Plutonium hand warmer

A ball of mass $m = 1\,\text{kg}$ has a radius $R$ , so that $m = \rho \cdot \frac{4}{3} \pi R^3 \quad \Rightarrow \quad R = \sqrt[3]{\frac{3 m}{4 \pi \rho}} \approx 2.29\,\text{cm}$ The heat source density equals $\dot q \approx 570\, \frac{\text{W}}{\text{kg}} \cdot 19,800 \,\frac{\text{kg}}{\text{m}^3} \approx 1.13 \cdot 10^7 \frac{\text{W}}{\text{m}^3}$ The heat equation $- \lambda \Delta T(r) = - \lambda \left(T''(r) + \frac{2}{r} T'(r)\right) = \dot q$ can be solved by $T(r) = \alpha r^2 + T_0$ . Substitution in the differential equation yields $\begin{aligned} & & \Delta T &= T''(r) + \frac{2}{r} T'(r) = 6 \alpha = - \frac{\dot q}{\lambda} \\ \Rightarrow & & T(r) &= - \frac{\dot q}{6 \lambda} r^2 + T_0 \\ \Rightarrow & & T_\text{ext} &= - \frac{\dot q}{6 \lambda} R^2 + T_0 \\ \Rightarrow & & T_0 &= T_\text{ext} + \frac{\dot q}{6 \lambda} R^2 \approx 167^\circ\,\text{C} \end{aligned}$