How can evaporative cooling air conditioning work?

[genekuli]

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?

 

[Janus]

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.

 

[genekuli]

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?

 

[russ_watters]

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.

 

[genekuli]

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?

 

[russ_watters]

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.

 

[genekuli]

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?

 

[russ_watters]

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.

 

[genekuli]

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?

 

[russ_watters]

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?

 

[russ_watters]

[edited post]
That last example wasn't the best because you already get that side of the equation.

 

[JT Smith]

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.

 

[russ_watters]

...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.

 

[genekuli]

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?

 

[russ_watters]

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.

 

[genekuli]

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?

 

[russ_watters]

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.

 

[hmmm27]

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]

 

[genekuli]

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?

 

[genekuli]

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

 

[hmmm27]

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.

 

[russ_watters]

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

 

[russ_watters]

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.

 

[genekuli]

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"

 

[genekuli]

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

 

[russ_watters]

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"

Yup. That's the entire issue at hand. That's why I don't think the nuts and bolts of which molecule evaporates and why is important. That's just words. The numbers/math tells the story that you are really asking to hear.

could you please re-state that in terms of the marble set example for all the people that lost theirs

I don't understand what you are asking.

 

[genekuli]

Yup. That's the entire issue at hand. That's why I don't think the nuts and bolts of which molecule evaporates and why is important. That's just words. The numbers/math tells the story that you are really asking to hear.

I don't understand what you are asking.

this is the marble example i gave:
"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?"
could you give your explanation of the original question in terms of marbles, PLEASE

 

[256bits]

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?

'Hot' is not a physically describable state.'

To be able to leave the liquid water surface the bonds have already been broken, as described by Russ
For example, at 20 C, water has an enthalpy of 84 kJ/kg, but water vapor 2537 kJ/kg.

The water ends up with less heat content, and the air more.

Since temperature is a measure of the 'average' kinetic energy of molecules, the liquid water temperature will decrease with evaporation, by loosing the more energetic molecules. The escaped water vapour will have a temperature of the average temperature of the liquid water, not a higher temperature. They have to be able to overcome the surface bonds and this result is the same kinetic energy distribution as the liquid water. I think that was mentioned by hmmm27.

So you can lower the temperature of the water, and also the air, by evaporation.

 

[russ_watters]

this is the marble example i gave:
"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?"
could you give your explanation of the original question in terms of marbles, PLEASE

I don't think the explanation in terms of marbles makes any sense at all because they aren't bonded together. So they are missing the key piece of the puzzle.

 

[genekuli]

I don't think the explanation in terms of marbles makes any sense at all because they aren't bonded together. So they are missing the key piece of the puzzle.

please feel free to add some goo to the marbles in the example

 

[russ_watters]

please feel free to add some goo to the marbles in the example

I honestly don't see a way to make this analogy work.

 

[genekuli]

i need this simplified,
would this be correct:
the hot (vibrating) marbles that escape the set lose energy to breaking the goo bond on the way.
they then have low vibration upon mixing with the air marbles and so do not increase the air marbles' vibration.
is that it?

 

[Dale]

the explanation given in the video has the hottest water molecules going into the air

Hmm, that doesn’t make sense. The water is one temperature, so there aren’t hotter and colder water molecules. At a given temperature the molecules have a distribution of kinetic energy, but that doesn’t make some hotter than others.

 

[russ_watters]

Hmm, that doesn’t make sense. The water is one temperature, so there aren’t hotter and colder water molecules. At a given temperature the molecules have a distribution of kinetic energy, but that doesn’t make some hotter than others.

The argument being made is that if molecules that are higher on that bell curve leave, the bell curve shifts in the opposite direction. If the temperature is somewhat a function of the average kinetic energy, then the shifting of the curve down is a drop in temperature.

I'm pretty sure I learned something like this in high school, but had forgotten about it. It makes sense in a self-contained way, but to me it lacks quantification. I don't see how you would calculate the temperature drop of the water based on this description. And it seems incomplete; it implies the energy of the bond (enthalpy of vaporization) is irrelevant to the temperature change in the water as it evaporates.

 

[russ_watters]

i need this simplified,
would this be correct:
the hot (vibrating) marbles that escape the set lose energy to breaking the goo bond on the way.
they then have low vibration upon mixing with the air marbles and so do not increase the air marbles' vibration.
is that it?

That really makes no sense to me. It doesn't capture the chemical energy involved (vague reference to "goo bond" is not really meaningful). To me, the entire problem you had from the beginning was treating temperature as the only component of this problem, and ignoring the chemical bond energy. Here you have an analogy that does the same thing. Maybe someone else can craft a useful analogy from that, but I'm not seeing it.

 

[Dale]

The argument being made is that if molecules that are higher on that bell curve leave, the bell curve shifts in the opposite direction. If the temperature is somewhat a function of the average kinetic energy, then the shifting of the curve down is a drop in temperature.

