3933 watts
At 100 C (boiling point of water), it's density is 0.9584 g/cm^3. The volume of water lost is pi * 12.5^2 * 10 = 4908.738521 cm^3
The mass of water boiled off is 4908.738521 * 0.9584 = 4704.534999 grams.
Rounding to 4 significant figures gives me 4705 grams of water.
The heat of vaporization for water is 2257 J/g. So the total energy applied is
2257 J/g * 4705 g = 10619185 J
Now we need to divide that by how many seconds we've spent boiling water. That would be 45 * 60 = 2700 seconds.
Finally, the rate of heat transfer in Joules per second will be the total number of joules divided by the total number of seconds. So
10619185 J / 2700 s = 3933 J/s = 3933 (kg m^2/s^2)/s = 3933 (kg m^2/s^3)
= 3933 watts
