There are too many unknowns to model the situation with any accuracy.
If you know the power being pumped into your pan then you can easily calculate the amount of steam generated from the latent heat of vaporisation of water. You can calculate the power being generated by your cooker from the flow rate of the gas, or the current if it's an electric cooker, but it's anyone's guess what percentage of this is lost to the environment and how much ends up in the pan.
For any significant flow rate of steam I would guess that turbulent mixing will ensure the steam temperature is roughly constant throughout the pan. The temperature will be whatever the boiling point of water is at the internal pressure. The steam flow rate through the hole in the pan is fairly straightforward to calculate from first principles, though since steam is so important industrially I'd guess some Googling will find tables and empirical equations for flow rate.
Response to comment:
Let the mass of water lost per second be $m$, then the power applied to the water in the pan is just:
$$ W = mL $$
where $L$ is the latent heat of vaporisation of water. This will be equal to the power generated by your cooker times some unknown factor less than unity to allow for heat loss to the environment.
Response to second comment:
The temperature of the water will be close to the boiling point because any water hotter than the boiling point turns to steam, and the latent heat required will cool the water again. The steam in the layer immediately above the water will be at the same temperature as the water because it's in thermal contact with it.
If the steam above the water is hotter than the water you have to ask what is heating it. The only things I can think of that could heat the steam are the pan walls and lid. However the pan is only being heated from the bottom, and heat flow by conduction though the pan walls is a lot slower than heating/cooling by convection between the pan walls and the water in the pan. Therefore I would guess that the pan walls and lid are also close to the boiling point of water - actually they will probably be slightly cooler because they will lose heat to the surrounding air.
So I would guess that as long as there is enough water in the pan the steam in the pan will be close to the temperature of the water.
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.
Best Answer
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.