A pot of water is boiling under one atmosphere of pressure. Assume that heat enters the pot only through its bottom, which is copper and rests on a heating element. In two minutes, the mass of water boiled away is m = 2.5 kg. The radius of the pot bottom is R = 8.5 cm and the thickness is L = 2.0 mm. What is the temperature of the heating element in contact with the pot?

Answer 1

The specific latent heat of evaporation of water at 100 C is:

L= 2260 kJ/kg

The amount of heat that entered the pot in two minutes is thus:

Q = 2.5 kg * 2260 kJ/kg = 5650 kJ

Dividing this by the time gives you the power:

Q/(2 minutes) = 5650 kJ/(120 s) =

47.1 kW

If you divide this by the surface area, you get the heat flux:

47.1 kW/[pi (8.5 cm)^2] =

2.07*10^6 W/m^2

This heat flux is delivered to the water via thermal conduction, it is therefore equal to minus the gradient of the temperature in the copper inbetween the heating element and the water, times the heat conduction coefficient of copper.

If we take the heat conduction coefficient of copper to be

lambda = 385 W/(m K), then we have:

lambda (T_heating element - 100 C)/(2 mm) = 2.07*10^6 W/m^2 ------->

T_heating element = 110.8 C

Answer 2

Thank you. Your answer is correct, but I get a different answer when I try to calculate that last equation myself. I must be in kelvin or something.