# 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 $r$. – Maxim Umansky Dec 31 '13 at 7:06

## 1 Answer

No calculations are necessary, the First Law of Thermodynamics tells us the unheated end will never reach 800 degrees Celsius. As long you as there is heat loss along the length of the wire the unheated end can never reach the temperature of the heated end.