Boiling Order (Thermodynamics) Three pots made of the same material are placed one inside the other and all pots are filled with water to the same level.  If heat is applied to the bottom of the outermost pot what will be the boiling order of the water in each pot and why?
PS: Clarified with the person who asked me this question--The pots are suspended inside one another and so there is no metal to metal contact with the inner pots.
 A: Let's assume that the inner pots are suspended within the containing pots, so that heat can only be transferred to them through the water in the containing pot, and that they are all open above so that steam/water vapour can escape equally well from all of them.
The water in the outermost pot cannot exceed the boiling point, so it can only heat the outer surface of the contained pot to the boiling point. Since heat can only move through a solid by flowing through it from a higher temperature to a lower temperature, heat can only be delivered into the water in the inner pot while it is at a lower temperature than the water in the surrounding pot.
As water heats up, it loses heat through evaporation, so as the water in the inner pot heats up it will eventually lose as much heat through evaporation as is supplied from the containing pot.
The system would reach steady state with the outer pot boiling, the intermediate pot close to boiling, and the innermost not as close to boiling as the intermediate one. Only the water in the outermost pot would actually be boiling. The temperature differences may be very small, but enough to make the difference between boiling and not boiling.
A: Because it would take a long time for water to boil, and heat and metal are good conductors of electricity, by my estimation, all the water would boil at the same time. The best way to find out is to do the experiment your self or model it on software...
A: If we assume the pots are suspended (i.e. the bottom/sides of an inner pot is not touching an outer pot), the answer is: it depends.
In particular, it depends on the nature of the heat source. If you put the pot on an extremely hot surface with a good heat capacity (so the heat is consistent), the bottommost pot would flash-boil very quickly, as the metal is a very good conductor, while while the water would insulate the inner pots. Additionally, the escaping steam will be at a temperature of about 100C, so the inner pot(s) are in thermal contact with temperatures much lower than the burner at the bottom.
If instead you heated the bottom pot so slowly that the temperature of all three pots and all three partitions of water were in thermal equilibrium, then they would all boil at the same time.  
As an aside: this is related to the reason some dishes such as fondue are prepared with a double-boiler. The bottom pot is filled with water, which eventually will boil. Since the water boils at a consistent temperature (~100C), the bottom of the inner pot is never heated beyond this temperature, preventing the ingredients from burning.
A: If the bottoms are not touching and heat input rate is significant (non-equilibrium), the outer pot boils, while heating the middle pot that finally boils, while heating the inner pot that finally boils.  If heat input is very slow, allowing equilibrium, they all boil together near enough.  Each inside must present nucleation centers (e.g., surface roughness) to prevent superheating.
If the bottoms are in intimate contact, only the outer pot loses heat by surface area.  The inner pot boils first (smallest volume, too), then the middle pot, then the outer pot.  Do I get an A for the assignment?
