Temperature of steam rising off boiling water When boiling water on a stove, will the temperature of the steam vary significantly with the temperature of the burner? 
Person A's argument: So, once individual water molecules reach 100C/212F, they become vapor. The water molecules in the pot are <100C; the water molecules in the air are >100C. Generally, the only way to heat up the water vapor to significantly more than 100C would be to trap the water vapor. In a big kitchen, the water vapor rises rather quickly and gets sufficiently far away from the burner. Within the first couple seconds that the molecule becomes vapor, the vapor may still be close enough to the burner to become slightly more than 100C (101C?), but generally, no matter what the temperature of the burner, the water molecules will escape at 100C and won't reach a temperature significantly above 100C, given a large room.
Person B's argument: 
With a hotter burner, the water in the pot is hotter and as a result the water molecules that become steam - and bubble up from the bottom of the pot - transfer less heat to the surrounding water on their way to the top of the pot and leave as hotter steam.
Or do persons A and B just have a poor grasp of physics?
 A: Both A and B are slightly wrong.   The 'boiling point' of water is
the temperature at which steam and liquid exist at equilibrium,
and the roiling boil of a pot of water on the stove indicates a
lack of equilibrium.   Each steam bubble, expanding as it rises
from the bottom of the pot, is accumulating vapor from the surrounding
liquid (not staying a constant volume).
So, A is wrong to think that there is an equilibrium-temperature
indication in the boiling pot.  A single molecule can become vapor
only at the water surface, or by doing work against surface tension
and water pressure by expanding the diameter of a bubble.  If
the work is done leaving uncondensed water vapor, it must
have been hotter than 'the boiling point'.
And, if B is naiive in thinking that the temperature outside the 
pot is important in determining the temperature inside.
The evaporation of water
is a heat sink more than capable of cooling the metal, it might just
be that higher outer temperature turns a boil with four streams of bubbles into a similar boil with eight streams of bubbles.   More
heat doesn't guarantee higher temperature, just higher heat flow.
As for 'significantly higher' temperature of the bubbles, that calls for judgment.  The observation of small bubbles expanding as they rise, means
there is significance, because it's observable.    
A: In order for there to be a phase change from liquid to vapor, water must release latent heat of 100 degrees centigrade.  A hotter burner will not raise the latent heat temperature of water vapor at phase change (the boiling point) of the water.
However, if the burner is large enough to heat the entire room and the room is closed, it could raise the temperature of water vapor already in the air by transferring sensible heat from the burner to the air in the room.
As the temperature of the room rises, the saturated vapor pressure inside the room would also rise, and molecules of water vapor in the air would move with greater kinetic energy, raising their temperature.  In order for the burner to raise the temperature of water vapor in the ambient air, the room would have to be closed.  Otherwise, the saturated vapor pressure would not rise and neither would the temperature of the water vapor.
A: All the answers are partially right, with some errors.  Water has three very important properties that make it interesting.  The molecule is electrically polar, with one side being slightly +, the other side being sliggtly -.  This gives water a large amount inherent order and intermolecular attraction, and an extremely high surface tension.  This provides a very high specific heat, requiring the largest amount of energy to change temperature, which also effects its density and natural convection, and an extremely high flash energy to turn it to steam, once it is at the boiling point for the pressure it is under.  Water in a boiling pot will actually average between about 185 and 195 depending on room conditions, because of waters high specific heat and natural convection from temperature and density variations, between the heated bottom, and cooling top.  As the pot becomes warmer the natural convection will give way to water in contact with bottom absorbing enough energy to flash.  A natural way of speeding up the energy transfer from bottom to top.
A: Person A is essentially right, with the exception that the water molecules in the air are not at 100°C or higher, but at room temperature, since they're (supposedly) in equilibrium with the other components of air. Remember that even at room temperature water turns into vapor: this evaporation is what makes puddles on the sidewalk eventually disappear after the rain.
Supporting person A are measurements made in kitchen experiments that attest the water temperature won't exceed 100°C even very close to the bottom surface.
If you preheat the pot to at least some 200°C and throw a bit of water in, you'll get Leidenfrost effect, i.e., the drops will float for a while over a cushion of vapor. With more water, it'll splatter around with vapor that might be in part above 100°C close enough to the hot surface.
A: The boiling pot of water acts as a temperature regulator once the water begins to boil. As long as the burner doesn't have any additional means of transferring heat into the steam above the water, it does not increase in temperature but actually decreases rapidly as you'll often note condensation on the upper part of the pot.
If you wanted to create steam above $100^o C$ then you need to apply additional heating above the water with another burner/heater.
