How can evaporative cooling air conditioning work?

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If evaporative cooling (Such as sweating) is due to the escape of the hottest molecules into the air thereby lowering the total average temperature of the water then that means that that hot water molecule has gone into the air and has made the air hotter, so then how does evaporative cooling air conditioning work when these hot water molecules that have escaped the water now make the air hotter because the water became cooler?
 

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  • #2
Janus
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Sweating and evaporative cooling works because of "latent heat", When water goes from liquid to vapor, it takes energy to make the transition between the states, but it doesn't rise in temp when it does so. All the energy supplied is used in making the transition from liquid to gas. That energy comes from the environment. I the case of sweat, it comes from you giving up body heat, for example.
 
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  • #3
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Sweating and evaporative cooling works because of "latent heat", When water goes from liquid to vapor, it takes energy to make the transition between the states, but it doesn't rise in temp when it does so. All the energy supplied is used in making the transition from liquid to gas. That energy comes from the environment. I the case of sweat, it comes from you giving up body heat, for example.
so then is this physicist's explanation not correct?
 
  • #4
russ_watters
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so then is this physicist's explanation not correct?
No, it's fine. Unless you have the wrong time stamp selected, it doesn't say that the air gets hotter.

What you are missing here is just that "hotter" is not a single parameter. There are two kinds of energy/temperature in this context, and if one goes up, that does not mean the other does too -- and often they act opposite each other. In this case, the total energy (enthalpy) goes up, but the sensible temperature goes down while the latent temperature/enthalpy (dew point) goes up.
 
  • #5
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No, it's fine. Unless you have the wrong time stamp selected, it doesn't say that the air gets hotter.

What you are missing here is just that "hotter" is not a single parameter. There are two kinds of energy/temperature in this context, and if one goes up, that does not mean the other does too -- and often they act opposite each other. In this case, the total energy (enthalpy) goes up, but the sensible temperature goes down while the latent temperature/enthalpy (dew point) goes up.
you said: "it [the video] doesn't say that the air gets hotter.". but the explanation given in the video has the hottest water molecules going into the air; so surely that must make the air warmer, right? I mean it made the skin cooler by leaving and now is part of the air, making the air warmer, right?
 
  • #6
russ_watters
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you said: "it [the video] doesn't say that the air gets hotter.". but the explanation given in the video has the hottest water molecules going into the air; so surely that must make the air warmer, right? I mean it made the skin cooler by leaving and now is part of the air, making the air warmer, right?
No. Re-read what I(we) said about there being more than one type of temperature/heat.

You can also read about it here: https://climate.ncsu.edu/edu/Heat#:~:text=Latent and sensible heat are,with no change in phase.
 
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  • #7
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thank you, yea i did reread, but i still am missing something, could you explain that part more in this specific scenario of how it can cool one body (skin) but not make the other (air) warmer?
 
  • #8
russ_watters
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thank you, yea i did reread, but i still am missing something, could you explain that part more in this specific scenario of how it can cool one body (skin) but not make the other (air) warmer?
Turning water into water vapor consumes energy. This removes energy from the air, making it cooler (in temperature).

You're probably still thinking about this in terms of sensible temperature. A 10C mass of water evaporating into a 5C mass of air does not raise the temperature of the air because the water in vapor form isn't at 10C anymore. It consumes energy in the evaporation process, making it cooler in temperature.
 
  • #9
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right, so the evaporation cools the body of water and the air because the change of state consumes the energy. but in the video above it is more explained as the hottest water molecules leaving the water/sweat and therefore causing the total temp of the water/sweat to decrease. so how does that fit in with the first explanation?
 
  • #10
russ_watters
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right, so the evaporation cools the body of water and the air because the change of state consumes the energy. but in the video above it is more explained as the hottest water molecules leaving the water/sweat and therefore causing the total temp of the water/sweat to decrease. so how does that fit in with the first explanation?
How doesn't it? What contradiction do you think you see? The molecules that evaporate have to be "hot" because it takes energy to evaporate. This is like asking why you need to turn on your stove to boil water. There's no contradiction here.

You're repeating yourself here, which I think means that you think the words you are saying mean something different from what they actually mean. Can you try explaining in more detail where your understanding is/the contradiction you think you see?
 
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  • #11
russ_watters
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[edited post]
That last example wasn't the best because you already get that side of the equation.
 
  • #12
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right, so the evaporation cools the body of water and the air because the change of state consumes the energy. but in the video above it is more explained as the hottest water molecules leaving the water/sweat and therefore causing the total temp of the water/sweat to decrease. so how does that fit in with the first explanation?

It's not a very good video. While the guy doesn't say anything wrong he glosses over the fact that escaping the attraction of the other water molecules takes energy away from (slows down) the escaping molecule. It's such a basic element of evaporative cooling that it's a shame he never got around to saying it.
 
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  • #13
russ_watters
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....escaping the attraction of the other water molecules takes energy away from (slows down) the escaping molecule.
That's a good point/good way to say it; The act of evaporating lowers the temperature of the escaping water vapor.
 
