Answer:
The rate of the heat transfer from the pan is 3.93kW
Explanation:
3933 watts At 100 C (boiling point of water),
density is 0.9584 g/cm^3
diameter = 25cm
The volume of water lost = πr²h
V = π * 12.5² * 10 = 4908.738521 cm³
The mass of water boiled off = density * volume
mass = 4908.738521 * 0.9584
= 4704.534999 grams.
Rounding to 4 significant figures = 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 or 3.93kW
The rate of the heat transfer from the pan is 3.93kW
