as I got out of the shower

Originally posted by sonhouse
IR might work but remember, they are specific to a particular material.

Not an issue as we want to measure the temperature of the can before and after shaking and see if it really cools. (same material)

We know this for a fact because of our use of IR thermometers, those hand held units that looks like a toy gun.
Do you have one now perhaps? If so, we basically want the following:
1. Measure the temperature of a spray can that is standing on its own and has been there for a while.
2. Shake it. (preferably without touching it too much with bare hands, such as lifting it up by the plastic lid). But keep in mind that hands can only warm it, so if cooling is observed, touching it is not responsible).
3. Measure the temperature again. Preferably in exactly the same way the first reading was obtained. See if the temperature actually went down.

Originally posted by twhitehead
Not an issue as we want to measure the temperature of the can before and after shaking and see if it really cools. (same material)

[b]We know this for a fact because of our use of IR thermometers, those hand held units that looks like a toy gun.
Do you have one now perhaps? If so, we basically want the following:
1. Measure the temperature of a s ...[text shortened]... exactly the same way the first reading was obtained. See if the temperature actually went down.[/b]

Here at my company, we have two, one good one that has the materials offset chart of temperature changes and a cheapo from somewhere, same thing but no offsets. I have one of those at home🙂 We also have spray cans of various stuff, like paint and WD 40 and such, could be I could do that experiment here.

Originally posted by twhitehead
Actually heat loss is directly proportional to temperature difference. Our sense of hot or cold actually is a measure of temperature, but it measures the temperature of the substance directly contacting our skin.

[b]So I think the second explanation is much more plausible.
It does sound plausible, but doing my own testing, I am not convinced. If I ...[text shortened]... n the bonds are broken. This is no different from a chemical reaction in which bonds are broken.[/b]

It's better to talk about internal energy than heat energy. The potential energy of the individual atoms contributes to the internal energy of the and so you can't really leave it out. The system has work done on it in the shaking and is heated by the shaker's hand, so it's internal energy can only have risen. However, the system has two components the liquid and the vapour. The liquid has lost a molecule, since it's the higher energy ones that can escape the average energy per molecule in the liquid drops. In escaping the gas the molecule's climbed out of a potential well and so has lost kinetic energy but has gained potential energy. The vapour and the liquid are in thermal contact and I don't believe can be at vastly different temperatures. This leads me to think that it is not an evaporation effect, consider the following:

Suppose the effect is caused by evaporation. Since we started off with the can in thermal equilibrium with the room and the can is apparently colder now heat should be flowing into it from the room. The heat flowing in should cause more of the liquid to evaporate and the whole thing should go into thermal runaway. So the evaporation hypothesis seems to lead to a ludicrous conclusion, liquid phases would be unstable against evaporation. This means the can must be getting hotter or staying the same temperature and not getting colder. But it's still colder than a hand and I think that that is why it feels cold, I'm believing googlefudge's heat transfer/physiology explanation at the moment.

Originally posted by DeepThought
It's better to talk about internal energy than heat energy. The potential energy of the individual atoms contributes to the internal energy of the and so you can't really leave it out. The system has work done on it in the shaking and is heated by the shaker's hand, so it's internal energy can only have risen. However, the system has two components th ...[text shortened]... y it feels cold, I'm believing googlefudge's heat transfer/physiology explanation at the moment.

Ok, I got out the cheapo IR gun, and did notice a temperature difference between the bare metal of a couple of pressurized cans V the painted part of the cans, about one degree F. I had it set to F. Anyway, on to the experiment. Untouched, pointing at the metal part of the can, temperature indicated 74 degrees F. Shake can for about ten seconds, using top of can to limit contact with human skin, put can down, re-measure, temperature now 73 degrees F.

Did that with another pressurized can, different brand, before shake temperature, 73.5 degrees F. After shaking, 72.5 degrees F.

So the can does get cooler. Why, up to youse guys to figure out🙂

In both tests, I aimed the laser for the metal part of the can so the material was the same in both tests.

