 Solution

For an ideal blackbody surface completely surrounded by an isothermal environment, the Stefan–Boltzmann law gives emitted power $A\sigma T_s^4$ and absorbed power $A\sigma T_e^4$. The net outward radiative heat transfer is thereforeThe gross emitted power is about $1022\mathrm{W}$, but about $837\mathrm{W}$ is absorbed back from the surroundings. The difference, rather than the emitted term alone, is the relevant energy loss.