= Solution
Initially and finally the block and lake are at the same <temperature>. After the falling block has come to rest, its lost <gravitational potential energy> has dissipated into <internal energy> of the lake and surroundings. In the idealization that all $mgh$ is ultimately thermalized in the lake at $T_0=283.15\,\mathrm K$,
$$
Q=mgh=(0.5)(9.81)(200)=981\,\mathrm J.
$$
The block returns to its original thermal state, so its net <entropy> change is zero. Mechanical <potential energy> carries no thermal <entropy> that must be subtracted. Thus
$$
\boxed{\Delta S_{\mathrm{Universe}}=\frac{mgh}{T_0}=3.46\,\mathrm{J\,K^{-1}}.}
$$
Use of $g=9.80\,\mathrm{m\,s^{-2}}$ gives the same three-significant-figure answer. Splashing and any temporary heating do not change this final-state balance if their mechanical <energy> is eventually dissipated in the same reservoir.
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