Ok, actually thinking about what is going on, how 'bout this:

The outside world in my test was about 75 degrees or so. I measured 74 on the metal surface of the can and a degree less after shaking since the change in temperature is so small, the amount of energy exchanged in each shake would be on the order of a BTU or less per shake. (One BTU= one degree F change in temperature of one pint of water, no matter if it takes one microsecond or one hour) Which amounts to about 1/3 joule. (1/3 of a watt if the change occurred in one second)

Could it be something as simple as the inside being somewhat insulating from the outside world and therefore somewhat cooler, would lead to a temperature gradient inside to outside. So in this case, shaking would send some of the cooler stuff on the inside to the outside of the can.

My prediction: If this is true, you keep doing that and say, after ten such shakings and quiet moments, the temperature change would gradually decrease till the inside and outside were the same temperature and therefore no amount of shaking would change the temperature of the outside of the can.

I don't have time to do that experiment however. I would think that would take an hour at least, maybe 2 or 3 to get to a point of diminishing returns in that case.

Another thought, one I that was already pointed out: If this was true, it wouldn't matter if the can was pressurized or not, that being a red herring. It could be the case that any fluid inside a can, pressurized or not, would experience the same internal to external temperature gradient and therefore respond exactly the same as a pressurized can.

So a soda can filled with water and the top lid covered, in that case, would show the same effect if it was undisturbed in a room long enough for some kind of thermal equilibrium to have taken place. If left alone say for one day, there would still be a small temperature gradient, but if left alone for a year, the temperature gradient would be all but zero degrees.

Originally posted by sonhouse
Ok, I got out the cheapo IR gun, and did notice a temperature difference between the bare metal of a couple of pressurized cans V the painted part of the cans, about one degree F. I had it set to F. Anyway, on to the experiment. Untouched, pointing at the metal part of the can, temperature indicated 74 degrees F. Shake can for about ten seconds, using top o ...[text shortened]... sts, I aimed the laser for the metal part of the can so the material was the same in both tests.

What's the precision of the IR gun and was the point on the tin in contact with the liquid or the gas in the can?

I'm wondering about my thermal runaway objection. What decides whether evaporation is possible is whether the vapour is saturated and that depends on the temperature alone. That provides a break to stop the entire liquid content turning into vapour (in a closed container, in the open air it's a different story), but still leaves evaporation driven explanations not working. Since we're talking about a temperature drop the saturated vapour pressure can only have fallen. So unless the tin has an unsaturated vapour above it and by shaking the tin we saturate it I don't see how there can be an evaporation driven effect.

Originally posted by DeepThought
This leads me to think that it is not an evaporation effect, consider the following:
Suppose the effect is caused by evaporation. Since we started off with the can in thermal equilibrium with the room and the can is apparently colder now heat should be flowing into it from the room. The heat flowing in should cause more of the liquid to evaporate and the whole thing should go into thermal runaway.

No, that doesn't make sense. You are ignoring pressure and also the likelihood that the can started with fully saturated gas in equilibrium.

So the evaporation hypothesis seems to lead to a ludicrous conclusion, liquid phases would be unstable against evaporation.
Are you denying the well known scientific fact that evaporation causes cooling? It is not clear.

I also have to point out that without a sealed can, liquid phases are unstable. That is why water evaporates. You don't get a runaway effect though, but I am not sure where you got that from.

There is no doubt whatsoever that evaporation causes cooling. There is also no doubt that shaking the can should cause evaporation. What is in question is whether or not the cooling effect from evaporation is significant and whether that is what we are feeling.

Originally posted by twhitehead
No, that doesn't make sense. You are ignoring pressure and also the likelihood that the can started with fully saturated gas in equilibrium.

[b]So the evaporation hypothesis seems to lead to a ludicrous conclusion, liquid phases would be unstable against evaporation.
Are you denying the well known scientific fact that evaporation causes cooling? I ...[text shortened]... er evaporates. You don't get a runaway effect though, but I am not sure where you got that from.[/b]

The can is closed, external air pressure is irrelevant. The pressure in the tin depends on temperature only. Typically air isn't saturated with water vapour so water will evaporate. But in a closed system like that at a given temperature the liquid will evaporate until the saturated vapour pressure is reached. Once it's reached the liquid and vapour phases are in equillibrium. If you lower the temperature then some of the vapour will condense until a new equillibrium is reached and the pressure in the tin will be lower. If you raise it then more liquid will evaporate and the pressure in the tin will be higher. But shaking it will not drive further evaporation.

