On new year's eve, after a splendid red and an assortment of sumptuous repasts, I made a bold remark which, on further consideration, may turn out to be incorrect. Unless! Unless I can concoct an impressive scientific explanation - for which I will need your input
- Temperature is the average kinetic energy of a "bulk" liquid
- Particles within the liquid will have a range of kinetic energies according to kinetic molecular theory (Maxwell-Boltzmann curve etc.)
- Water boils at 100 deg Celsius (101.3kpa etc)
- Evaporation can happen at any temperature (>273K)
- Individual particles don't have temperature but rather kinetic energy
Assuming we could measure the kinetic energy of multiple evaporating water molecules just as they left the surface of liquid water (at standard temp and pressure) over time, would the average kinetic energy of the sample of evaporating water molecules equal the average kinetic energy of boiling water (at 101.3kpa)? That is, if the evaporating water molecules had a temperature - would it be approximately one hundred degrees. Or, as I may have put it at the time, at a molecular level, do my drying undies effectively boil? (Note: I understand that the bulk water is at room temperature)
If I knew the speed of evaporating water molecules I could calculate their energy and compare this to the average energy of boiling water and see if they were similar - but I can't see how to estimate the speed of evaporating water molecules?
...I have read "Is it true that an evaporating molecule has the same kinetic energy as a molecule in a pot of boiling water?" on this site. I don't think it answers this question.
Diagram for discussion
If I assume that Boltzmann's distribution works for liquids (I get that it's meant for ideal gases) and assume 9 degrees of freedom for water molecules then:
This suggests (if it's even remotely correct) that a very small number of molecules in room temperature water are moving very fast and therefore are at very high temperatures? It's not a relationship per se but, if correct, it does affirm the initial idea that evaporating water molecules are "hot"...even boiling?
Thanks for the input and careful consideration. The chart below is my attempt to summarise my thoughts inspired by your comments. It's rendered in excel from the equation shown and accords well with the chart for water included in Boltzmann distribution for water which didn't extend far enough on the x axis for what we're trying to show here. Thanks for the heads up re energy distribution rather than speed - much easier to understand.
The equation comes from BC Campus Molecular Speeds I have ignored degrees of freedom effects in this and the other equations on the chart.
Clearly the energy of evaporating water molecules (let's say liquid immediately before take off), at room temperature (25 deg) is significantly higher than the average energy of boiling water.
What can be said about the little molecules about to liberate themselves from my drying undies then? The original question precipitated from the idea that molecules evaporating from washing 'boiled'.
- They are at similar energy, a little greater in fact, than the energy of molecules about to jump from a pot of boiling water (44kJ vs 41kJ, 7% difference), and massively more energetic than the average energy of boiling water (4.7kJ)
- If we could measure the temperature of a bunch of them they would be at an energy equivalent temperature some ten fold greater (3527K/373K) than the average temperature of boiling water (T = 2E/3Nk, bc campus university physics internal energy), but close to the equivalent (theoretical) temperature of the 'boiling' molecules. I know we can't measure their temperature as it is related to the average energy of the bulk liquid - but theoretically... (there's a case of beer in this)
- The proportion of molecules ready to let loose from my y-fronts is much smaller per unit of room temperature water than it would be for boiling water (see the area under the curve right of the 41kJ and 44kJ points)
So...to all intents and purposes, the undie vaporizing molecules are doing what the boiling water vaporizing molecules are doing ....boiling (but they're not at 100 degrees, and we can't really measure their temperature - maybe half a case of beer then?). We don't feel the heat of the highly excited little blighters because it's a (relatively) small number of individual molecules within a bulk liquid at an average temperature of 25 degrees.
In coming to the answer from a kinetic energy perspective, I found this paper Velocity of a droplet evaporated from waterwhich measures and models the speed of protonated water molecules evaporated from nano droplets. Speeds of non-protonated molecules in bulk liquid water will vary but it was encouraging to see that the speeds (say 2000m/s) were of an order of magnitude commensurate with with Boltzmann's distribution for water in Boltzmann distribution for water. On this curve, the experimental speeds would also be out in the flat area of the curve equivalent to my energy curve above.