# is this heat calculation equation correct?

If I have a line of copper wire (lets say $\textrm{1 meter}$ long, $\textrm{1 mm}$ thick) and one end is a flattened disk of copper about the size of a quarter, and I apply a lot of heat to it (I'm talking $800\,^{\circ}\textrm{C}$) will the entire line be heated to the same degree? I mean what temperature will the unheated end be after, say, a minute? Can it too reach $800\,^{\circ}\textrm{C}$ degrees over time?

I'm going to start by asking the question, what would happen if the cool end were at $400\,^{\circ}\textrm{C}$ ? In this case, the rate of heat flow from the hot to the cool end would be \begin{align} \frac{k A}{l}\Delta T &= \frac{400\,\textrm{W/mK}\cdot \pi\,(0.0005\,\textrm{m})^2}{\textrm{1 m}} \cdot 400\,\textrm{K} \\ &= 0.1257\,\textrm{W} \end{align} The radiative transfer from the copper to the surroundings, which I'll call air at $20\,^{\circ}\textrm{C}$, will follow from the Stefan-Boltzmann Law. For the copper, the radiative flux is \begin{align} \sigma \, T^4 &= \left( 5.67 \times 10^{-8}\,\textrm{W/m}^2 \textrm{K}^4 \right) \left(673\,\textrm{K}\right)^4 \\ &= 11632\,\textrm{W/m}^2 \end{align} For the back flux from the air (disregarding any convection), you have $\left( 5.67 \times 10^{-8}\,\textrm{W/m}^2 \textrm{K}^4 \right) \left(293\,\textrm{K}\right)^4 = 418\,\textrm{W/m}^2,$ so if the disk has a radius of $1\,\textrm{cm}$ and is two-sided, its surface area is $2\,\pi \,\left(0.01 \,\textrm{m}\right)^2 = 6.28 \times 10^{-4}\,\textrm{m}^2,$ and the $\textrm{NET}$ radiative loss is about $\left( 11214\,\textrm{W/m}^2 \right) \left( 6.28 \times 10^{-4}\,\textrm{m}^2 = 7.04\,\textrm{watts}. \right)$ Evidently the radiative loss would be a lot more than the conductive gain, so the equilibrium temperature of the "disk" end is going to be considerably lower than $400\,^{\circ}\textrm{C}$.

Next I tried a formal solution, but I didn't like the result(!) so I'll just see what happens if the disk temperature is $100\,^{\circ}\textrm{C}$:Radiative loss $= 5.67 \times 10^{-8}\cdot \left( 373^4 - 293^4 \right) = 680\,\textrm{W/m}^2$, total of $0.427\,\textrm{watts}$. Conductive gain $= 400\,\pi\, 0.0005{^2} \cdot 700 = 0.220\,\textrm{watts}.$ So, as a whole, if the copper wire were thicker(lets say ten times so, placing it at $1\,\textrm{cm}$), would it enable it to reach $800\,^{\circ}\textrm{C}$ as a whole?

• For given parameters (L=1 m, r=0.5e-3 m), the side surface of the wire $2 \pi r L$ ~ 3e-3 m$^2$ is larger than the disk area ~ 0.6e-3 m$^2$ so the disk is not relevant. A thicker wire would be at higher temperature since the heat flux along the wire scales as $r^2$ and heat losses through the side surface scale as $r$. Dec 31, 2013 at 7:06