Past exam of the physics B course of the University of Cambridge / 2019 / Paper 2 / Section A / A1 / b / 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 $T_0=283.15\mathrm{K}$,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. ThusUse of $g=9.80\mathrm{m}{\mathrm{s}}^{-2}$ 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.