So I am doing a lot of research on evaporative coolers and I have come to a question I can't find much info on.

I am currently a mechatronics engineering student that has yet to take thermodynamics so my understanding is anything but perfect; however, thermo is my favorite subject.

That being said I am wondering about where the energy is going when increasing the temperature of the water.

I know that water's specific heat is approximately 4.179 kJ/kg * kelvin and the enthalpy of vaporization is 2441 kJ/kg at 25 degrees C.

I could be incorrect in this assumption but assuming the water evaporates at any temperature between 0-100 degrees Celcius then when is the enthalpy of vaporization actually factored in? only when trying to get the temp from 100 degrees C to 101 C? or is it factored in all along the traverse from 0-100 C?

Say that you supply a source of water with the required energy to evaporate 1 kg of water that is 25 degrees C, which would be 2441 kJ as stated above. Does this action mean that 1 kg of water will evaporate from the source and there will be no change in the temperature of the water? or there will be a temperature change in the water but will come back to equilibrium at 25 degrees C after that 1 kg of water evaporates? if the energy is applied but slowly then would the temperature not increase and cause slightly increased evaporation than if no energy were applied and if applied quickly the temp will increase but return to equilibrium after evaporation?

Any insight is appreciated, thanks!


1 Answer 1


When heating water that's simultaneously evaporating, the details of the temperature profile depend on the kinetics of the various processes, in addition to the thermodynamics.

The diffusion of thermal energy from heating depends on the geometry, heat flux you apply, and thermal diffusivity of the material, among other potential factors. The evaporation rate depends on the temperature, surface area, relative humidity of the adjacent gas, and enthalpy of vaporization, among other potential factors.

Thus, it's possible for water at 25°C to evaporate and consequently cool down, with the cooling rate mitigated by gentle heating you provide; it's also possible for energetic heating to bring the water above 25°C even as it's simultaneously evaporating. Finally, the water could maintain a steady 25°C while evaporating; this is typical when the environment's at 25°C and there's very efficient heat transfer.

Note the existence of negative feedback: a higher temperature generally increases the evaporation rate exponentially, which pulls much more thermal energy out of the material and tends to cool it down.

Is this close to what you're asking about?

  • $\begingroup$ It certainly helps. My main goal is to make an evaporative cooler, even though many designs are already present, because working through the design and development will solidify the concepts and facts in my understanding. I am having a difficult time trying to understand how much energy is transferred through standard evaporation say at room temperature. If say 1 kg of water was evaporated from a container without any supplied energy to facilitate evaporation would it have, over whatever time it took for that kg to evaporate, absorbed the 2441 kJ's? $\endgroup$
    – Thermr
    Jul 29, 2020 at 17:58
  • $\begingroup$ Yes; the heat of vaporization is the energy is takes to break the bonds between water molecules. That's the miracle of the evaporative cooler; the cooling is spontaneous because the evaporation is spontaneous, and then the humidified air just blows away to be replaced with drier air. $\endgroup$ Jul 30, 2020 at 3:46
  • $\begingroup$ In cooling, the enthalpy diagram is used:youtube.com/watch?v=QqEbN4dDp8M $\endgroup$
    – The Tiler
    Feb 18, 2022 at 12:56
  • $\begingroup$ For more details, see also :engproguides.com/pressure-enthalpy-diagram.html $\endgroup$
    – The Tiler
    Feb 18, 2022 at 13:13

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