as I got out of the shower

Originally posted by DeepThought
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.

Tell that to a coke bottle (shaking it causes gas to come out of solution, why?).

Here is another experiment to try. Find an empty spray can. Cut off the top. Put some water in it (about a quarter full or less). Leave it to stand. come back and either shake it (if you can cover the top) or stir it till the water covers the sides. See if it feels colder to the touch when shaken or stirred. If the conduction proponents are correct then it should feel identical to the original spray can does when shaken.

Originally posted by twhitehead
Here is another experiment to try. Find an empty spray can. Cut off the top. Put some water in it (about a quarter full or less). Leave it to stand. come back and either shake it (if you can cover the top) or stir it till the water covers the sides. See if it feels colder to the touch when shaken or stirred. If the conduction proponents are correct then it should feel identical to the original spray can does when shaken.

Wasabi is sold in re-sealable metal cans here, so I will put various liquids inside and do 'the shake'. If it gets too cold, I can eat the Wasabi and get warmed up again.

Originally posted by twhitehead
Tell that to a coke bottle (shaking it causes gas to come out of solution, why?).

You've got a good point with that. The problem with coke is that it's a liquid containing a dissolved gas and nucleation points are a significant factor. I've been thinking in terms of a tin of lighter fluid because it's just butane and so there's only one component to worry about. My intuition is that shaking the tin is to sort settling problems out rather than to increase the amount of gas already evaporated. Anyway, after thinking about it I've got a theory for you, but there's problems with it.

Molecules evaporate from a thin layer near the surface whose thickness is of the order of the mean free path within the liquid. They have to have enough kinetic energy to climb out of the potential well. This means this surface layer has fewer evaporation candidates than the body of the fluid. As they escape the liquid they are either replaced by condensing molecules or diffusional drift of molecules from the bulk of the fluid, a proportion of whom will have escape velocity. The net result is that the surface layer of the fluid should be slightly colder than the body of the fluid if the vapour is not saturated.

The consequence of this is that shaking the can brings fluid from the bulk of the liquid into contact with the vapour. So, there is the potential for more evaporation. However there are two difficulties with this possible explanation. Firstly I'd expect it to be a tiny effect, secondly I assume that the liquid and vapour have had time to come into equilibrium with one another and so we shouldn't see the effect.

Originally posted by DeepThought
You've got a good point with that. The problem with coke is that it's a liquid containing a dissolved gas and nucleation points are a significant factor. I've been thinking in terms of a tin of lighter fluid because it's just butane and so there's only one component to worry about. My intuition is that shaking the tin is to sort settling problems out ...[text shortened]... pour have had time to come into equilibrium with one another and so we shouldn't see the effect.

I didn't read this thread carefully, but shaking the can should decrease the pressure in the can from bernoullis principal. For fluid flows increased velocity is decreased pressure. In high velocity flows the pressure can decrease below the vapor pressure, and dissolved gases escape potentially causing damage to the pumps, conduit, etc... The same thing happens with the soda can. Shake it, dissolved gasses escape, then are re-dissolved when the shaking stops. Then just apply the ideal gas law.

T2 = P2/P1*T1

For this scenario P2/P1 < 1, hence T2<T1

Originally posted by joe shmo
I didn't read this thread carefully, but shaking the can should decrease the pressure in the can from bernoullis principal. For fluid flows increased velocity is decreased pressure. In high velocity flows the pressure can decrease below the vapor pressure, and dissolved gases escape potentially causing damage to the pumps, conduit, etc... The same thing ha ...[text shortened]... Then just apply the ideal gas law.

T2 = P2/P1*T1

For this scenario P2/P1 < 1, hence T2<T1

Ok., this gives us a mechanism to drive evaporation that I'm happy with. But the change in pressure wouldn't explain the cooling of the tin. Consider a tin of butane. Half the contents by volume are butane and the other half butane vapour. Initially the pressure inside is 170 kPa (this is the saturated vapour pressure of butane at zero Celcius). The density of butane is 2.43 Kg/m^3 at 15 celcius so from Bernoulli's equation:

P + v^2/2 = constant.

