Is more water lost when it boils in a larger pot? Suppose you have two otherwise-identical pots of different radii on two identical burners, both large enough to heat the entire bottoms of both pots. If you want exactly 1L of boiling water (say, 373.15K) in each pot, to which pot should you add more water?
Essentially, my question is about which effect is greater: the increased evaporation for a larger surface area vs. the increased rate of heat transfer over that surface area in the bigger pot (decreasing the time required to boil the water, at which point measurement ends). Would a pot large enough to heat each molecule of water at the same time be less water-efficient than a pot 1 molecule wide, which can only allow 1 molecule of water to evaporate at a time?
 A: I came up with the following rough qualitative argument. Throughout I assume that both pots are surrounded by identical air, and in particular, identically humid air. This last assumption could be problematic, but without it the answer will depend strongly on the details of the room etc.
The evaporation rate (kg / time)  will depend linearly on the number of molecules exposed to the air (moles), which will itself be linear in the surface area exposed to the air (m^2): $M/t \propto A_\mathrm{top}$, $M$ being the total mass of the water, $A_\mathrm{top}$ being the area of the top of the pot.
To bring the pot to boiling we need to deliver a certain amount of energy (Joules) to the pot from the burner. The total energy needed is linear in the total mass (kg) of the water. With temperature held constant the mass is linear in the volume (m^3): $Q \propto M \propto V$.
The heat transfer rate (Joules/s) obeys $Q/t \sim A_\mathrm{bot} \Delta T$, where $A_\mathrm{bot}$ is the surface area of the bottom of the pot and $\Delta T$ the temperature difference between the burner and the water. I assume the pot is slowly stirred so that convective transfer through the water is essentially instantaneous. Dropping this assumption complicates things considerably, but I'll try and relax it later.
It will take about $t_\mathrm{boil} \sim Q/ A_\mathrm{bot} \Delta T$ seconds to bring the pot to boiling. The mass of water lost during this time will then follow $M \propto A_\mathrm{top} t_\mathrm{boil} \sim A_\mathrm{top} Q / A_\mathrm{bot} \Delta T$. For a cylindrical pot with $A_\mathrm{top} = A_\mathrm{bot}$ the area should thus drop out, and both pots should require equal amounts of water to have equal volume at the boiling point (at the boiling point, as at any constant temperature, the volume of water in the pot should depend on the mass in the same way for both pots).
Roughly speaking, the rate of convective heat transfer between two fluid elements with an interface area $A$ is also linear in $A$, so this argument should hold to low order even if convection starts to be important. I'm not sure what will happen once say convective currents and asymmetries start to be the dominant effect, so this won't be true of any two pots probably. For example the ocean would probably evaporate differently than a column of water the size of your pinky finger with the volume of the ocean.
Edit:
I failed to take into account heat loss from the tops of the pots. Given that both pots are surrounded at all times by "the same air", this changes the heat transfer rate to $Q/t = A_\mathrm{bot} \Delta T_1 - A_\mathrm{top} \Delta T_2$, where $\Delta T_1$ ($\Delta T_2$) is between the burner and the water (between the air and the water), and so the area will still drop out of the quotient. It seems to me it still will, even if the possibly time-dependent humidity and temperature of the air are taken into account.
A: Very qualitative. Assuming liquid/vapour equilibrium is not reached, that is air is sufficiently dry, in a given time more water molecules will escape from the larger pot. This effect should take place independently from heating the container or not.
Bottom counts only if the pot are not cylinder but bottle shaped. I would conclude that you need less energy to finally attain 1 l boiling water if the recipient is small. I assume no heat loss as well as burner are identical in term of heat transfer (electrical plate at same T and fitting the bottom ). 
