Final answer:
Without specific heat transfer details, it's impossible to accurately calculate the entropy change for the scenarios described. The entropy change for a rock and lake at equilibrium is zero, and for a hotter rock, entropy increases, but a precise value would require more information.
Step-by-step explanation:
The student's question involves calculating the change in entropy of the universe when a rock is dropped into a lake from a significant height.
The scenarios given are (a) where the rock and the lake are at the same temperature and (b) where the rock is at a higher temperature than the lake.
Assuming air friction is negligible, when the rock at the same temperature as the lake (20°C) is dropped from a height of 1.0×103 m, it will gain kinetic energy, which is then converted to thermal energy upon impact.
However, since the rock and the lake have the same initial temperature, the system is in thermal equilibrium; therefore the entropy change of the universe is zero.
For scenario (b), if the temperature of the rock is 100°C when dropped, its kinetic energy upon impact will still be converted to thermal energy, which will now be transferred to the cooler lake.
This transfer of heat increases the entropy of the lake (and overall universe) due to the dissipation of thermal energy. However, without specific details about the heat capacity of water and the heat transfer involved, it is not possible to compute the exact entropy change.
To solve such problems, we typically need to apply thermodynamics concepts and equations related to entropy change, heat transfer, and specific heat capacity.
The provided answer choices, however, do not necessarily correspond to a complete understanding or correct calculation of the scenarios posed without this additional information.