Shaking the bottle vigorously is going to make the fluid move at about 10 m/s (my argument is sensitive to this so feel free to dispute this number), so assuming this is also true of the vapour we get a pressure drop of about 124 Pascals which is approximately an eighth of a percent. So I don't think the ideal gas law explains the drop in temperature (what we feel is the temperature of the fluid), which will be of the order of 1/3 of a degree Kelvin and not persist after we stop shaking the container and the pressure in the tin becomes higher than the normal vapour pressure of the substance. I need to work out what the drop in temperature from evaporation is to see if it's realistically an explanation.

Suppose the fastest one could move the vapour is 100 m/s then my earlier answer gets multiplied by 100 and the pressure drop is about 10% which would produce a thirty degree drop in temperature.

Originally posted by DeepThought
Ok., this gives us a mechanism to drive evaporation that I'm happy with. But the change in pressure wouldn't explain the cooling of the tin. Consider a tin of butane. Half the contents by volume are butane and the other half butane vapour. Initially the pressure inside is 170 kPa (this is the saturated vapour pressure of butane at zero Celcius). The ...[text shortened]... 100 and the pressure drop is about 10% which would produce a thirty degree drop in temperature.

Well, strictly speaking I don't think the Bernoulli derivation is giving the full scale of the pressure drop. The shaking motion (or the flow motion) varies in direction and time. The Bernoulli assumes steady flow, which is not the case.

-d/ds ( p + Gamma*z ) = pho*( V dV/ds + dV/dt ) Euler Equation (all partial derivatives)

The Euler equation is the better starting point for these parameters but there are obvious barriers to its quantification.

I 'think' the very high localized accelerations could have significant impact on the predictions in the same direction (decreasing pressure with increased acceleration), coupled with the models sensitivity to shaking velocity. How much so? Unfortunately, it seems to me the only path is to solve the equation... or hope an expert in fluid dynamics chimes in.

Originally posted by DeepThought
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 ...[text shortened]... and by shaking the tin we saturate it I don't see how there can be an evaporation driven effect.

The precision is 0.1 degree F and the point of contact was at about the 50% point and I aimed right at the bare metal part of the can, where the crimping takes place, for some reason there was no paint on either can in that place so I was hoping to make a similar test in both cases.

The idea of evaporation inside a can, even if it takes place, wouldn't a total temperature change violate the law of conservation of energy? It would seem if internal evaporation lowered the temperature of a closed system you could extract energy from that system say by putting two cans hooked to a Peltier converter, shake one can, it gets cooler, energy comes from the converter.

I guess you would argue that shaking introduces mechanical energy too so maybe I am all wet🙂

Originally posted by sonhouse
The precision is 0.1 degree F and the point of contact was at about the 50% point and I aimed right at the bare metal part of the can, where the crimping takes place, for some reason there was no paint on either can in that place so I was hoping to make a similar test in both cases.

The idea of evaporation inside a can, even if it takes place, wouldn't ...[text shortened]...

I guess you would argue that shaking introduces mechanical energy too so maybe I am all wet🙂

O.k. that sounds reasonably thorough and the reading is well within the claimed precision of the instrument. With my tin of butane the cooling is noticeable and definitely more than 1 Fahrenheit (but butane is volatile so it's not surprising I get a bigger effect than from a spray-paint tin or deodorant). Also as a check I picked up the tin by the bottom and that area of the can became colder. So I no longer think that googlefudge's physiological explanation fits.

You can dump large amount of energy into a mixture of ice and water without the temperature changing. During evaporation the molecules of evaporating butane (or whatever) have to climb a potential well. So the vapour gains energy because there are more particles, but once they're out of the liquid they've slowed down so the only reason the vapour's internal energy has increased is because the number of particles has increased, not because their average energy has. The liquid loses particles and hence energy. Because it's lost its hottest ones it's temperature drops. Overall energy is conserved. Basically you're assuming a linear relationship between temperature and energy in a situation you shouldn't be. I don't think that means you can pick up two of them turn the whole thing into a perpetual motion machine.

Originally posted by sonhouse
The idea of evaporation inside a can, even if it takes place, wouldn't a total temperature change violate the law of conservation of energy?

I had missed the post where you reported the results and only just seen it. So it appears that there is a genuine temperature drop. I can't think of any possible way in which your readings could have gone wrong to the point of falsely reporting a drop.

