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?
[Physics] In evaporative cooling does the water and the air both cool down? Wouldn’t that violate energy conservation
thermodynamics
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these high speed particles hit us and hence evaporate our sweat
Indeed it's evaporation that cools us down, but the particles that hit us are not "high speed" particles in any sense (the warming effect of the fan is very small) and evaporation happens not because of their speed.
Evaporation always happens when there is a water surface with not enough water vapour next to it (nothing to do with non-water portions of air, only partial water vapour pressure matters).
To understand how a normal fan cools you down, consider evaporation rate with and without a fan. Without a fan you'll have a lot of water vapour next to your skin because of earlier evaporation, which slows evaporation down. A fan creates air flow that blows a lot of that water vapour away from your skin, therefore making evaporation go faster.
I understand some of the heat that the fan throws out is absorbed by the water which in turn heats up and cools down the air.
By now it should be clear that that is not at all how it works. By injecting water drops into the airflow you create a lot of opportunity for evaporation by creating large water surface with (hopefully) not enough water vapour around it. Thus evaporation happens and air cools down.
The precise meaning of "enough water vapour" is given by a value called equilibrium vapour pressure. (see Vapour pressure of water) When water vapour pressure reaches that level we say that Relative humidity reached 100% and evaporation stops.
With this it's easy to understand the limitations of this DIY AC:
- It won't work well if relative humidity in the room is high;
- By evaporation it raises air humidity so it will stop working after a while unless you have a window open to let humid air out and bring dry air in.
Firstly, it would be better to use actual, accurately measured numbers than the human 'experience': the human body is a poor thermometer and the mind plays tricks on us.
But that hot water droplets lose heat and thus cool down in cooler air is an established fact and a consequence of the laws of thermodynamics.
Regarding your three first bullet points, despite some limitations you point out, those do not mean a hot droplet of water doesn't cool in air: it does and partly in accordance Newton's cooling law.
As regards work done by the droplet (overcoming the viscous drag), if anything that would lead to heat generation, not cooling (but the effect is truly minuscule).
Kinetic or potential energy of the droplet have no effect on the droplet's temperature. Temperature is simply a measure of the average speed of the molecules of the water and that is not affected by these energies. Spinning water in an ultra-centrifuge does not make its temperature rise, for instance.
You have however overlooked one major cause of heat loss: evaporation. Your shower 'steams up' because hot water evaporates and that costs energy, known as the Enthalpy of vaporisation.
Millions of tons of water are cooled this way everyday in power plants world wide: the cooling towers drop hottish water from the top of the towers and evaporative heat cools down the water (the evaporated water escapes as steam clouds).
If your shower has been in operation for a long time and the bathroom's temperature is equal to the shower head's water temperature and the air is saturated with water vapour, then no cooling of the shower water would take place.
Best Answer
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.