evaporationthermal-radiationwater
While reading the introduction to Feynman's lectures, it's mentioned how a glass of water cools down through evaporation, when some molecules get a bit extra energy and break free. If it's not a closed system, energy will be gradually taken away from the cup, hence blowing at the soup helps move those molecules away so that they don't reenter the surface.
But I thought that all bodies also radiate heat? Does a cup of water also emit low frequency radiation, or is my understanding incorrect?
There are several ways to reduce the rate of evaporation of water. However, first it is important to understand the factors influencing the rate of evaporation:
1) Amount of humidity in the air 2) Speed of air flowing past the water droplets 3) Intramolecular bonding strength 4) Pressure
There are more such as surface area, density, etc. but since we are considering the same drop of same substance these can be ignored. (temperature is there too, obviously)
So what we can do: Make the air around the drops more humid. Put the drops in a closed environment, to reduce the speed of air flowing past it. For example, water droplets inside a room will evaporate slower than water droplets placed outside. There certainly are some additives to help increase the intramolecular bonding strength of a substance. Also, the substance the water droplets are placed on also makes a difference on the rate of evaporation. Decreasing your altitude will also decrease the rate of evaporation.
Very interesting question.
As you wrote yourself in your Edit it is hard to describe water via the ideal gas model.
You have to introduce at least two important improvements of your ideal gas:
Because of $V_d$ a certain alignment of water molecules is favoured. The rotation of the molecules which increases with Temperature $T$ "works against this" alignment. This state averaged $V_d$ is called Keesom-Interaction. Let's call it $V_k$ and note that $V_k \propto \frac{1}{T}$.
Up to this point we introduced a model that predicts qualitatively stronger hydrogen bonding and more regular alignment for deeper temperatures.
Now let's take a step back from liquid water and look at ice. In this lattice we practically eliminated the rotational degrees of freedom of the individual molecules (and gained 3 modes of vibration each). This means if we want to minimize $V_d$ we bring the water molecules into a certain structure and they stay this way.
It is important to realize that (in difference to e.g. salt lattices) minimizing the distance between atoms (and in the macroscopic scale the density) alone does not minimize $V_d$. Taking the angle dependencies of $V_d$ into account can increase the Volume as happening in the case of water/ice.
The cold but liquid water between 0° and 4° in the upper layers of your lake is not cold enough to form a lattice but has more locally ordered structures than the water at 4° or above. Besides the Keesom - Interactions $V_k$ are stronger
It is important to realize that (in difference to e.g. salt lattices) minimizing the distance between atoms (and in the macroscopic scale the density) alone does not minimize $V_d$. Taking the angle dependencies of $V_d$ into account can increase the Volume as happening in the case of water/ice.
The cold but liquid water between 0° and 4° in the upper layers of your lake is not cold enough to form a lattice but has more locally ordered structures than the water with 4° or higher, because the Keesom - Interactions $V_k$ are stronger and more important. This local ordering because of $V_k$ also explains the lower density compared to the water at 4°. (Taking angle dependency into account again).
Now after all this introduction what happens with your water molecule that comes with its high velocity to an upper layer. As you wrote it can easily overcome the gravitational field. Somehow it looses its translational energy and other molecules now move or/and rotate faster. We also can reverse this process and look at a slow molecule from the 0° layer that goes down and becomes accelerated in the 4° layer. Macroscopically this heats up the upper layer and cools down the lower layer until equilibrium is reached.
The famous lake in the winter is not a closed system. You constantly heat from the floor and cool it from the air. This is comparable to a "normal liquid" where you constantly cool from the floor and heat from the air. Also in this case you would get two layers. If you heat and cool with the same rate it is possible to get a steady state that does not change, although it is not in equilibrium.
The important question is, what happens if you stop heating or cooling. Here your questions becomes important and leads to the prediction that water will equilibrate quite fast. To say it in another way, we expect water to have a high thermal conductivity. Now if you look at this table.
You get some nice data for thermal conductivity. I made a plot out of it which makes this point pretty clear if you keep in mind that the red bar shows the mean and the leftmost point represents water:
Perhaps it is also nice to note that the substance with $\lambda = 0.5\rm \, W\, m^{-1}\, K^{-1}$ is ammonia. This compound also has significant dipole dipole interaction.
Best Answer
Yes, all matter above absolute zero emits radiation. To quote wiki:
This continuous release of energy would eventually cool the source to a lower and lower temperature except your glass of water is in contact with a heat reservoir (the room) which compensates for the energy loss.