What you want are steam tables. These define the relationship between vapour pressure of steam and temperature.
http://www.connel.com/freeware/steam.shtml
Energy dumped to atmosphere will be related to air flow rate, temperature of inlet air and inlet air relative humidity. With a perfect evaporator (which you won't get in reality), the total heat loss will be the latent heat of evaporation multiplied by the density of water vapour at the inlet temperature, multiplied by (1-inlet humidity fraction).
The problem here is that the temperature limit of an evaporator isn't set by the ideal situation. For instance, if you double the surface area used for evaporation if all else is held constant the mass of water evaporating will be doubled too. This will then increase the heat dumped to atmosphere, and drop the temperature.
You do then end up with a set of differential equations which define the relationship between vapour pressure, water temperature, droplet size, etc. Unfortunately, this doesn't give you the equilibrium temperature. The shape of the bong will also affect things - as the vapour pressure will not be constant within the bong, changing the rate at which water evaporates in different parts.
This means you end up with two coupled sets of differential equations - which you are very unlikely to be able to solve analytically. You'll have to derive both sets of equations (probably not that difficult), and then set up a solver to find the numerical solution. That is probably what the cooling tower programs you found do.
Just to complicate things, you'll also have heat transfer between the air and the liquid droplets, water/air and bong, etc.