Heat Losses from water pan due to vaporisation before boiling I'm studying heat transfer from an electric heater to a cooking pan (Stainless steel) filled with water; specifically, my problem is related to water vaporisation, as I've already found good models to predict radiative and convection losses.
The point is that vaporisation losses are easy to calculate after the water reaches the boiling point, as mass of water vaporised times enthalpy of vaporisation of water at atmospheric pressure (which is 2257 kJ/kg). But what happens before boiling?
My experiments show that a considerable amount of water already evaporates starting from a temperature of 70°C. In fact, it is well known that water evaporates even at ambient temperature due to mass diffusion, and as the temperature increases the rate of evaporation is also increased.
But since the water is not saturated, I guess I cannot calculate this loss by means of any enthalpy of vaporisation. How should I calculate the heat removed from the pan by the water molecules escaping from the upper surface at temperatures lower than boiling?
Thank you very much
 A: At the interface, the air is saturated with water vapor at the interface temperature.  So you know the partial pressure of the water vapor at the interface (if you know the interface temperature).  The water then diffuses into the room air above, where the bulk partial pressure of water is less than the saturation vapor pressure at the interface.  There is also convective transport of the air away from the interface, and this air carries the water vapor away from the interface.  So this has to be included in your model.  There is a heat flux from the bulk of the water to the interface (i.e., at temperature gradient in the water below the interface), and there is a heat flux in the air above away from the interface.  The difference between these two heat fluxes is equal to the heat of vaporization times the vaporization rate.  There may also be natural convection currents in the water below the interface to enhance the rate of heat transfer.  This is just a rough outline of what is happening, but all these things need to be considered in formulating a model of the heat and mass transfer in your system.
A: As you said, heat loss due to vaporization happens at any temperature. You can reasonably assume that the air right above the liquid is trying to remain saturated - and if you put a lid on the pot, that will indeed be the case. In that case, the heat loss due to vaporization will be quite low.
If you don't have a lid on the pot, the rate of heat loss due to vaporization becomes extremely hard to calculate, because it depends on the relative saturation of the air at the interface, which depends on the details of the air flow. Tables have been published for rates of heat loss as a function of wind velocity; see for example my earlier answer on heating swimming pools for some details. This shows that the rate of evaporation strongly depends on wind velocity: you are familiar with the concept of wind chill of which this is one manifestation.
The actual heat per unit mass removed can be determined from tables of heat of vaporization as a function of temperature - see for example https://en.wikipedia.org/wiki/Enthalpy_of_vaporization from which I reproduce this figure:

This shows a clear dependence of the heat of vaporization on temperature: not surprising, since you expect a molecule that escapes the liquid at a lower temperature to carry a relatively higher fraction of energy with it. I fitted (by eye) a straight line to the data between 260 and 360 K (roughly from freezing to boiling); the overlay (zoomed in) looked like this:

The coefficients of that straight line were
$$y = 3195 - 2.5T$$
which gives a value of 2262 J/g at 373 K - pretty close to various published values. I recommend using this relationship to estimate the heat of vaporization at temperatures below boiling - it should be close. The main problem is knowing how much liquid evaporates, and that's much harder.