The argument is fine, but the terminology is the problem. At a single temperature there is a distribution of energy. To call the molecules at the high end of the curve “hot” means that at a single temperature there is a distribution of temperatures!

This is exactly the problem the OP is having. He has been falsely told that there are hot molecules and so his conclusion that the air should get hotter is reasonable given that misinformation.

But the molecules are all at a single temperature, and molecules at a single temperature have a distribution of energies. The high energy molecules can leave, but that does not increase temperature because they are not hotter. I think it is exactly this that is causing the OP’s confusion.

 

[jartsa]

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?

Let's say a water bucket has been standing in a room for a long time with a lid on.

Now we open the lid.

The first small puff of steam coming out is at the same temperature as the water and the room.

The second small puff of steam coming out is at the same temperature as the water, which has cooled very slightly, because of a few of the faster molecules leaving.

The faster molecules become average molecules in the process of leaving the liquid.

 

[genekuli]

The argument is fine, but the terminology is the problem. At a single temperature there is a distribution of energy. To call the molecules at the high end of the curve “hot” means that at a single temperature there is a distribution of temperatures!

This is exactly the problem the OP is having. He has been falsely told that there are hot molecules and so his conclusion that the air should get hotter is reasonable given that misinformation.

But the molecules are all at a single temperature, and molecules at a single temperature have a distribution of energies. The high energy molecules can leave, but that does not increase temperature because they are not hotter. I think it is exactly this that is causing the OP’s confusion.

right, so then i don't understand when you say the escaping molecules have more energy but are the same temperature, part of your explanation? if they have more energy, that energy is heat energy? no? so they are hotter, no? or what energy would it be if not heat?

 

[russ_watters]

The argument is fine, but the terminology is the problem. At a single temperature there is a distribution of energy. To call the molecules at the high end of the curve “hot” means that at a single temperature there is a distribution of temperatures!

I don't think that's the OP's confusion -- he said he understands why the water gets cooler, just not why the air gets cooler.

And to be honest the whole "one molecule can't have a temperature" thing has always confused me too. It begs the question of how many it takes before we can call it an "average energy". But we can easily avoid this by discussing bundles of molecules instead of individual molecules. Then, I see nothing wrong with dividing a distribution into two distributions. If you have 1 kg of water at 20 C and 1 kg of water at 30 C and you mix them together to get 2 kg of water at 25 C, the reverse should be possible -- if the distribution is wide enough and you can select the molecules.

I just don't think that's what's happening.

This is exactly the problem the OP is having. He has been falsely told that there are hot molecules and so his conclusion that the air should get hotter is reasonable given that misinformation.

But the molecules are all at a single temperature, and molecules at a single temperature have a distribution of energies. The high energy molecules can leave, but that does not increase temperature because they are not hotter. I think it is exactly this that is causing the OP’s confusion.

"hot molecules" or "higher energy molecules"? Because the molecules that leave are higher energy and do increase the average energy of the air. Just not in the way he thinks (via latent heat, not sensible heat).

Similar to above, if I take 1 kg of air at 20C and 1 kg of water vapor at 30C I get a mixture at about 23C. That's what the OP thinks is happening, but it isn't because the water vapor temperature isn't above the air temperature.

 

[russ_watters]

right, so then i don't understand when you say the escaping molecules have more energy but are the same temperature, part of your explanation? if they have more energy, that energy is heat energy? no? so they are hotter, no? or what energy would it be if not heat?

I have tried to tell you several times that there is more than one kind of heat/energy -- and even, there's more than one kind of temperature. It's what you've been missing since the beginning, I've said it several times, but it isn't getting through.

No, having more energy doesn't necessarily mean being at a higher temperature.

At this point, I think you simply don't want to believe it.

But if it makes you feel better about it, I don't think I mentioned the name of that other temperature before: it's called Dew Point Temperature. And it does rise in this situation.

So, you can increase the enthalpy of the vapor mixture by increasing its sensible/dry bulb temperature or by increasing its dew point temperature.

So, does that help?

 

[genekuli]

I have tried to tell you several times that there is more than one kind of heat/energy -- and even, there's more than one kind of temperature. It's what you've been missing since the beginning, I've said it several times, but it isn't getting through.

No, having more energy doesn't necessarily mean being at a higher temperature.

At this point, I think you simply don't want to believe it.

But if it makes you feel better about it, I don't think I mentioned the name of that other temperature before: it's called Dew Point Temperature. And it does rise in this situation.

So, you can increase the enthalpy of the vapor mixture by increasing its sensible/dry bulb temperature or by increasing its dew point temperature.

So, does that help?

you said: "having more energy doesn't necessarily mean being at a higher temperature." so please tell me specifically what this energy is called if it is not called heat energy?

