Solution
= Solution
If the <temperatures>, exposed <area>, and surroundings remain fixed, the net radiated <energy> over one day is
$$
\boxed{E=P_{\rm net}(86400\,\mathrm s)\simeq1.60\times10^7\,\mathrm J.}
$$
Using the supplied conversion gives
$$
\boxed{E\simeq3.82\times10^6\,\mathrm{cal}=3.82\times10^3\,\mathrm{kcal}.}
$$
A dietary Calorie is one <kilocalorie>, so the result is about 3820 dietary Calories, rather than 3.82 million dietary Calories.