The latent heat of vaporization is given by

$$\lambda = 2.501 - 0.002361*T_s$$

with the unit $[M*J*kg^{-1}]$, where the temperature is given in degree Celsius. This means that 2.5 million joules (J) are required to evaporate one kilogram of water. The latent heat of vaporization decreases slightly with increasing temperature. When the latent heat of vaporization is used as a constant and not as a function of temperature, the symbol $L$ is used.

A note on energy unit conversion. To convert cal to joule (J), simply multiply the value in cal with 4.1868 to obtain the value in joule. If you want to convert from Joule to cal, multiply with $1/4.1868=0.23885$.

To convert from joule (J) to kWh multiply with $2.7778*10^{-7}$. In order to convert from kWh to joule (J), multiply with $3.6*10^{6}$.

On the article on energy budgets for lake evaporation a value of 590 cal/g is given. According to the conversion formula above this corresponds to $590*4.1868*1000=2,470,212$. This corresponds more or less to the above value. A value of $597.5 cal/g$ would be needed to get a value of $2.501 MJ/kg$.