= Solution
Let $Q_c$ be <heat> extracted from the cold reservoir, $Q_h$ the <heat> delivered to the surroundings and $W$ the supplied work. The <entropy> inequality for a cyclic <refrigerator> is
$$
\frac{Q_h}{T_h}-\frac{Q_c}{T_c}\geq0.
$$
Equality is the reversible <Carnot refrigerator> limit. Therefore
$$
\boxed{Q_{h,\min}=\frac{300}{4}\times1\,\mathrm J=75\,\mathrm J.}
$$
<energy> conservation gives $W_{\min}=Q_h-Q_c=74\,\mathrm J$. The requested rejected <heat> is \b[75 J], while \b[74 J] is the minimum work; both have to be distinguished. Any <entropy> production increases the required rejection at these fixed reservoir <temperatures>.
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