# How much heat is required to evaporate water

I was wondering how to calculate the mass of water that evaporates if I were to expose the water to some amount of heat given by Q. $$Q=ml$$ ($$l=2.26\times10^3kJkg^-1$$) is the standard way to calculate the amount of water that vaporizes. However this method only applies to boiling water at $$100^\circ C$$, so are there alternatives to calculating this at lower temperatures where evaporation occurs?

Thank you!

Yes. Get yourself a set of steam tables (which you can get by Googling "water steam tables." The steam tables give the heat of vaporization at all temperatures from 0 C to the critical temperature.

Incidentally, it looks to me like the value you gave for the heat of vaporization is high by a factor of 1000.

• so without a reference table we wouldnt be able to calculate it numerically? p.s. thank you for pointing my mistake out! The heat of vapourisation is high by a factor of 1000 – Lucas Tan Jul 11 '20 at 5:07
• It depends on what data we're allowed to have.ll. If we knew the heat capacity in the ideal gas limit of low density as a function of temperature and we knew the equation of state for water (P,V,T), we could calculate it. This is basically how the steam tables were created. – Chet Miller Jul 11 '20 at 11:35
• For pressures low enough for water vapor to approach ideal gas behavior (say, up to a few atmospheres), if you know the heat capacity of the liquid and the vapor, and also the heat of vaporization at one temperature, you can get the heat of vaporization at higher and lower temperatures simply by using Hess' Law. – Chet Miller Jul 11 '20 at 20:36
• So seems like I would need 1. the equation of state for water and 2. Hess' Law (if ideal gas), thank you! – Lucas Tan Jul 12 '20 at 10:50
• For the case of an ideal gas, you would not need to know the PVT behavior; all you would need would be the heat capacities vs temperature of the liquid and vapor. – Chet Miller Jul 12 '20 at 11:41

The Clausius-Clapeyron relation lets you calculate the heat of vaporisation for water at any temperature using some standard values that you might need to remember. Using $$Q=ml$$ with $$l$$ of that particular temperature lets you calculate the heat required under the physical conditions used to calculate $$l$$ in the first place.

One way to get a better approximation is as follows:

You want: energy to change $$1$$ kilogram of water at temperature T to $$1$$ kilogram of water vapor at the same T...

Substitute the following path to achieve the same overall change:

1. Warm the kilogram of water from T to $$100^{\circ}$$C. Use the standard specific heat for liquid water; you will be adding heat energy

2. Convert the kilogram of $$100^{\circ}$$C liquid water to water vapor at $$100^{\circ}$$C. Use the standard latent heat of vaporization for water; you will be adding heat energy

3. Cool the the kilogram of $$100^{\circ}$$C water vapor from $$100^{\circ}$$C to T. Use the standard specific heat of water vapor; you will be removing heat energy.

Note that you have moved, in a roundabout way, to the same final state. If you add up all the energy added and removed in this route, you will arrive at the enrgy change you want.

• However we notice that water from clothes evaporate without needing to reach boiling point. So how can this be explained? From another answer in this thread, I gathered that latent heat of vaporization changes with different temperatures, hence water can evaporate at that specific temperature without needing to be heated to 100 degrees celsius. Would I be correct in saying that? – Lucas Tan Jul 12 '20 at 10:48

What happens is the hest we supply to the water help to increase it's temperature, and the change of state. So your total heat will be used as $$Q=mS\delta T+ml$$ so as you said water was at 100°C so there would no temperature change, so whole heat convert the water state to vapour from liquid.

• but we see find that water from clothes evaporate without needing to reach boiling point, how can this be explained? – Lucas Tan Jul 11 '20 at 5:06