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According to another answer, the rate at which water evaporates depends on

the temperature of both air and water, the humidity of the air, and the size of the surface exposed.

The air temperature, humidity, and surface area are all essentially constant. The water temperature is too–roughly 100ºC, right?

So am I correct in thinking that whether I set my stove to medium or high, as long as the water is boiling, the evaporation rate is constant?

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You are mixing two phenomena: boiling that occurs within the bulk and indeed only at 100°C at atmospheric pressure, and evaporation that can happen at any temperature and is a surface phenomenon.

However, if you are strictly speaking about boiling, the boiling rate depends on the energy you input into your system. If you put your stove from medium to high, your water will stay at 100°C but will boil faster (by this I mean that more water gets converted into vapor per second) because you input more energy per second into your system.

In fact, the enthalpy of vaporization of water is 2257 kJ/kg, which means that you need 2257 kJ of energy to vaporize 1 kg of water at 100°C and atmospheric pressure. Imagine you want to boil 1 kg of water with a 2 kW heating plate. The minimum time required to boil the entire quantity would be

$$t = \frac{2257\ kJ}{2\ kW} \approx 1000\ \text{seconds}$$ which is about fifteen minutes. In fact you need more time than that because of all the energy losses through radiation, diffusion in surrounding materials...

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    $\begingroup$ +1. You might even add a note about enthalpy of vaporization which for water is 2257 kJ/kg. This means that if you have a kilogram of water you need to add 2257 kJ of energy to vaporize it all-- and the faster you add that energy the faster you boil it off. $\endgroup$ – pentane Mar 9 '16 at 18:59
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    $\begingroup$ Thanks for suggestion, just edited it. Fell free to add whatever you like :) $\endgroup$ – Dimitri Mar 10 '16 at 10:18

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