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?
