The latent heat of vaporization of water is what we normally use to calculate the heat transfer that occurs at 373 K when liquid water transits to vapor phase. But there are curves for latent heat of vaporization at different temperatures. What is meant by latent heat of vaporization at a temperature different from its boiling point?
Liquids evaporate at any temperature - not just at their boiling point. This is the reason, for example, why wearing a wet shirt on a windy day makes you so cold: the water evaporates, and in the process "takes some heat with it".
The explanation for this is simple when you think about statistical thermodynamics. You have a lot of molecules whose energies follow a statistical distribution. The fastest of these molecules can escape the forces holding the liquid together; but in the process, they take away "more than the average amount of energy". It's like the smartest person leaving the room: the average of the room just got a little dumber.
Now the rate at which this happens is a strong function of the temperature (and the relative humidity of the air): however, it does happen at all temperatures. Therefore, you can express the average "surplus energy" that each molecule takes with it as a function of temperature - and when you scale this to unit mass, you get the latent heat of evaporation. At any temperature.
Note this appears to be in contradiction to LDC3's answer. One of us is wrong... I believe heat of vaporization is a perfectly valid concept when you are not at the boiling point.
The curve represents the boiling point of water at different pressures and temperatures. If you are heating water at the pressure selected, it will be at the temperature indicated by the curve when the water boils.
The latent heat of vapourization is always at the boiling point. The pressure and temperature determine where this point is.