I think the interesting aspect here is that the evaporation is greater than what can be explained by heat alone.
Similar to the photoelectric effect. Similar say to an enzyme.
All these environmental changes to the reaction lower the ‘action’ energy making the reaction vastly more efficient or possible in an environment that it wasn’t possible in previously.
The embodied energy of a phase transition does not pay attention to what path you took to cross it. There’s no “threshold” between phases of water.
For H20 to move from liquid water to vapor, energy must be added. There’s no catalyst.
So either we’ve discovered some new physics since I last studied thermodynamics, or this isn’t an accurate analogy.
Just a wild guess though. Haven't yet read the article.
I think that is a wrong assumption. Liquids will naturally evaporate even with 0 external energy, assuming there is not too much pressure in the surrounding atmosphere.
But thermodynamics still hold; the water vapor is still in a more energetic state than liquid water or ice.
Basically, some amount of a liquid will move to the higher energy state, and other parts will move to a lower energy state. The energy to vaporize some of the liquid doesn't need to come from something external to the liquid.
Momentum is mass x velocity; what’s the mass of a photon?
https://en.wikipedia.org/wiki/Photon#Relativistic_energy_and...
https://www.youtube.com/watch?v=bvzr2HbbPC8
(Maxwell's equations are consistent with relativity)
...or another way of looking at it (that I presume Boltzmann would agree with). If your had a single black body mass at some temperature greater than absolute zero in an otherwise empty universe, it would radiate away heat and thus cool off. The cooler body means the individual atoms in the mass have less energy and less momentum. If momentum is conserved, then that momentum must have been carried away from the mass in the mass-less radiation. Another neat thing is that light can also have angular momentum.
> Momentum is mass x velocity; what’s the mass of a photon?
Photons have zero mass. What's your point?
For photon,
p = hλ