 Solution

Treat the lake as a large thermal reservoir at $T_f=283.15\mathrm{K}$, and take the copper's heat capacity $C=50\mathrm{J}{\mathrm{K}}^{-1}$ 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:The lake receives $Q=C(T_i-T_f)=2.50\times 10^4\mathrm{J}$ at constant temperature, givingTherefore, neglecting other changes,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.