Past exam of the physics B course of the University of Cambridge / 2018 / Paper 2 / Section C / C12

Origin of diffusion. In a material near local thermal equilibrium, microscopic carriers transfer energy from hotter regions to colder regions. On distances long compared with their mean free path and times long compared with relaxation times, the local constitutive law is Fourier's law,Combining this temperature-gradient-driven flux with local conservation of energy produces a diffusion equation. This description assumes diffusive thermal conduction; it is not generally valid for ballistic transport, and flow would add advective transport.

Energy balance with heating. Choose any fixed volume $V$, let $c$ be its uniform volumetric heat capacity and $H$ its heat generation rate per unit volume. The rate of increase of internal energy equals heat generation minus outward heat flux:Since $V$ is arbitrary, for constant thermal conductivityThe ratio $\kappa /c$ is the thermal diffusivity in ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. Here $c$ is in $\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$, $\kappa$ in $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and $H$ in $\mathrm{W}{\mathrm{m}}^{-3}$, so both right-hand terms have units $\mathrm{K}{\mathrm{s}}^{-1}$. If thermal conductivity varies with position or temperature, the divergence $\nabla \cdot (\kappa \nabla T)$ must be retained rather than replacing it by $\kappa {\nabla }^{2}T$.

Distinguish electrical current density ${\mathbf{J}}_e$ from heat flux ${\mathbf{J}}_Q$. Uniform current across the circular section givesby Ohm's law. The local Joule heating rate is electrical work per unit volume and time:For a segment of length $\ell$, integration gives $H\pi a^2\ell =I^2\ell /(\sigma \pi a^2)=I^{2}R$, confirming the usual resistor power. The model takes electrical conductivity constant and ignores temperature-induced current redistribution.

In a long wire away from its ends, the steady temperature is independent of axial position and azimuth. The heat equation reduces toMultiplying by $\rho$ and integrating givesA finite temperature and nonsingular flux at the axis require $C_1=0$. Applying $T(a)=T_0$ determines $C_2=T_0+H{a}^2/(4\kappa )$. ThereforeThis is the steady temperature of a uniformly heated cylindrical wire. Its hottest point is at the centre, with $T(0)-T_0=I^2/(4{\pi }^2{a}^2\kappa \sigma )$. The outward heat flux is $J_Q(\rho )=-\kappa T^{\prime }(\rho )=H\rho /2$, so the power crossing a cylindrical surface per unit length, $2\pi \rho J_Q(\rho )=H\pi {\rho }^2$, equals the heating generated inside it.

Replacing the fixed surface temperature by Newton's law of cooling changes the boundary condition but not the radial derivative in steady state. At the surface,HenceHere $\gamma$ is a surface heat transfer coefficient in $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$, unrelated to the surface tension symbol used in C10. An independent energy-balance check uses a length $\ell$: the electrical input $I^2\ell /(\sigma \pi a^2)$ must equal the surface loss $2\pi a\ell \gamma [T(a)-T_{\mathrm{air}}]$, which gives the same answer.

The complete temperature profile isThermal conductivity controls the internal gradient, but cancels from the surface-temperature rise because all generated power must eventually cross the prescribed surface. This assumes positive constant $\gamma$, uniform electrical conductivity and negligible end or additional radiative losses; temperature-dependent material properties could make the heating and boundary problem nonlinear.