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!