Some cooking pan whose bottom is made out of copper has a radius 0.1 m and thickness 0.003 m. The water in the pan boils at 1.7e-3 kg/s. How much higher is temperature of the outside surface of the pan's bottom compared to the inside surface?

Answer 1

The heat transfer rate is

Power = 1.7 g/s * 540 cal/g
= 918 cal/s
= 3841 J/s

Power = k*A*dT/dx,
where k is the thermal conductivity of copper (which you need to look up) and A is the pan's bottom area, 0.03142 m^2.

Solve for the temperature gradient, dT/dx = Power/(k*A).

The temperature difference between inside and outside surfaces of the copper is the copper thickness x, multiplied by dT/dx.
T2 - T1 = x*Power/(k*A)