Chemistry - Thermodynamics - II Question with Solution | TestHub

Explain the question:

Determine the nature of entropy changes for the system and surroundings when ice and hot water are mixed in a thermally insulated container.

Concept:

1. For a system in a thermally insulated container, no heat is exchanged with the surroundings, so $q_{surr}=0$, which implies $\Delta S_{surr}=0$.

2. Mixing substances at different temperatures is a spontaneous irreversible process, which always leads to a net increase in entropy of the isolated system ($\Delta S_{system}>0$) according to the Second Law of Thermodynamics.

Solution:

Surroundings: The insulation prevents any heat transfer between the system (ice + water) and the outside world. Since

the entropy of the surroundings remains unchanged: $\Delta S_{surr}=0$.

System: Inside the container, heat spontaneously flows from the hot water to the ice. The ice gains entropy by melting and warming up, while the hot water loses entropy as it cools. For a spontaneous process in an isolated system, the total entropy change must be positive.

Calculations show that the entropy gain of the ice ($+11.04J/K$ for melting and $\approx +3.16J/K$ for warming to $\approx 24$℃) exceeds the entropy loss of the water ($\approx -12.63J/K$ for cooling to $\approx 24$℃), resulting in a net $\Delta S_{system}>0$.

Final Answer: Option (1)