 

[256bits]

right, so then i don't understand when you say the escaping molecules have more energy but are the same temperature, part of your explanation? if they have more energy, that energy is heat energy? no? so they are hotter, no? or what energy would it be if not heat?

You pick a reference point to describe the energy content of a mass of molecules.
For water, we pick 0.01 C, where the enthapy is designated as being 0.

To gain any other HIGHER temperature, or to vapourize, one has to add energy, so that is where the talk about having more energy come from.
https://www.engineeringtoolbox.com/water-properties-d_1508.html

You will experience a more severe burn from steam than water, even though they may be at the same temperature, due to the higher energy content at the vapourized state. The steam will condense on your skin releasing heat 'of condensation' to the liquid state, while the liquid water will only have conduction available to transfer heat.

 

[genekuli]

You pick a reference point to describe the energy content of a mass of molecules.
For water, we pick 0.01 C, where the enthapy is designated as being 0.

To gain any other HIGHER temperature, or to vapourize, one has to add energy, so that is where the talk about having more energy come from.
https://www.engineeringtoolbox.com/water-properties-d_1508.html

and that energy would be called heat energy, no?

 

[256bits]

and that energy would be called heat energy, no?

I edited , and added more .

 

[russ_watters]

you said: "having more energy doesn't necessarily mean being at a higher temperature." so please tell me specifically what this energy is called if it is not called heat energy?

It is "heat energy". That's still not specific enough though. Specifically, it is latent heat energy. Because, again, there is more than one kind.

Your mistake is still thinking heat = temperature. It does not.

Every post, I'm saying the same thing over and over in different ways.

Here's another description of the process:

Evaporative cooling takes place along lines of constant wet bulb temperature or enthalpy. This is because there is no change in the amount of energy in the air. The energy is merely converted from sensible energy to latent energy. The moisture content of the air increases as the water is evaporated which results in an increase in relative humidity along a line of constant wet bulb temperature.

http://www.coolbreeze.co.za/psyevap.htm

 

[genekuli]

right, so the energy of the heat of the water, breaks the H bond, and that kinetic heat energy (sensible energy) changes to potential energy (latent energy) and therefore does not transfer the energy to the air, as the E stays in the H bond potential. is THIS right?

 

[jbriggs444]

i need this simplified,
would this be correct:
the hot (vibrating) marbles that escape the set lose energy to breaking the goo bond on the way.
they then have low vibration upon mixing with the air marbles and so do not increase the air marbles' vibration.
is that it?

That matches the way I think of it. The molecules that leave the liquid phase expend energy breaking free of the liquid and, so, no longer have the high kinetic energy they started with.

 

[sophiecentaur]

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?

Your second paragraph is more like the correct message.
Firstly, it's not a good idea to talk about 'hot' and 'cold' molecules; they are fast or slow. Temperature is a bulk quantity.
The description of the system seems to be failing to stress the total dependence of the effect on change of state. The escaping molecules of water are not going as fast as the temperature of the liquid would imply; they are only just managing to escape the Potential Energy of the surface, having lost most of their Kinetic Energy. ````Only by blowing them away with a fan can you manage to remove that energy permanently from the liquid (cooling it down).

This can't go on forever in a closed room because, eventually, the air can become saturated with water vapour. Neither does the system work at all well when outside air is used but when it's already very humid. You end up with cold wet air in the room which actually feels disgusting. In hot dry locations, the system is fine and cheap to buy but not in UK.

 

[Dale]

right, so then i don't understand when you say the escaping molecules have more energy but are the same temperature, part of your explanation? if they have more energy, that energy is heat energy? no? so they are hotter, no? or what energy would it be if not heat?

This indicates a misunderstanding about thermal energy. Thermal energy is not something that even makes sense for individual molecules. Thermal energy is only something that a large ensemble of molecules have. Thermal energy is energy distributed randomly in internal degrees of freedom.

What these individual molecules have is kinetic energy. Inside a large group of molecules, all at a single temperature, there are molecules with a distribution of kinetic energy. When you are looking at the details at the scale of the individual degrees of freedom then what is distributed is no longer thermal energy, it is kinetic energy or whatever other degree of freedom you are looking at.

They are not hotter because they are in thermal equilibrium with the rest of the water. They have more kinetic energy. Kinetic energy and thermal energy are not the same thing.

 

[Dale]

he said he understands why the water gets cooler, just not why the air gets cooler

Right, and that misunderstanding is due to the fact that he incorrectly believes the air is getting hot molecules.

Then, I see nothing wrong with dividing a distribution into two distributions.

Yes, but it is a distribution of kinetic energies at a fixed temperature. It is not a distribution of temperature. The faster molecules are not hotter, they are in thermal equilibrium with the rest.

"hot molecules" or "higher energy molecules"? Because the molecules that leave are higher energy and do increase the average energy of the air.

Yes, energy and temperature are not the same thing. They are high energy molecules, they are not hot molecules.