= Solution
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
$$
q_v=\rho h=4.375\times10^{-7}\,\mathrm{W\,m^{-3}}.
$$
Total radiogenic <power> equals the surface <blackbody radiation> loss. The <Stefan–Boltzmann law> gives
$$
\frac{4\pi R^3}{3}q_v=4\pi R^2\sigma_{\rm SB}T_S^4,\qquad
\boxed{T_S=\left(\frac{q_vR}{3\sigma_{\rm SB}}\right)^{1/4}=22.52\,\mathrm K.}
$$
For <steady radiogenic heating of a conducting sphere>, the interior <heat equation> is
$$
\frac1{r^2}\frac d{dr}\left(r^2\kappa\frac{dT}{dr}\right)+q_v=0.
$$
Regularity at the centre removes the $1/r^2$ integration term and yields $dT/dr=-q_vr/(3\kappa)$. Matching $T(R)=T_S$ gives
$$
T(r)=T_S+\frac{q_v}{6\kappa}(R^2-r^2),\qquad
\boxed{T_C=T_S+\frac{q_vR^2}{6\kappa}=371.4\,\mathrm K.}
$$
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 $T_{\rm bg}$ replaces $T_S^4$ by $T_S^4-T_{\rm bg}^4$ in the surface balance. The calculation neglects convection, stored transient <heat>, changing radioactive abundances, phase changes and any substantial external irradiation.
Back to article page
