Recall the Stefan-Boltzmann formula:
$\begin{align*}\frac{\Delta Q}{\Delta t} = \sigma A T^4\end{align*}$
Here $\Delta Q$ / $\Delta t$ is the rate at which energy is radiated from the blackbody, $\sigma$ is the Stefan-Boltzmann constant, $A$ is the area of the blackbody, and $T$ is the temperature of the blackbody in $^{\circ}\mbox{K}$ . So, doubling $T$ increases the power radiated by the blackbody by a factor of $16$ . Assuming the heat capacity of the water is constant over the relevant temperature range, this means that the temperature would increase $8 ^{\circ}\mbox{K}$ instead of $0.5 ^{\circ}\mbox{K}$ . Therefore, answer (C) is correct.

Solution to 1992 Problem 14