This is a question about evaporative cooling as used in residential evaporative cooling appliances. This type of cooling uses the heat in the ambient outside air to evaporate water and remove the heat from the air, then push the cooled air inside. The equation to predict the temperature of the resulting air after it's given up its heat to evaporate the water is as follows:

$$T_{output} = T_{dry} - (T_{dry} - T_{wet}) * \epsilon$$

where $T_{output}$ is the output air temperature, $T_{dry}$ is the air temperature of the dry bulb, $T_{wet}$ is the air temperature of the wet bulb, and $\epsilon$ is the cooling efficiency.

For example, on a very dry summer day (dry bulb 95 degrees, wet bulb 60 degrees) my evaporative cooler with 90% efficient media is capable of cooling the air to 63.5 degrees.

However, this equation does not seem to take into account the temperature of the water itself. Does it matter? Intuitively, it would seem to make sense to me that hotter water would be easier to evaporate, since it's closer to its boiling point. Or maybe colder water is better because it will absorb more heat from the air? Or maybe it's a wash because the same amount of heat is required, but with hotter water, more is needed because it will evaporate faster? Help me understand this.


2 Answers 2


No, the temperature of the water is not that important for the performance of an evaporative cooler. This is basically because the energy needed to increase the temperature of liquid water (its specific heat capacity) is very small compared to the energy needed to evaporate the same amount of water (its enthalpy of evaporation).

At room temperature the specific heat of liquid water is 4.18 J/(g$\cdot$K) while the ethalpy of evaporation is 44.0 kJ/mol. Since the molar mass of water is roughly 18 g/mol, this means that approximately 585 times as much energy is needed to evaporate an amount of water as to increase the temperature of the same amount of water by 1 K. This means that even if the water starts at freezing temperature, is heated to 40 $^\circ$C (104 $^\circ$F) and then evaporates, less than 7% of the energy absorbed by it is used for increasing the temperature.

The temperature of the water might effect the rate at which evaporation occurs, but I guess evaporative coolers are designed to achieve 100% relative humidity in the wet bulb regardless of the temperature of the water, so this probably doesn't affect the performance.


After evaporating the water is at the same temperature as the ambient air. The heat removed from the air is whatever it takes from the starting condition of the water to get to water vapor at that temperature. In particular, cooling the water will require the air to heat it, so each degree C the water is colder will require about an additional 1 cal/g=4.184 Joules/g from the air.

  • $\begingroup$ So you're saying that the colder the water, the more heat that can be theoretically removed from the air by evaporating it? $\endgroup$
    – iLikeDirt
    Jun 4, 2014 at 22:19
  • $\begingroup$ That is correct. You are just looking for the energy difference between the starting and ending states. $\endgroup$ Jun 4, 2014 at 22:20
  • $\begingroup$ This answer does not take into account the enthalpy of evaporation, which accounts for most of the heat absorption. $\endgroup$
    – jkej
    Jun 5, 2014 at 0:56
  • $\begingroup$ @jkej: I realize it accounts for most of the heat absorption, but we were not asked to calculate the absolute heat absorption, just the change when the source water is cooled. The heat absorption of evaporation is 43.99 kJ/mol at 25 C and 43.35 at 40 C, so to heat 25 C water to 40 C and evaporate it (assuming that is ambient) will take 15*4.18*18+43.35=44.48 kJ/mol. If you evaporate it first, you will get the same result, but I don't have a figure for the heat capacity of water vapor handy. $\endgroup$ Jun 5, 2014 at 2:40

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