Posted by **Daniel** on Friday, April 22, 2011 at 3:34am.

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?

- Physics -
**drwls**, Friday, April 22, 2011 at 6:18am
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)

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