Evaporation Clarification

[cskib21]

I have been reading a bunch of different explanations of evaporation- and now I am confused. Hopefully someone can help clear this up in my head so that I can sleep tonight. I am interested in what is happening at the molecular level as well. Let's take a look at a body of water.

In a body of water, due to molecules bumping into each other, some molecules "acquire" enough energy to break through the surface of the water and become a vapor. The high energy molecule leaves the body of water with a lower average kinetic energy. Since this high energy molecule is transferred to the air, shouldn't the air go up in temperature?

Or does the air temp go down because more energy from the air is being transferred to the body of water than from the water to the air...?

(This is the first of a few conceptual problems I am having with this concept..)

 

[DrClaude]

In a body of water, due to molecules bumping into each other, some molecules "acquire" enough energy to break through the surface of the water and become a vapor. The high energy molecule leaves the body of water with a lower average kinetic energy. Since this high energy molecule is transferred to the air, shouldn't the air go up in temperature?

Yes. What you describe is part of evaporative cooling, which is the main factor leading to the cooling of your open cup of coffee.

If you start from water and air at the same temperature, the change in temperature of the air due to a single molecule of water will be negligible. Also, other processes will result in a rethermalization of the air and the water, such that both are basically always at the same temperature. The overall temperature will be lower than the initial temperature as some of the heat will have gone into changing the liquid water into water vapor.

 

[cskib21]

Yes. What you describe is part of evaporative cooling, which is the main factor leading to the cooling of your open cup of coffee.

If you start from water and air at the same temperature, the change in temperature of the air due to a single molecule of water will be negligible. Also, other processes will result in a rethermalization of the air and the water, such that both are basically always at the same temperature. The overall temperature will be lower than the initial temperature as some of the heat will have gone into changing the liquid water into water vapor.

So I also thought you needed a certain amount of external latent heat to turn liquid into vapor. But this latent heat can also come from the internal collisions of the molecules? Does the latent heat represent the energy needed to break through the barrier and become a vapor or for some sort of chemical transformation?

 

[DrClaude]

The latent heat needed is mostly needed to pull apart the molecules. There is significant intermolecular bonding energy (especially in water because of hydrogen bonding) in the liquid, as the molecules are closer together.

 

[cskib21]

The latent heat needed is mostly needed to pull apart the molecules. There is significant intermolecular bonding energy (especially in water because of hydrogen bonding) in the liquid, as the molecules are closer together.

Ok. So in terms of the evaporative-cooling process, heat is transferred from the air to the water, which gives the water the latent heat necessary for some of the molecules to break free and become vapor. Typically this results in a lower temperature air with higher relative humidity, and a slightly higher temperature body of water (net energy loss) with a bit less mass... Right?

 

[CWatters]

The energy need not come from the air but the surroundings in general. When water evaporates from the back of your hand your hand gets colder.

 

[cskib21]

The energy need not come from the air but the surroundings in general. When water evaporates from the back of your hand your hand gets colder.

But in terms of say, a swamp cooler, the majority of the energy is coming from the air- which is the goal so that air temp drops right?

In terms of the effects of fan/air speed on the evaporation rate, my understanding is that it can have two effects:
1) the air (which has already had water evaporate "into the mixture") that is moving away from the water allows air with less relative humidity to move over the body of water, which allows water to evaporate quicker. Increasing fan speed more or less "refreshes" the rate at which air with less relative humidity can move over the water.

2) The faster the air velocity, the less pressure over the water, which makes it easier for high energy molecules to become vapor..

Sound right?

 

[DrClaude]

2) The faster the air velocity, the less pressure over the water, which makes it easier for high energy molecules to become vapor..

The difference in pressure is completely negligible. The main obstacle to a water molecule escaping from the liquid phase is surface tension, not atmospheric pressure.

 

[cskib21]

The difference in pressure is completely negligible. The main obstacle to a water molecule escaping from the liquid phase is surface tension, not atmospheric pressure.

I am new to surface tension but I just read a description and I guess that makes sense. However, it would seem to me that evaporation would happen at a faster rate when there is more water in the air, because the water molecules in the body of water would have an attraction with the molecules in the air which would then help to "pull them" through the surface barrier into vapor.

This is counter-intuitive to blowing a fan over water to increase the rate of evaporation though. I guess the point of the fan then would be not to remove "high humidity air" but instead to just add heat to the water?

 

[DrClaude]

I am new to surface tension but I just read a description and I guess that makes sense. However, it would seem to me that evaporation would happen at a faster rate when there is more water in the air, because the water molecules in the body of water would have an attraction with the molecules in the air which would then help to "pull them" through the surface barrier into vapor.

The attraction is minuscule. You have to see it from a statistical point of view. Some molecules of water break out of the liquid phase, some water molecules in the vapor phase go back into the liquid. In the vapor phase, the water molecules move about randomly, bumping into other gas molecules, and from time to time one will have the right direction and speed to hit back the liquid surface and join the liquid phase again. The two rates are independent, and will depend on temperature and, for the rate of molecules going back into the liquid phase, the amount of water molecules in the vapor phase. At equilibrium, the two rates would be equal. If you constantly replace the layer of air above the liquid with air of lower humidity, you skew the balance by reducing the rate of reentry in the liquid. Hence, the evaporation will be faster.