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First off some constants - The design has a base area of .0491 m

^{2}, or a diameter of 25 cm. The base material of the bottom of the pan before adding thermoelectrics is Stainless, with a thermal conductivity of 50.2 W/m*k. Heat is flowing from an open flame through the pot to water. We will assume it is boiling at 100C or 373K. The pan bottom is 5 mm thick.

I began with dQ/dt=m*Cp*dT/dt, where Cp is the specific heat if water, m is the mass of water (1kg for simplicity), T is temperature, and t is time. Through a messy derivation, I came to a final equation for t, the time it takes the water to boil by integrating from 20C to 100C, starting at room temp and heating to boiling. That equations was:

t=(ln(dT/To))/(((k/L)*10

^{-2})/Cp)

Where dT is the difference in temperature between the outside air under the pot and boiling water (assumed to be 80), To is room temperature, L is thickness, k is the conductivity of stainless, and 10

^{-2}is from unit conversions. This equation assumes a closed system (I know its not) and gives boiling times around 50 seconds. This value seems reasonable since with actual heat losses this number will rise to real world values.

My professor however would like me to calculate the outside temperature of the pot using a known heat flow. He has provided sources stating a fire can output 6 kW, of which 25% goes to the pot, or 1.5 kW. Using q=k*L(T

_{out}-T

_{in}) where T

^{in}is boiling water at 100C, this gives an absurd outside temperature of 6345 C.

Can anyone see a way to revise either of these methods???