= Solution
Treat the lake as a large <thermal reservoir> at $T_f=283.15\,\mathrm K$, and take the copper's <heat capacity> $C=50\,\mathrm{J\,K^{-1}}$ constant over the cooling interval from $T_i=783.15\,\mathrm K$. Although the actual cooling is irreversible, <entropy> is a state function, so its change can be calculated along a hypothetical reversible path:
$$
\Delta S_{\rm block}=\int_{T_i}^{T_f}\frac{C}{T}\,dT
=C\ln\left(\frac{T_f}{T_i}\right)
=\boxed{-50.9\,\mathrm{J\,K^{-1}}.}
$$
The lake receives $Q=C(T_i-T_f)=2.50\times10^4\,\mathrm J$ at constant <temperature>, giving
$$
\Delta S_{\rm lake}=\frac{Q}{T_f}=88.3\,\mathrm{J\,K^{-1}}.
$$
Therefore, neglecting other changes,
$$
\boxed{\Delta S_{\rm Universe}=\Delta S_{\rm block}+\Delta S_{\rm lake}
=37.4\,\mathrm{J\,K^{-1}}>0.}
$$
The positive total is the <entropy> generated by transferring <heat> across a finite <temperature> difference. The copper's <entropy> decreases, but the colder lake gains a greater amount.
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