  • #14
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the video said the water molecule "is" hot, that is why it escapes the H bonds. when it leaves the sweat, it decreases the total heat in the sweat.
it didn't say: The act of evaporating lowers the temperature of the escaping water vapor.
those are two different explanations/phenomenons right?
 
  • #15
russ_watters
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the video said the water molecule "is" hot, that is why it escapes the H bonds. when it leaves the sweat, it decreases the total heat in the sweat.
it didn't say: The act of evaporating lowers the temperature of the escaping water vapor.
those are two different explanations/phenomenons right?
Kind of, or maybe it can be said that it's a "what" and a "why". The fact that it's the hottest molecules that escape, to me, isn't really all that important here. It's not involved in the math of it at all. In the math you [have to] assume the temperatures are uniform.
 
  • #16
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so why does that not work in reverse, that is, if one pours water into water and creates H bonds, would that not then create heat? just as it looses heat in the case of breaking the H bonds?
 
  • #17
russ_watters
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so why does that not work in reverse, that is, if one pours water into water and creates H bonds, would that not then create heat? just as it looses heat in the case of breaking the H bonds?
Pouring water into water doesn't create many new H bonds. Condensing water from the air does, though.
 
  • #18
hmmm27
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Hmm... I think I could use a refresher, as well.

It's easy enough to see how the water cools down : water molecules that happen to be on the surface, if they're hot enough, simply fly off.

A mister on a fan is the same, massive surface area of the water, instant saturation, temperature drop, all'round, plus the mist lands on your skin, saving you the bother of having to sweat.

But, long term ? Those molecules are packed with heat : won't they simply heat up the air molecules, themselves losing very little temperature ?

[edit: or is it something to do with the posts I missed while typing this one out : specifically what Russ just said]
 
  • #19
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but so basically the simple video explanation of the decrease in temperature resulting from the hotter molecules leaving is not as simple as a simple example of a set of marbles of different temperatures decreasing in total temperature due to the hotter ones leaving the set. as is implied by the video.
it is more like that the hotter marbles cool as they leave the set. correct?
 
  • #20
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Hmm... I think I could use a refresher, as well.

It's easy enough to see how the water cools down : water molecules that happen to be on the surface, if they're hot enough, simply fly off.

A mister on a fan is the same, massive surface area of the water, instant saturation, temperature drop, all'round, plus the mist lands on your skin, saving you the bother of having to sweat.

But, long term ? Those molecules are packed with heat : won't they simply heat up the air molecules, themselves losing very little temperature ?

[edit: or is it something to do with the posts I missed while typing this one out : specifically what Russ just said]
yea but apparently not
 
  • #21
hmmm27
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yea but apparently not
Well, that makes sense : the individual water molecules flying around are just the same as the air molecules, getting hotter or colder (changing energy levels) when they bounce off another molecule... until one water molecule meets up with another water molecule and they join exothermically.
 
  • #22
russ_watters
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It's easy enough to see how the water cools down : water molecules that happen to be on the surface, if they're hot enough, simply fly off.
Well, if we really do want to get into the nuts and bolts of this, I'm not sure that's actually true (that that effect is what cools the water). It ignores the energy lost to the broken bond. Are we to assume all of the energy of breaking the bond is taken from the molecule that leaves? Doesn't make sense to me. Selection of which molecule is going to escape and the value of the cooling done aren't necessarily the same thing.

But I'd need someone with more of a chemistry background to explain that part...
But, long term ? Those molecules are packed with heat : won't they simply heat up the air molecules, themselves losing very little temperature ?
No.

And yes, this is what I was saying before; heat/energy is not the same thing as temperature. You can see this in a steam table or psychrometric chart. For example, at 20 C, water has an enthalpy of 84 kJ/kg, but water vapor 2537 kJ/kg. Same temperature, much different amount of energy. Yep, the steam is "packed with heat", but it won't raise the temperature of air at that is also at 20C.
https://www.engineeringtoolbox.com/saturated-steam-properties-d_101.html
 
  • #23
russ_watters
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but so basically the simple video explanation of the decrease in temperature resulting from the hotter molecules leaving is not as simple as a simple example of a set of marbles of different temperatures decreasing in total temperature due to the hotter ones leaving the set. as is implied by the video.
it is more like that the hotter marbles cool as they leave the set. correct?
Whether or not the explanation for the [liquid] water is true (and I'm not sure it actually is), the energy of the bond is the key to this entire issue.

Note: for evaporative cooling devices in real life, the entire system is at the same final temperature, so the issue of "where" the cooling is done never even comes up. You just account for the energy for the system in equilibrium.
 
  • #24
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ok, so that blows my mind; "20 C, water has an enthalpy of 84 kJ/kg, but water vapor 2537 kJ/kg. Same temperature, much different amount of energy"
 
  • #25
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could you please re-state that in terms of the marble set example for all the people that lost theirs.
that is the original question and answer, in marbles metaphor
 

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