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[MadMojoMonkey]

It takes 40.66 kJ/mol of water to turn it from 100 degree water into 100 degree water vapor.
1 mol of H2O is (2*1g/mol + 1*16g/mol = ) 18 g/mol. Since 1 cc of water has a mass of 1 g, a mol of water is 18 cc.
So evaporating 18 cc of water will require the water to absorb 40.66 kJ of energy

The density of dry air is 1.225 kg/m^3.
Your 8 m^3 room has (8 m^3 * 1.225 kg/m^3 = ) 9.8 kg of air in it.

The specific heat of air is 1.005 kJ/kg*K.
You want to change the temperature of a 9.8 kg of air, so 1.005 kJ/kg*K * 9.8 kg = 9.849 kJ/K.

You want to lower the temperature by 10 C (which equals 10 K), so 9.849 kJ/K * 10 K = 98.49 kJ

This is a lottle more than double the energy required to evaporate 18cc of water, so evaporating a bit more than 36 cc, maybe 40 cc of water should do the trick.
This is a 1-time cost. Once you cool the room, you don't have to do so again... so long as there are no heat sources in the room, and it is perfectly insulated from heat transfer through the walls / doors.


In practice, you sitting there doing no exercise is close to ~75 - 100 Watts of power produced. Most of that power is heating the room.
That's another ~0.1 kJ per second that you need to dissipate. To do this, you need to evaporate another ~27 cc per minute.
Plus the computer is also a heat source.

[OngBonga]

It takes 40.66 kJ/mol of water...

Thanks for this analysis, I'll get back to you when the weather perks up again and I put this into action!

[MadMojoMonkey]

Thanks for this analysis, I'll get back to you when the weather perks up again and I put this into action!

Keep in mind that you don't want that evaporated water in the room. It bears the energy and keeping that energy out is kinda the point.

It's important that the water is evaporated by energy taken from the room. Adding an energy source to evaporate the water kinda eliminates the point.

Keep me posted. You'll want a cool room while you setup your pilot wave detector.