[redleader]

I think [my] previous explanation was a bit off. Its more accurate to think of a bong as a system in equalibrium. For every bit of heat that enters the coolant, the same amount leaves instantly.

In this case there is no change in heat at all. The added energy from the CPU simply pushes a really tiny bit of vapor out of the liquid. Because the system is in equalibrium, the input heat has no effect on temperature.

So in this case, the vapor has the same temp as the water because no change in temperature is possible without throwing off the equailbrium. In this case the vapor temp just depends on how warm the coolant is and thus how effective the bong.

This is a logical outcome. A small bong would not be able to transfer enough heat at a low temp to maintain equalibrium, so temps would rise until the vapor pressure of the coolant was sufficent to match the incoming heat. A larger bong would the evaporate coolant more effectively, thus requiring a less favorable vapor pressure (and thus cooler water/water vapor).

Edit: Added my to first sentance.