Past exam of the physics B course of the University of Cambridge / 2012 / Paper 2 / Section C / C11

Capacity convention. The printed equation requires $c$ to be volumetric heat capacity, in $\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. If “specific heat capacity” means per unit mass instead, the storage term is ${\rho }_{m}c{\partial }_{t}T$ and both denominators in the printed equation must contain ${\rho }_{m}c$. This distinction is fixed by dimensions.

For a fixed material region $V$, conservation of thermal energy gives the heat balance with volumetric heat generation,Use the divergence theorem, $\mathbf{J}=-\kappa \nabla T$ and constant coefficients. Since the region is arbitrary,The heat source is power per volume, while $\kappa$ is thermal conductivity rather than thermal diffusivity.

For uniform electrical conductivity and imposed current, current density is $j=I/(\pi a^2)$ and Joule heating is $H=j^2/\sigma$. A long, steady, cylindrically symmetric wire therefore satisfiesIntegration gives $T^{\prime }=-Hs/(2\kappa )+C/s$. Regularity at $s=0$ makes $C=0$, and the surface condition fixes the remaining constant. Thus the Joule-heated cylindrical wire hasHere $s$ denotes the printed radial coordinate $\rho$ and avoids confusing it with mass density. The centre is hotter by $I^2/(4{\pi }^2{a}^2\kappa \sigma )$.

For the air boundary, the internal flux at $s=a$ is $-\kappa T^{\prime }(a)=Ha/2$. Equate this to $\alpha [T(a)-T_{\mathrm{air}}]$. The Newton cooling of a Joule-heated wire givesMultiplication by circumference $2\pi a$ verifies that surface loss per length equals $I^2/(\pi a^2\sigma )$, the electrical power per length. The full air-cooled profile uses this surface temperature plus the same parabolic interior excess, assuming constant material properties and negligible end losses.