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ID: past-exam-of-the-thermal-physics-course-of-the-university-of-oxford/2016/5/solution

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Past exam of the thermal physics course of the University of Oxford / 2016 / 5 / Solution by

Codex  0 2026-10-25

Assume a stationary temperature profile, spatially uniform mass density and radioactive heat generation rate, constant thermal conductivity, isotropic solid thermal conduction, and blackbody emission into effectively cold surroundings. The volumetric source is

q_{v} = ρ h = 4.375 \times 1 0^{- 7} W m^{- 3} .

Total radiogenic power equals the surface blackbody radiation loss. The Stefan–Boltzmann law gives

\frac{4 π R ^{3}}{3} q_{v} = 4 π R^{2} σ_{SB} T_{S}^{4}, T_{S} = (\frac{q _{v} R}{3 σ _{SB}})^{1/4} = 22.52 K .

For steady radiogenic heating of a conducting sphere, the interior heat equation is

\frac{1}{r ^{2}} \frac{d}{d r} (r^{2} κ \frac{d T}{d r}) + q_{v} = 0.

Regularity at the centre removes the

1/ r^{2}

integration term and yields

d T / d r = - q_{v} r / (3 κ)

. Matching

T (R) = T_{S}

gives

T (r) = T_{S} + \frac{q _{v}}{6 κ} (R^{2} - r^{2}), T_{C} = T_{S} + \frac{q _{v} R ^{2}}{6 κ} = 371.4 K .

The high central temperature is compatible with a cold surface because finite thermal conductivity requires a large temperature gradient to carry the accumulated interior heat. A nonzero background radiation temperature

T_{bg}

replaces

T_{S}^{4}

by

T_{S}^{4} - T_{bg}^{4}

in the surface balance. The calculation neglects convection, stored transient heat, changing radioactive abundances, phase changes and any substantial external irradiation.

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