Let's assume a one litre $1000{\,\rm W}$ electric kettle, filled with $0.5$ kilograms of water at $20^\circ \mathrm{C}$:
It takes 4.2 joules to warm one gram of water one degree Celsius.
So, to warm the $500$ grams of water $80$ degrees from $20$ to $100$ takes $168,000$ joules. The kettle will supply $1000$ joules per second, so it'll take $168$ seconds for the kettle to come to a boil.
During this time, the $0.5$ litres of air will expand by a factor of $\frac{373}{293}$, to a volume of $0.637$ litres. So in the almost three minutes of heating, only $0.137$ litres of air will be forced out through the whistle spout.
Now we're at the boiling point. It takes $2,280$ joules to vaporize $1$ gram of water. So the kilowatt heater will vaporize $0.439$ grams of water each second!
Those $0.439$ grams of water vapor will occupy around $0.750$ litres at $100^\circ$ Celsius. So this much gas will be forced out the spout each second...
$0.137$ litres of air in three minutes, vs $0.750$ litres of steam each second.. That explains the difference...
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
when you boil water, you convert it into water vapor, which leaves the pot and mixes with the atmosphere. If you boil the pot long enough, eventually all the water in it is converted to vapor and leaves. the pot is then empty.