Why do steam bubbles increase in size as they rise? In the following video (a customer's review of a glass kettle), we can observe water boiling: http://youtu.be/jByY5I7Xk7w?t=2m55s
As the kettle starts to boil at around 2:55, we can see large steam bubbles being formed at the bottom, where the heating element is, and these bubbles shrink as they rise. Presumably this is because they are coming into contact with cooler water. Then we get a crazy convection current for a bit before the element switches off again.
After the chaotic motion has died down (and the fluid is presumably very well mixed) we see small steam bubbles being forming at the bottom, which grow as they rise.  I can think of two possible explanations for this, and I'm curious as to which is correct:


*

*The water is superheated. Nucleation sites exist on the bottom of the kettle, so that's where steam bubbles form. Steam is produced at the interface between steam and water, which causes the bubbles to grow as they rise.

*The pressure at the bottom is slightly higher than at the top. Assuming a depth of 15cm, the boiling point at the bottom of the water is about $100.3^\circ \mathrm{C}$, compared to $100.0^\circ \mathrm{C}$ at the top. Bubbles form at the bottom because the heating element is still slightly hotter than $100.3^\circ \mathrm{C}$, and as they rise they drag hot water up into the slightly lower-pressure area, where it turns to steam because its boiling point lowers, and this increases the size of the bubble.
In particular I'm interested in whether the second of these explanations plays a role. If it doesn't happen in a boiling kettle, is there any situation in which it does?
 A: The discussion of nucleation sites is very much to the point.  Water at atmospheric pressure without nucleation sites will theoretically boil only at $320.7 {}^o C$.  The bubbles act as nucleation sites that reduce the energy required for evaporation.  In the case of a bubble, the effective contact angle between the superheated liquid and the bubble surface is $180 {}^o$ which reduces the superheat needed to evaporate the water to $0$.  Interestingly, there is actually an impediment to bubble growth caused by the reduced temperature of the vapor inside the bubble and a corresponding lower superheat boundary layer of liquid surrounding it.
FYI: My information is based on Collier and Thome pages 138 and 549.
In that text, an equation is given for the rate of bubble growth as:
$$ R = \sqrt{\frac{12\alpha_f}{\pi}} \frac{\rho_f c_{pf} \Delta T}{\rho_g i_{fg}}\mbox{Sn} t^{1/2}$$
where
$$\mbox{Sn} = \left[ 1-(y-x)\sqrt{\frac{\alpha_f}{D}}\left(\frac{c_{pf}}{i_{fg}}\right)\left(\frac{\partial T}{\partial x}\right)_p\right]^{-1}$$
and the variables are:
$R$ - rate of bubble growth
$\alpha_f$ - thermal diffusivity of liquid
$\rho_f$ - density of liquid
$c_{pf}$ - specific heat of liquid phase
$\Delta T$ - temperature difference
$\rho_g$ - gas density
$i_{fg}$ - latent heat of vaporization
$t$ - time
$D$ - molar diffusion coefficient
It's been a while since I looked at this in depth, but I think the $x$ and $y$ variables refer to position relative to a uniformly heated tube coaxial to the $y$ axis.  Honestly, I don't expect you to actually use this formula, but hopefully it will impress that there are people who have spent a great deal of time on this subject.  If you find it interesting, you might have a promising career in power plant boiler engineering in general or nuclear power plant engineering in particular.
