# Heat problem and unreasonable heat rates

## Main Question or Discussion Point

Here is the question verbatim:
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.
For the first part, finding the rate of conduction, the formula is:
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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For a start, you've got the diameter in centimetres but the thickness in metres...

Jess

Even if that change that to the correct units, the result energy rate is just huge. I'll edit the post to get a new answer.

0.5 MW isn't that big - and don't forget that the thermal conductivity of copper is unusually high. As far as I can tell from a quick dimensional analysis the units are all okay, so the answer should be correct.

Jess

Lets convert that to electric energy though to see what our connection looks like. Let's assume that this burner is electric.
Standard voltage for a range is 240V so I'll assume the stove is 240.
P = IV
I = P/V
I = 551349/240
I = 2297 amps

Now that seems way beyond unreasonable. An entire house only uses like 100 amps. 2300 is just... wow. I can't even imagine that kind of power.

Sorry to point out the obvious, but this isn't an electrical conductivity question, so that comparison just doesn't stand up.

The thing is, in this question the temperature gradient is high - 20,000 K/m, which is what gives the high rate of thermal conduction.

Jess

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Are you sure though? I mean 0.5MW is just huge. At the rate of 0.5MW, the water in the pot would evaporate at a rate of 146L per minute. It seems almost impossible to fill that requirement for energy.

The 0.5MW is only the heat conduction rate - it says nothing about how much of that heat is actually transferred to the water. There'll be losses and all sorts. I suspect that these points (plus some others that haven't occurred to me) will make up the second part of part (b)...

Jess

Integral
Staff Emeritus
Gold Member
I am not sure where you got your number for thermal conductivity, My CRC Handbood of Chem and Physcis gives it as 4.83W/(cmC). What are the units on your number? It is pretty meaningless without them
When I do your calculation I get 68W.

Redo you calculation, ensure that ALL of your units agree with the units of your thermal conductivity.

Integral
Staff Emeritus
Gold Member
Originally posted by Jess
Sorry to point out the obvious, but this isn't an electrical conductivity question, so that comparison just doesn't stand up.

The thing is, in this question the temperature gradient is high - 20,000 K/m, which is what gives the high rate of thermal conduction.

Jess
It certianly does, the Wattage going into the pot cannot be greater then that produced by the heating element. Perhaps you need to refresh your basic concepts.

This was an execellent test to show that the calculated number was off.

Originally posted by Integral

Redo you calculation, ensure that ALL of your units agree with the units of your thermal conductivity.
Dimensionally at least, 0.5MW is correct

Jess

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Originally posted by Integral
I am not sure where you got your number for thermal conductivity, My CRC Handbood of Chem and Physcis gives it as 4.83W/(cmC). What are the units on your number? It is pretty meaningless without them
When I do your calculation I get 68W.
I found values ranging from 385 W /m K to 400 W/ m K. So I suspect this is the difference between mks and cgs systems of units.

I don't get 68W, even when I use cgs units - unless I use cm for the thickness and m for the diameter, interestingly.

Jess

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krab
quote: 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.
For the first part, finding the rate of conduction, the formula is: Q/t = kAT/d the conductivity of copper is given in a table on a previous page: 390 so Q/t = (390)((0.15^2))(150 - 100) / (0.0025) Q/t = 551349W (an astronomical heat transfer rate)
So the rate is unreasonable. The problem asks you to explain why. You've done the right calculations: they indicate that sustaining a temperature difference of 50 degrees with a typical 1 kW input power is not possible.

Integral
Staff Emeritus
Gold Member
You are correct, I had m in the numerator, cm in the denominator.

The 4.83W/(Cm C) is for Copper at 100C so should be a very good one for this problem.

The result comes from the temperature differential. This number corresponds to the power input required to maintain a 50C temperature differential across the bottom of the pot.

Watch what happens to a burner on an electric stove. When you turn it on high, it turns red, as soon as you place a pot of water on it, the color vanishes. This is because the temperature of the burner has dropped to that of the bottom the pan, with a temperature difference that is determined by the power available. This problem shows that if you had a .5MW stove you could maintain a 50C differential. It is more likely that your stove produces .5KW So the differential for that same pot in your kitchen will be closer to .05C