Heat of vaporization is related to enthalpy change, while dew point is related to free energy change, i.e. enthalpy plus entropy. That's why they are very different concerning relative humidity.
The enthalpy of a gas is more-or-less independent of pressure or partial pressure, because gas molecules don't really interact with each other. At insanely-high pressures there would be some effect on enthalpy of course, but the effect at everyday pressures is very low. Pressure mainly affects a gas via entropy not enthalpy.
The enthalpy of a liquid is somewhat dependent on total pressure: A high pressure will push the molecules closer together and therefore change their interaction energies. But obviously the enthalpy of the liquid doesn't depend on what the gas partial pressures are, it can only depend on the liquid's own total internal pressure.
So the answer is: Heat of vaporization, being related to enthalpy not entropy, has essentially no dependence on relative humidity. (given a constant total air pressure)
-- UPDATE --
Oops, whenever I wrote "enthalpy" I should have said "enthalpy per molecule" or "enthalpy per mole" ["molar enthalpy"]. You can check for yourself that the enthalpy per molecule of an ideal gas is independent of pressure or partial pressure. For a real-world gas, it's approximately independent. The "per mole" quantities are what matter for dew point etc.
You need to know the rate of heat given by the flame to the water.
Suppose the flame transfers $h$ kJ/s to the water. The latent heat of evaporation of water is $2260\ \mathrm{kJ/kg}$.
For energy balance, the heat given to the water must be equal to the amount of heat required to convert water into steam.
$$
h = \dot{m} \times 2260 \\
\therefore \dot{m} = \frac{h}{2260}\ \mathrm{kg/s}
$$
If you say the rate of heat transfer doesn't matter (ignore the flame and assume the water is at a certain temperature and stays at that temperature throughout the experiment),
$$
\dot{m} = \frac{\Theta A (x_\mathrm s - x)}{3600}\ \mathrm{kg/s}
$$
Where
$\Theta = (25 + 19 v)$ is the evaporation coefficient ($\mathrm{kg/(m^2\cdot h)}$). This is an empirical equation, so you can't derive it from first principles.
$v$ is the velocity of air just above the surface of the water ($\mathrm{m/s}$)
$A$ is the surface area of the water ($\mathrm{m^2}$)
$x_\mathrm s$ is the humidity ratio in saturated air at the same temperature as the water surface ($\mathrm{kg}$ $\mathrm{H_2O}$ in $\mathrm{kg}$ dry air)
$x$ is the humidity ratio in the air ($\mathrm{kg}$ $\mathrm{H_2O}$ in $\mathrm{kg}$ dry air)
It's fairly straightforward to find $x$ and $x_\mathrm s$ from the relative humidity and the Mollier chart.
Best Answer
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.