= Solution
\b[(a)] Treat the lake as a <heat bath> at fixed <temperature> $T_0=288.15\,\mathrm K$ and the block's <heat capacity> as constant. Its initial <temperature> is $T_i=473.15\,\mathrm K$, and $C=mc=355\,\mathrm{J\,K^{-1}}$. Although the cooling is irreversible, <entropy> is a <state function>, so evaluate the block's change along an imagined <reversible thermodynamic process>:
$$
\Delta S_{\rm block}=\int_{T_i}^{T_0}\frac{C\,dT}{T}=C\ln\frac{T_0}{T_i}=-176.1\,\mathrm{J\,K^{-1}}.
$$
The lake receives <heat> $Q=C(T_i-T_0)=65675\,\mathrm J$, and therefore
$$
\Delta S_{\rm lake}=\frac{Q}{T_0}=227.9\,\mathrm{J\,K^{-1}}.
$$
Adding the two contributions gives
$$
\boxed{\Delta S_{\rm Universe}=C\left[\ln\frac{T_0}{T_i}+\frac{T_i-T_0}{T_0}\right]=51.9\,\mathrm{J\,K^{-1}}.}
$$
The positive result is consistent with the <Second law of thermodynamics> for <heat transfer> across a finite <temperature> difference. The calculation uses $c=710\,\mathrm{J\,kg^{-1}\,K^{-1}}$ exactly as printed in both the original PDF and the transcription; it does not replace the supplied value by a tabulated value for lead.
\b[(b)] The block starts and eventually finishes at the lake's <temperature>, so its net <entropy> change is approximately zero. Neglecting buoyancy, air resistance and the final immersion depth, the lost <gravitational potential energy> $mgh$ becomes <internal energy> in the lake after the motion and waves have dissipated. The lake remains at $T_0$, giving
$$
\boxed{\Delta S_{\rm Universe}\simeq\frac{mgh}{T_0}=\frac{(0.5\,\mathrm{kg})(9.81\,\mathrm{m\,s^{-2}})(200\,\mathrm m)}{288.15\,\mathrm K}=3.40\,\mathrm{J\,K^{-1}}.}
$$
Temporary heating of the block or temporary storage in <kinetic energy> does not change this final <energy> and <entropy> balance once <thermal equilibrium> is restored.
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