A hot penny is dropped into cold water inside a polystyrene foam cup. Assuming negligible heat loss to the atmosphere and the cup,Group of answer choicesthe decrease in entropy of the penny is more than the increase in entropy of the water.∣ ΔSpenny ∣ > ∣ ΔSwater ∣the entropy of both the penny and the water increases.the decrease in entropy of the penny is less than the increase in entropy of the water.∣ ΔSpenny ∣ < ∣ ΔSwater ∣the decrease in entropy of the penny is equal to the increase in entropy of the water.∣ ΔSpenny ∣ = ∣ ΔSwater ∣

The correct answer is "the decrease in entropy of the penny is less than the increase in entropy of the water. ∣ ΔSpenny ∣ < ∣ ΔSwater ∣".

Here's why:

Entropy is a measure of disorder or randomness in a system. When a hot penny is dropped into cold water, heat will flow from the penny (higher temperature) to the water (lower temperature) until thermal equilibrium is reached. This is in accordance with the second law of thermodynamics.

As the penny cools down, it loses heat and its entropy decreases. However, as the water heats up, it gains heat and its entropy increases.

In an isolated system like this, the total entropy should increase or, at the very least, remain the same (if the process is reversible). Therefore, the increase in entropy of the water must be greater than the decrease in entropy of the penny. Hence, ∣ ΔSpenny ∣ < ∣ ΔSwater ∣.