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? Thanks in advance.