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?
I am aware that are similar posts about this topic, but there were no satisfactory answers in them.
Thank you!
 A: 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.
A: 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.
A: 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:

*

*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


*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


*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.
A: 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.
