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?

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.

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.

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)...

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.

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.

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.

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