# Confusion about energy taken to evaporate water with air instead of heat

So I know that water will take up 4200 J/Kg°C (specific heat capacity) if we heat it. Then when it reaches 100°C, it will take 2260kJ/Kg (latent heat of vaporisation) until all the water is gas. Then it takes the steam's specific heat capacity at its pressure as it continues to increase in temperature.

But what happens when, say, a powerful fan, evaporates water from the surface of the water? Here there is no "energy" being provided to the water, so it (the bulk of it, at least) remains at the same temperature as opposed to evaporation following boiling. But if the water at the surface manages to evaporate, that means some energy must be taken from the surroundings!

It helps to use examples. Suppose 5 kg of water evaporated by boiling and another 5 kg evaporated by a powerful fan.

Can I say that in the fan situation too, the water at the surface took 4200kJ/Kg°C until it reached 100°C; then it took in 2260Kg°C until it all became steam?

Does this mean both cases take in the same amount of energy?

• Water doesn't need to be at 100 C to evaporate, and you don't need to supply a heat of vaporization to evaporate. Consider water in a cup that you leave out for days, it will soon evaporate, even more so depending on the humidity of the room. Even at room temperature, water is evaporating from a surface due to mass diffusion from a high concentration of water vapor near the surface, to a low concentration throughout the room. Commented Apr 29, 2020 at 12:27
• I'd just like to add: By blowing the water with a fan, you are more quickly breaking up the thin layer of water vapor near the surface, so that more water molecules diffuse across the liquid vapor interface. The bulk of the liquid loses energy equal to the heat of vaporization during this process, but it is quickly equilibrated with the room if the amount of water evaporating is small. Commented Apr 29, 2020 at 12:29
• The fan blowing reduces the air pressure surrounding the water. And less air pressure means faster evaporation. Commented Apr 14, 2023 at 5:57

In both cases, the water takes in the energy. It is not exactly the same amount of energy for evaporation and boiling, but it's similar.

Energy is conserved. This means that the energy that goes into a system is equal to the change in energy of a system. In this case, we plan to make all energy transfers as heat (and none as work), so we can write the equation $$Q = E_f - E_i.$$

$$Q$$ is the heat that enters a system. $$E_f$$ is its final energy. $$E_i$$ is its initial energy.

First let's consider boiling $$1 \;\mathrm{g}$$ of water from room temperature until it's completely boiled away. The initial state is has some thermal energy, $$E_{\mathrm{liquid};0}$$, for $$1 \;\mathrm{g}$$ of liquid water at room temperature. The final state has thermal energy $$E_{\mathrm{vapor};b}$$ for $$1 \;\mathrm{g}$$ of water vapor at the boiling point. So the heat input is $$Q_{\mathrm{boil}} = E_{\mathrm{vapor};b} - E_{\mathrm{liquid},0}.$$

Next let's consider evaporating $$1 \;\mathrm{g}$$ of room temperature liquid water completely into water vapor. In this process, we can assume that the air temperature remains the same, and that the water vapor is in thermal equilibrium with the air. So the initial state is the same, but the final state is water vapor at room temperature, not water vapor at the boiling point. So the heat input is $$Q_{\mathrm{evaporate}} = E_{\mathrm{vapor},0} - E_{\mathrm{liquid},0}.$$

The heat input is not the same in the two cases. It takes less heat to evaporate water than to boil it because the final state is lower energy. (Water vapor at a lower temperature has less energy.) However, the energy needed to evaporate the water is still significant, and comparable in magnitude to the energy needed to boil the water. This is because the energy needed to heat the water vapor is relatively small compared to the heat of vaporization, which is energy needed to break the hydrogen bonds between the water molecules.

During boiling, the heat will have to be supplied by some external source in order to heat up the water. During evaporation, the heat can come simply from the surroundings, e.g. from the remaining water, from the air, etc. The result is that when the water evaporates, the surroundings cool. This is the mechanism behind sweat cooling you. You sweat, the sweat evaporates, taking heat away from the remaining sweat and from your body, and you feel cooler. Another, more dramatic consequence is that a puddle of water can freeze when the air temperature is above freezing, if the air is dry. In that case, some water can evaporate into the air, cooling the remaining water until it freezes. For a given air temperature and humidity, water can be cooled in this way down approximately to the wet bulb temperature.

But what happens when, say, a powerful fan, evaporates water from the surface of the water? Here there is no "energy" being provided to the water, so it (the bulk of it, at least) remains at the same temperature as opposed to evaporation following boiling. But if the water at the surface manages to evaporate, that means some energy must be taken from the surroundings!

Although the temperature of a liquid is a measure of the average kinetic energy of the molecules of the water, individual molecules can have velocities (and thus kinetic energy) above and below the average.

Evaporation only occurs at the surface of the water. It is due to the fact that that some of the water molecules at the surface have higher velocities (they are more energetic). These molecules may have sufficient energy to overcome the intermolecular attractive forces between molecules at the surface, allowing them to fly off and become gaseous H$$_2$$O molecules above the surface. The average kinetic energy of the molecules remaining at the surface becomes lower, cooling the water at the surface. This is called evaporative cooling.

A powerful fan blowing air across the surface removes water vapor near the surface (lowers the vapor pressure) making it easier for the more energetic water molecules at the surface to escape. In effect, the fan increases the rate of evaporation at the surface.

Hope this helps.

• Yes. It DID help! Commented Apr 30, 2020 at 8:24

Bob D. wrote a great answer. I just wanted to add a link to Feynman's lectures of physics, where at the first lecture he talks exactly about this:

https://www.feynmanlectures.caltech.edu/I_01.html

See section 1.3 for you question and answer :)

(The reason I'm adding a link is that I think you should read the whole thing - maybe you will get more insights)

• Thanks for the link! Cleared out much of the doubts. Commented Apr 30, 2020 at 8:23

That there is not energy provided to the water to evaporate is incorrect. Take sweating for example. If you evaporate water off the surface of your skin, it cools you because the energy to evaporate that water came from your body. It is the same in evaporating from the surface of water. As the higher velocity molecules escape from the water surface in greater proportion than those recondensing in it, the net temperature decreases in the water, because a higher than average particles escaped thus lowering the net energy in the water. In slower evaporation from a water surface, the temperature doesn't need to decrease in the water because the energy of evaporation is conducted in from the environment somewhere else. Otherwise all those evaporative coolers we have all over the place wouldn't work.