# Does the temperature of water determine how much heat will be removed from air used to evaporate it?

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.

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.