#### ShawnD

Science Advisor

- 638

- 1

Here is the question verbatim:

Q/t = kA[del]T/d

the conductivity of copper is given in a table on a previous page: 390 so

Q/t = (390)([pi](0.15^2))(150 - 100) / (0.0025)

Q/t = 551349W (an astronomical heat transfer rate)

Without even going to part B I can see that this value isn't right. What I was thinking is that maybe this value I just calculated out is a MAXIMUM heat transfer rate which can happen and it assumes that infinite energy can be supplied to the system.

I was looking at it sort of like a little D size battery. The battery is 1.5v but if I put a copper wire from 1 side to the other which has a resistance of almost nothing, the current is not even close to the hundreds of amps that would be expected from I = v/r. Just like the D cell, the burner can only supply so much energy to the system. If the burner actually was able to transfer 551349W of power to the copper pot, it would.

Is my reasoning correct or is there something else I'm missing?

For the first part, finding the rate of conduction, the formula is:A copper teakettle with a circular bottom 30cm in diameter has a uniform thickness of 2.5mm. It sits on a burner whose temperature is 150C. (a) If the teakettle is full of boiling water, what is the rate of heat conduction through its bottom? (b) Assuming that the heat from the burner is the only heat output, how much water is boiled away in 5.0 min? Is your answer reasonable? If not, explain why.

Q/t = kA[del]T/d

the conductivity of copper is given in a table on a previous page: 390 so

Q/t = (390)([pi](0.15^2))(150 - 100) / (0.0025)

Q/t = 551349W (an astronomical heat transfer rate)

Without even going to part B I can see that this value isn't right. What I was thinking is that maybe this value I just calculated out is a MAXIMUM heat transfer rate which can happen and it assumes that infinite energy can be supplied to the system.

I was looking at it sort of like a little D size battery. The battery is 1.5v but if I put a copper wire from 1 side to the other which has a resistance of almost nothing, the current is not even close to the hundreds of amps that would be expected from I = v/r. Just like the D cell, the burner can only supply so much energy to the system. If the burner actually was able to transfer 551349W of power to the copper pot, it would.

Is my reasoning correct or is there something else I'm missing?

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