closer to 35-40 celsius $\endgroup$ Commented May 1, 2021 at 20:52
  • 1
    $\begingroup$ Of course, if you're getting cold water by cooling it in your fridge, then overall you're probably increasing the temperature of your house, since the heat that the fridge takes out of the water is just dumped into the house, with some extra from the energy needed to run the fridge. $\endgroup$ Commented May 1, 2021 at 22:30
  • 1
    $\begingroup$ The fridge's duty cycle increases if it's loaded by tap temperature water so it does produce more heat when it contains warmer things. $\endgroup$
    – alessandro
    Commented May 2, 2021 at 17:36
  • 1
    $\begingroup$ @silverrahul your fridge doesn't magically output the same amount of heat regardless of what's put into it. If your fridge is cooling more of its contents then it will produce more heat. Also water from the tap is generally colder than the temp of the air since it's generally underground, which fluctuates in temperature less. $\endgroup$
    – Kat
    Commented May 2, 2021 at 18:05

You've intuited two possibly relevant counteracting mechanism (of which none, one, or both might be negligible): colder water takes more energy to heat up, and colder water evaporates slower.

As a first stab at the problem, let's assume that the water evaporates at its well-defined supply temperature and that the rate of water evaporation is essentially the same at both temperatures. This is a good approximation if the water evaporates quickly under forced convection of hot, dry air—a pretty good description of a swamp cooler fed from a (room-temperature or cold) water flow. Here, the difference in water chemical potential (essentially, the concentration) between the liquid water and dry air in conjunction with a very high relative surface area drives rapid evaporation, and the water temperature doesn't have time to equilibrate at room temperature.

The enthalpy of vaporization (also called the heat of vaporization) reflects the amount of "heat" needed to achieve evaporation, corresponding to the cooling effect applied to the surroundings. The amount reflects evaporation/vaporization at the given temperature, from liquid water to gaseous water. It does not require boiling.

The parameter expressed here is the specific enthalpy (i.e., per unit mass). If the water in both temperature cases is fed at a constant rate and evaporates quickly, then the per-mass enthalpy can act as a surrogate for the per-time enthalpy, giving us an effective cooling power rate. Try comparing the rate at which you need to replenish water. If the rate isn't lower for the colder water, then the above assumption is valid and the specific enthalpy translates easily into a cooling power. If the rates are different, then the specific enthalpy should be adjusted according to the mass rate to again obtain a cooling power.

The relative slope around 0°C to 50°C is clearly slight. We can estimate the slope from the difference between the heat capacity of liquid water and steam, about $4.2\,\frac{\mathrm{kJ}}{\mathrm{kg\,K}}$ and $1.9\,\frac{\mathrm{kJ}}{\mathrm{kg\,K}}$, or about $2.3\,\frac{\mathrm{kJ}}{\mathrm{kg\,K}}$. So by switching from water at 40°C to 10°C, for example, you'd increase the cooling effect by about $70\,\frac{\mathrm{kJ}}{\mathrm{kg\,K}}$, an improvement of about 3%. Put another way, the energy it takes to heat or cool a kilogram of liquid water is generally much less than its latent heat (the heat required to vaporize it) of thousands of kilojoules. This would account for the difference you experienced as being essentially imperceptible.

Alternatively, it may be the case that you're providing a large tub of water that evaporates slowly. Note that water away from room temperature will tend to equilibrate toward room temperature, minimizing the influence of the original temperature (while also providing a separate mechanism of cooling). This would also contribute to the difference for different temperatures being imperceptible. In any case, the coupled heat–mass transfer problem becomes more complex, as you may need to consider the exposed area, temperature distribution inside the water container, and boundary conditions that mediate the heat and mass transfer. (In other words, the kinetic process of evaporation depends on more parameters than the thermodynamic energy balance.) Do you have this information? You may wish to migrate to the Engineering Stack Exchange site, which might be more useful in terms of, say, providing recommendations from manufacturers, who have studied this problem in detail.

  • $\begingroup$ So, although the effect is miniscule, you are saying that the net effect will be that in theory, the colder water will provide better cooling effect ? My thinking was that since hotter water needs less heat from air to evaporate , hence it would be better, since it can evaporate easier. But you seem to suggest precisely the opposite, that since colder air needs more heat to evaporate , it is better. Also, you are not taking time into account i.e. is their a difference between how fast , say water at 10°C absorbs Q kJ , vs how fast water at 35°C absorbs Q kJ. Because that matters in this case. $\endgroup$ Commented May 1, 2021 at 21:07
  • $\begingroup$ Also, how doe it compare to how fast , water absorbs Q kj, during the latent heat phase $\endgroup$ Commented May 1, 2021 at 21:13
  • $\begingroup$ In fact i was thinking about this , and i dont think it matters how MUCH heat the water can absorb before evaporating. Because of how the cooler operates, what matters is how fast can the water absorb heat. So , the heat of vaporization plot seems irrelevant. I need to know how fast will the water absorb heat at each of the temperatures. Can you find any data, plots etc. regarding that ? $\endgroup$ Commented May 1, 2021 at 21:18
  • $\begingroup$ " If the water............evaporates quickly " This assumption is the crux of my question. If water evaporates quickly in both cases, then it is a no brainer, obviously, colder water will absorb more heat. And i agree, the other factors and engineering considerations are also an issue. I guess my question, is more about just the physics of cooling. In other words, with air blowing through them, would water at 10 degrees absorb heat faster than water at 30 degrees ? also putting it in those terms , the answer seems a simple yes. Is it indeed yes, for this specific theoretical question? $\endgroup$ Commented May 1, 2021 at 21:53
  • $\begingroup$ Is there something special about the "evaporation point" where it absorbs heat faster than it does while it is at 10 degrees or at 30 degrees ? $\endgroup$ Commented May 1, 2021 at 21:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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