Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 75 / 2 / c / Solution

A linear, or “three-thirds”, boundary heat transfer closure needs its own dimensional conductance $L>0$, so $q=L\Delta$. One cannot simply keep the dimensions of the previous nonlinear coefficient $K$. With equal conductances at both boundaries, the mean-temperature equation becomesIntegrating this linear ordinary differential equation givesThe relaxation time is $\rho c_{p}h/(2L)$. If the conductances differed, conservation of energy would instead select their weighted boundary temperature; the arithmetic mean follows from the assumed upper/lower symmetry.