In evaporative cooling does the water and the air both cool down? Wouldn't that violate energy conservation? I remember the explanation of a swamp cooler is that it exploits the heat of vaporization of water. The air passing over the water gives up energy enough to evaporate some of the water thus the air is cooled and conditioned with extra water. However the explanation for sweating that I just saw on khan academy seems to contradict that. So basically khan draws a picture of some sweat droplets and a zoomed in version with some hydrogen bonded water. He goes into an explanation of average kinetic energy of the molecules being the temperature, so if one of the moleucles near the surface of the droplet has a way above average energy it can break the hydrogen bond and overcome the air pressure and "evaporate." But in that picture the energy of the water is reduced by simply removing one molecule with higher kinetic energy, and thus the average kinetic energy is reduced. How can the sweat cool down, and the air in a swamp cooler cool down? Are they different processes?
 A: The air plays a different role in each situation. With the swamp cooler, the warm air is what warms up the water, allowing it to evaporate more quickly and transfer thermal energy out of the air and use it to drive the phase change. Both the air and the water cool down by virtue of the fact that the air gave energy to the water, and the water lost thermal energy by changing phase. With sweating, the body warms the sweat, allowing it to evaporate. Basically, the air in the swamp cooler example corresponds to the skin in the sweating example, and the water is the same in each. Overcoming the hydrogen bonds is the same thing as exploiting the heat of vaporization of water. It doesn't violate conservation of energy because so much energy goes into the phase change.
A: When air is warmer, it has higher kinetic energy.
When this warm air contacts the water, it gives some of it's kinetic energy to the water.  This lowers the temperature of the air, and also evaporates some water with the energy it gives.
The thing is, the process of going from liquid to vapour doesn't require a change in temperature; so the energy of the system stays the same, while the air loses temperature and the water only changes phase (maintaining it's temperature).
It's worth noting that the water could still heat up before or after the phase change if it is cooler than the air to begin with.
A: A swamp cooler functions by having warm dry air pass through a large wetted pad that is porous with respect to the flowing air. Evaporation cools the pad, which in turn cools air passing through the pad. The end result: The air is cooled (and humidified). Perspiration functions by having preferably dry air evaporate droplets of water on the skin. Evaporation cools the droplet of water, which in turn cools the body.
Both a swamp cooler and perspiration take advantage of water's large latent heat of vaporization. Neither process works very well when the air is fairly humid.
With regard to the question raised in the title, having both the water and the air cool does not necessarily violate conservation of energy. You are ignoring latent heat of evaporation, which is rather significant for water. Conservation of energy dictates that the change in enthalpy of the water plus the change in enthalpy of the air equal the change in enthalpy due to evaporation.
