Well, everyone has bits and pieces of this whole deal, but here goes my quick sum-up on evaporative cooling. Hope it's helpful.
At room temperature, water will evaporate. It's a very slow process, but water vapor will leave the surface of a pool of water. The amount of water vapor coming off can be greatly improved by increasing the amount of surface area (spraying the water does this very well) and by blowing air over the surface to remove any water vapor lingering there.
The air isn't there to heat up the water to cause evaporation. It doesn't need to. The air merely aids the evaporation by giving water vapor a place to go.
Energy is conserved - in this case, enthalpy is conserved. Ambient air meets ambient water. Water evaporates. The water that evaporates has an increase in enthalpy due to its becoming a vapor. The liquid water left must decrease its enthalpy accordingly, and does so by going down in temperature. (Due to equilibrium, the water vapor will decrease in temperature as well, but you get the general idea.)
The enthalpy of vaporization is so high that it takes very little evaporation to cool a CPU.
As for the heatercore placement, the liquid in the bottom will give you the best heat transfer. Even that will be very poor, given the water at the bottom of the bong isn't moving very much. Koslov is on track with the liquid/liquid heat exchanger. If that's not an option, coiled copper tubing or the heatercore inside the bong would work about as well, I'd expect. Or hey, try a cheapo low-pressure $5 pump at the bottom just to swirl the water around the heatercore.
On another note I just finished my finals (except the ones that finished me
), one of which almost entirely on distillation and vaporization, so I'm intrigued by airspirit's idea in the other thread. Might mosey over there soon . . .
Alchemy