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Imagine two pots boiling on a stove. One is tall, while one is wide. Both contain the same volume of water.

What will be the difference in the rate of evaporation between the two containers?

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    $\begingroup$ A precise modelling isn't easy at all. For comparable size and shape and the same stove flame you could expect that water evaporates faster (absolute volume reduction) from the wider one. $\endgroup$ – Alchimista Jul 11 '19 at 8:57
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Boiling water

When you boil water on a stove, the temperature on the bottom of the pot is higher than the boiling point. Steam bubbles are generated, and they help to heat the water at the top of the pot (convection also helps, and conduction of the pot itself). Once you have a rolling boil, the rate is determined by how quickly you can supply more heat (almost all of it now goes into the phase transition). If you use the same burner, and mostly heat the pot (so no tiny pot on a big burner), at that stage the rate of boiling off should be the same.

Evaporating water

When the temperature of the water is below boiling, the mechanism is different. Rather than steam displacing the air above the water, single water molecules break away from the liquid and mix with the air. Evaporation rates depend on the partial pressure of water just above the surface (this influences condensation, i.e. the reverse reaction), the surface area, and the vapor pressure of water in the liquid state. Anything that lowers the partial pressure of water (such as "wind" bringing in dry air), increases the surface area (different pot geometry) or increases the vapor pressure (temperature) will increase net evaporation rates.

Beaker vs Erlenmeyer

Tall vs. wide pot changes the contact area with the stove top and the area of the liquid:gas interface at the same time. If you compare a beaker with an Erlenmeyer (same bottom area), it should show similar boiling rates (maybe a big mess with overboiling for the Erlenmeyer) but very different evaporation rates.

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  • $\begingroup$ See this answer for an example how phase transitions help in heat transfer. $\endgroup$ – Karsten Theis Jul 11 '19 at 14:04
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There are multiple effects which differ in both cases.

Firstly, convection (e.g. "wind") in the air will be much more efficient in the wide container. The tall container will limit the quantity of air flowing just above the surface. This, in turns, means that the water that does evaporate will remain for longer just above the surface of the water, and has more chances of returning to solution.

Secondly, the container itself will act as condenser. As the metal is typically cooler than the water vapour, the metal can cool down the vapour and turning it back into water, which conveniently flows back into the pot. The more exposed surface you have around the boiling water, the more important this effect will be.

You might also consider the power of the stove itself. This is hard to account for, but briefly, there might be some differences in the heat diffusion in the water.

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    $\begingroup$ Not clear if water is boiling or just evaporation is considered. You mention metal at room T than stove, etc. $\endgroup$ – Alchimista Jul 11 '19 at 8:52
  • $\begingroup$ Boiling is fast evaporation. Feel free to edit my answer if you want to be more precise. Good point about the metal, I edited this. $\endgroup$ – Raphaël Jul 11 '19 at 11:16
  • $\begingroup$ See my comment. It is really tricky and engineering oriented. Water in a baking tin might even not boil, and in a very extended ideal one can fast evaporate. It is really complicated, at least in my opinion. Obviously there are parts of truth in your answer, at least listing the parameters, or some of the parameters, involved. $\endgroup$ – Alchimista Jul 11 '19 at 13:56

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