= Solution
To obtain the <maximum work from two finite heat reservoirs>, operate a <reversible heat engine> between the bodies until their <temperatures> coincide. Include the <work> store in the isolated total system: no <heat> enters from outside, while the bodies can transfer <energy> to that store. With the stated constant <heat capacities>, reversibility requires
$$
C_1\ln\frac{T_f}{T_1}+C_2\ln\frac{T_f}{T_2}=0,
\qquad T_f=T_1^{C_1/(C_1+C_2)}T_2^{C_2/(C_1+C_2)}.
$$
Numerically, $T_f=450^{5/8}273^{3/8}=373.094\,\mathrm K$. The decrease of the bodies' <internal energy> is
$$
W_{\max}=C_1T_1+C_2T_2-(C_1+C_2)T_f=\boxed{2.1061\times10^4\,\mathrm J\simeq21.1\,\mathrm{kJ}.}
$$
Direct thermal equilibration without an engine would instead reach the arithmetic heat-capacity-weighted <temperature> $383.625\,\mathrm K$ and deliver no <work>. <entropy production> can only raise the final common <temperature> above the value for zero <entropy production> and reduce extractable <work>.
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