The only thing I can think of that would allow evaporation with a small temperature drop in this system is the possibility of settling. If the vapour is saturated near the liquid vapour boundary but unsaturated near the top of the tin then shaking the jar would cause some of the liquid to evaporate as it's brought into contact with the unsaturated vapour. This is plausible for things like propane and butane which have heavy molecular masses than nitrogen and oxygen.

In that case one could try putting a tin on it's side. That way the column of vapour above it is much shorter. Once it's had a chance to settle and get into equillibrium shake it and see if the effect goes away.

Originally posted by DeepThought
The can is closed, external air pressure is irrelevant.

Agreed.

The pressure in the tin depends on temperature only.
Only under stable conditions. Shaking the can should result in evaporation and an increase in internal pressure.

But shaking it will not drive further evaporation.
Evidence please.

An additional empirical test suggestion: try shaking nearly empty spray cans and nearly full ones and see whether that makes any difference to the perceived temperature on the outside of the can. I'm guessing that a nearly empty can will undergo less perceived temp. change. I'll get two cans of deo out of the pantry and use only one of them all this week and report back. (And no sarcastic remarks, please, about me smelling like a pansy this week, ok?)

Next question: if it were a matter of entropy, that is changing from an ordered state to less ordered state, and not evaporation, then wouldn't the same temp. change be perceived by shaking a bottle of oil and vinegar? I'm assuming the oil and vinegar have separated before you start shaking the container (as the propellant and the deo have separated in the deo can before you start shaking it).

Originally posted by moonbus
Next question: if it were a matter of entropy, that is changing from an ordered state to less ordered state, and not evaporation,

It isn't an either / or. Evaporation, as with all chemical reactions, also involves a change in entropy. I probably should not have bothered mentioning entropy from the beginning and simply stuck with the fact that bonds are being broken and thus potential energy is being created, which results in cooling. This would be no different if we instead had a known endothermic chemical reaction going on.
You will find though that entropy is involved:
https://en.wikipedia.org/wiki/Endothermic_process

Originally posted by twhitehead
It isn't an either / or. Evaporation, as with all chemical reactions, also involves a change in entropy. I probably should not have bothered mentioning entropy from the beginning and simply stuck with the fact that bonds are being broken and thus potential energy is being created, which results in cooling. This would be no different if we instead had a kn ...[text shortened]... ou will find though that entropy is involved:
https://en.wikipedia.org/wiki/Endothermic_process

I believe that you are massively overestimating this effect.

Our sense of temperature is hugely effected by thermal conduction rates, such that
objects at exactly the same temperature will feel wildly warmer or cooler dependent
on their thermal conductivity. Touch a plastic bottle vs a metal can at exactly the same
temperature and the can will feel significantly colder.

The size of any possible actual temperature change due to evaporation inside a pressurised
can is going to be tiny as compared with the subjective temperature as 'measured' by a
human.

If there were evaporation inside the can INCREASING the pressure then the gas temperature
would go up. The liquid might cool, but then you're dumping energy into it by shaking heating it
back up. All in all these effects look to me to be tiny, and undetectable by touch.

I thus think it vastly more likely that the can feeling cooler when shaken is due to improved
thermal transport caused by mixing the internal liquid rather than any actual temperature drop.

Originally posted by twhitehead
Agreed.

[b]The pressure in the tin depends on temperature only.
Only under stable conditions. Shaking the can should result in evaporation and an increase in internal pressure.

But shaking it will not drive further evaporation.
Evidence please.[/b]

There is no reason for agitation to produce evaporation in a saturated vapour. All agitation does is mix the vapour and liquid, it increases the surface area and so if the vapour is unsaturated it will evaporate more rapidly, but in a saturated vapour the rate of evaporation from the liquid is matched by condensation. If the vapour has settled in the can so that there's a layer of unsaturated vapour then one might get a temperature drop, but that will be a small effect, otherwise I don't think shaking the container should cause a temperature drop.

Originally posted by googlefudge
I believe that you are massively overestimating this effect.

And I disagree. Of course I am not certain which is why I strongly feel we should go about this like scientists and actually do experiments that tell us the truth. Until someone is able to get a genuine temperature reading before and after that can tell us whether or not the real temperature goes down (or up) we are just guessing (unless someone here actually knows how to calculate the heat of evaporation).