# Applying the Fourier law of heating for abrupt temperature spikes in a thermal circuit

The figure shows the tungsten filament with a constant diameter except for a piece of it which has half of the diameter as the rest of the wire. Assume the temperature is constant within each part and changes suddenly between the parts. If the temperature of the thick part is 2000K, the temperature of the thin part of the filament is as:

I considered applying Fourier law hereby assuming the temperature of the middle part as $$T$$ and the thick part taking the midpoint of the thin part than from the midpoint of the thin part to end thick rod. After that assuming steady-state and approximating derivative as finite difference (*)

$$\frac{dQ_{in} }{dt} = \frac{dQ_{out} }{dt}$$

By fouriers law:

$$\frac{dQ}{dt} = - k A \frac{dT}{dx}$$

Hence,

$$-k A \frac{\Delta T_{left}}{\Delta x} = -kA \frac{\Delta T_{right}}{\Delta x}$$

Now if I since I took the midpoint of the rod, the $$\Delta x$$ cancel, the areas cancel and conductivity cancels. This leaves me with:

$$\Delta T_{left} =\Delta T_{right}$$

Now, the temperature difference from middle part to right (I.e: $$\Delta T_{right}$$ )is given as $$2000-T$$ and for left part it is given as $$(\Delta T_{left})$$ which is $$T-2000$$:

$$T-2000 = 2000 -T$$

Hence,

$$T=2000$$

But... this is wrong?

Self-Research attempts:

In solutions from some websites, they equate the term which looks like a joules heating effect in current electricity to the Stefan Boltzmann law. I have written the equation they've used below:

$$( \frac{dQ}{dt})^2 R = \sigma A T^4$$

Where R is the thermal resistance

As far as I understand, the Stefan Boltzmann law is regarding radiation and the Fourier heat law is heat transfer through thermal conduction. I assume that they derived the equation of $$( \frac{dQ}{dt} )^2 R$$ using Fourier's law, this makes it even more confusing of how these two forms of heat transfer are equal.

*: I feel a bit uneasy doing this because it is written in the question that the temperature spike is sudden so the function must be discontinuous at that point

• Are you saying that the temperature of the two thick parts are both 2000 and you are wondering what the temperature of the thin part in the middle is? Oct 9, 2020 at 11:31
• Yes, that's the objective Oct 9, 2020 at 11:31
• Then I would say you have successfully accomplished your objective. What did you expect the temperature in the middle section to be? Oct 9, 2020 at 11:35
• Supposedly the correct answer is : $8^{\frac14} 2000$ as is shown in the solutions which I found from some website Oct 9, 2020 at 11:37
• Sorry. I have no idea what they are talking about. Oct 9, 2020 at 11:45

If the filament is operating as part of an electrical circuit, then the electrical current I has to be the same in all three sections. If the inherent resistivity of the wire is $$\rho$$, then the resistance of the entire thin section is $$\frac{\rho L}{A}$$, where L is the length of the thin section and A is its cross sectional area. So the rate of heat generation in the thin part is $$Q=I^2\frac{\rho L}{A}$$This has to be equal to the rate of heat removal:$$I^2\frac{\rho L}{A}=\sigma \pi DLT^4$$So, $$T^4=\frac{I^2\rho }{\sigma \pi D A}$$The product DA is 8 times larger in the thick section than in the thin section. So $$T^4$$ is 8 times larger in the thin section than the thick section.

• You had told that it was incorrect to say that the electric current heating is equivalent heating but then you wrote the same.. Oct 9, 2020 at 14:39
• In this case, instead of the internal energy of the filament increasing, the electrical energy enters and heat exits. Oct 9, 2020 at 14:41
• So, what was different previously which made this idea fail? Oct 9, 2020 at 14:44
• I don't recall the example you are citing. Oct 9, 2020 at 14:46
• You had written this : "They have assumed that the rate of heat generation from electrical current stays the same in the thin center section as in the two end sections. This can't be correct" in the comments of the original post Oct 9, 2020 at 14:47