I have a bit of confusion about a closed container scenario. First, I'll start with the open container. Say water is at some temperature, exposed to low humidity atmosphere, and begins to evaporate. The water that evaporates diffuses never to return to the container. To find the final temperature of the water is pretty easy: Assuming Heat of Vaporization stays somewhat constant ... (Heat of Vaporization) * (Mass of Water Evaporated) = (Final Temperature * Mass Initial - Initial Temperature * Initial Mass) * (Specific Heat). However, I'm having trouble with the closed container. Assume initially there is no water vapor at all, but still the same atmospheric pressure on the water (just to avoid boiling, though it doesn't really matter). Then assume a certain mass of water quickly evaporated and is now in vapor form so that the vapor is now at saturation. At this point to find the temperature difference in the liquid water the previous equation should be fine to use, however: won't there now be a small temperature difference between the vapor and the water? The vapor wasn't able to escape and is still in the closed container. So now won't there be some heat exchange in addition to the temperature lost due to evaporation? How would I quantify this to find the final temperature of the liquid water based on the initial conditions of the water being at a particular temperature and no initial vapor above it?