Friday

July 25, 2014

July 25, 2014

Posted by **<3** on Sunday, April 21, 2013 at 7:36pm.

- Physics -
**Count Iblis**, Sunday, April 21, 2013 at 8:03pmThe 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

- Physics -
**<3**, Tuesday, April 23, 2013 at 2:40amThank 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.

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