If the cooling is due to evaporation (which I think is by far the best explanation at this point), then it isn't a violation of the conservation of energy. If you left the can to sit for a while till the liquid starts condensing again, it will heat up. A fridge depends on this principle.
A liquid contains something similar to chemical bonds. They are weaker than the well known covalent, ionic and metallic bonds, and they can swap partners but they are very real. When evaporation takes place, the bonds are being broken and potential energy is being created.
Maybe 'bonds' isn't the right word:
https://en.wikipedia.org/wiki/Intermolecular_force

I have a deodorant can that if I just gently tilt one way and another feels colder without
any vigorous shaking.

EDIT: also I should note that I can detect no difference in temp between gently tipping the
can and vigorously shaking it.

I had a holiday in the south of France where the temps were in the mid 30's [C] and we
had an outdoor pool with a water temp of 30~32C.

The air felt burning hot, the pool felt freezing cold, at exactly the same temperature.

The effect of heat flow on our perception of temperature is HUGELY dominant, you need tens of
degrees of actual temperature difference to compensate.

None of the effects you are discussing come close.

Originally posted by googlefudge
I have a deodorant can that if I just gently tilt one way and another feels colder without
any vigorous shaking.

EDIT: also I should note that I can detect no difference in temp between gently tipping the
can and vigorously shaking it.

I had a holiday in the south of France where the temps were in the mid 30's [C] and we
had a ...[text shortened]... tual temperature difference to compensate.

None of the effects you are discussing come close.

No harm in going down the rabbit hole. Everyone can only learn more from doing so. For instance, I provided a potential model, Eulers Equation. Without solving that, what makes you confident that decreased pressure via this relationship, in combination with the ideal gas law will give negligible temperature changes? Is it applied wrong? What assumptions are made to negate the relationship?

Originally posted by joe shmo
Well, strictly speaking I don't think the Bernoulli derivation is giving the full scale of the pressure drop. The shaking motion (or the flow motion) varies in direction and time. The Bernoulli assumes steady flow, which is not the case.

-d/ds ( p + Gamma*z ) = pho*( V dV/ds + dV/dt ) Euler Equation (all partial derivatives)

The Euler equation is t ...[text shortened]... s to me the only path is to solve the equation... or hope an expert in fluid dynamics chimes in.

The shaking generates shock waves in the fluid. That should be enough to produce bubbles of vapour in the liquid and get the vapour supersaturated. I'm fairly happy with this description.

To get a handle on how big an effect we need. The energy lost by the liquid is LdN where L is the latent heat of vapourisation and dN is the number of moles of the substance evaporated. The energy lost by the liquid is NCdT where N is the number of moles at the end, C is the specific heat capacity and dT the change in temperature. This gives us:

dN/N = CdT/L

C = 98 J/mol/K
L = 21 kJ/mol/K

so:

dT = 214*(dN/N)

so if 1% of the fluid evaporates dT ~ 2.14 Kelvin. Which means we need around 2% of the liquid to evaporate for it to be noticeable.

Originally posted by googlefudge
I have a deodorant can that if I just gently tilt one way and another feels colder without
any vigorous shaking.

EDIT: also I should note that I can detect no difference in temp between gently tipping the
can and vigorously shaking it.

I was able to detect a difference.

The effect of heat flow on our perception of temperature is HUGELY dominant, you need tens of
degrees of actual temperature difference to compensate.
That is true if you are trying to measure absolute temperature. It is less important in this instance.

None of the effects you are discussing come close.
Which is why I suggested we go about it scientifically and not rely solely on the touch test. Sonhouse has reported that there is a real measurable temperature drop not dependent on human touch. This cannot so easily be explained away as being a conduction issue.

Originally posted by twhitehead
I was able to detect a difference.

[b]The effect of heat flow on our perception of temperature is HUGELY dominant, you need tens of
degrees of actual temperature difference to compensate.
That is true if you are trying to measure absolute temperature. It is less important in this instance.

None of the effects you are discussing come close. ...[text shortened]... t dependent on human touch. This cannot so easily be explained away as being a conduction issue.

If you had a bottle of butane that was totally full, wouldn't that show pretty much zero temperature decrease, if the effect was due to evaporation? If it was totally full, there would be no volume in which to evaporate and if there were, it would be a very small percentage of the total volume so that should be a test of the evaporation theory.

Conversely, if the evaporation theory is correct, then a bottle say only 1/\4 full should have a lower temperature when shaken than one say 3/4 full.