Does evaporation violate the second law? Suppose I leave a glass of water on a table at STP. The water molecules at the surface of the water can be assumed to obey the Maxwell-Boltzmann distribution (at the least they obey a distribution of speeds comparable to the MB distribution). Only the most energetic molecules, near the surface of the liquid, have enough kinetic energy to overcome the attractive inter-molecular forces present in the liquid. As these highly energetic molecules escape from the liquid (i.e., evaporate), the average kinetic energy of the molecules in the liquid decreases. The temperature of the water is nothing but a measure of the average kinetic energy of the molecules in that water and hence the waters temperature decreases. This is effectively the mechanism behind sweating.
But now, if only the most energetic molecules were able to escape the inter-molecular forces and end up as water vapour, does this not mean that the water vapour above the surface now has a higher temperature than that of the water beneath it since the vapour is composed only of the most energetic molecules and the remaining liquid contains only the least energetic molecules? If this is the case, then surely it is an example of energy flowing from a cold sink to a hot sink? I realise this is impossible and so my thinking is definitely incorrect although I can't seem to build an intuition for why I am incorrect. All I can do is simply state the second law and tell my self I'm wrong. But that's no way to understand so if anyone can help me out on this issue it would be most appreciated!
 A: 
But now, if only the most energetic molecules were able to escape the inter-molecular forces and end up as water vapour, does this not mean that the water vapour above the surface now has a higher temperature than that of the water beneath it since the vapour is composed only of the most energetic molecules and the remaining liquid contains only the least energetic molecules? [emph added]

No. Some energy went into breaking the intermolecular bonds of the liquid to obtain the gaseous state. At equilibrium, the temperatures, pressures, and chemical potentials (meaning the partial pressures) of the liquid and vapor are identical. This is actually a great example of the Second Law at work: gradients in the intensive variables (temperature, pressure, chemical potential) are eliminated through shifting and exchange of the corresponding extensive variables (entropy, volume, matter).
Note that this equilibrium temperature (of the liquid and gas) will be lower than the original temperature of the liquid. The reason is that the molecules in the liquid were bonded to each other to some degree, meaning that they were in a low-energy state relative to a gas. Upon evaporation, energy was required to break these bonds, and—assuming a simple isolated system—this energy could come only from the thermal energy of the substance. So yes, you might that the liquid originally at 25°C is now a liquid–gas mixture at 24.8°C.
A: There is no violation of the second law here -- you're neglecting that a glass of water is an open system.
In other words, it interacts with the environment not only via "heat transfer" but also via mass transfer as well.
So, while heat is transferred from a "colder/liquid" phase to a "hotter/vapor" phase, the entropy lost by the liquid water is more than offset by the entropy gained by the environment from the contribution of higher-energy water molecules that got vaporized away into it (away from the liquid phase). So ultimately there is a net increase in the "total entropy" of universe, fully consistent with the Second Law.
