I have a liquid (say, water) and a gas (say, air) at the same temperature. How will the temperature of either evolve over time? What's the steady-state of relative temperature of liquid and gas with respect to each other?

What's the right mathematical machinery to throw at this kind of question?

  • $\begingroup$ Are the water and air in a closed or open container? Is there any heat transfer with the environment? Is there any air flow (e.g., wind) over the container of water? The devil is in the details. $\endgroup$ – David White May 1 '19 at 0:02
  • $\begingroup$ I will be happy to just have something for the simplest-possible, "ideal" situation, things like wind and complex external behavior can always be added. My guess for the simplest would be: bisection of 3D space by an infinite horizontal plane, creating an infinite and infinitely deep lake and atmosphere, respectively. No wind, no convention, constant gravity, starting out with perfectly equal temperature etc. $\endgroup$ – Johannes Ernst May 3 '19 at 18:40

How will the temperature of either evolve over time?

First of all, lets assume the water is in an open container with respect to the atmosphere so that when evaporation occurs it doesn't increase the vapor pressure above the water.

The bulk temperature of the water is the average translational kinetic energy of the water molecules. The same applies to the bulk temperature of the air above the water.

But individual molecules of the water or the air can have translational kinetic energy above or below the average. In the case of the water, for those water molecules well below the surface, it doesn't matter because they are trapped within the liquid. But at the surface of the liquid those molecules that happen to have above average kinetic energy have the opportunity to escape the surface and become gaseous $H_{2}0$ in the air provided that the air is not saturated with water vapor. When they escape they take their higher than average kinetic energy with them, thus reducing the average kinetic energy of the remaining molecules as the surface. That means the temperature of the molecules at the surface of the liquid drops over time, ergo, evaporative cooling.

What's the steady-state of relative temperature of liquid and gas with respect to each other?

That depends on the mass of the air and the mass of the liquid, and also on how the mass of the liquid is distributed. Let's say the water is a deep lake. In this case, the drop in temperature at the surface of the lake due to evaporative cooling has little effect on the overall temperature of the lake. In other words, the steady state bulk temperature of the lake will be the same as the air.

But if the water is a thin layer on, say, some surface, then evaporation will result in its bulk temperature being less than the air (gas)

Hope this helps

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  • $\begingroup$ I appreciate this (qualitative) answer. I was wondering what the equations are, however, as I don't even know where to look for the right ones. I would expect the time evolution to be some kind of statistical partial diffeq that's probably hard to handle, so I settle for how to calculate the steady state, which should be simpler. $\endgroup$ – Johannes Ernst May 3 '19 at 18:37
  • $\begingroup$ @JohannesErnst There are no simple equations because you would need to narrow down some of the parameter. To give you an idea, the following link describes equations for the example of the evaporation of water from the surface of a swimming pool. engineeringtoolbox.com/evaporation-water-surface-d_690.html $\endgroup$ – Bob D May 3 '19 at 18:57

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