Question:
\text{The entropy change can be calculated by using the expression } \Delta S = \frac{q_{rev}}{T}
\text{When water freezes in a glass beaker, the correct statement among the following is:}
\text{1. } \Delta S \text{ (system) decreases but } \Delta S \text{ (surroundings) remains the same.}
\text{2. } \Delta S \text{ (system) increases but } \Delta S \text{ (surroundings) decreases.}
\text{3. } \Delta S \text{ (system) decreases but } \Delta S \text{ (surroundings) increases.}
\text{4. } \Delta S \text{ (system) decreases but } \Delta S \text{ (surroundings) also decreases.}
Solution:
$\text{Hint: Entropy is the degree of randomness.}$
$\text{To determine the entropy changes when water freezes, we need to analyze what happens to both the system (water) and the surroundings:}$
$\text{1. System (Water):}$
$\text{When water freezes, it transitions from a liquid state to a solid state (ice). In the liquid state, water molecules have more freedom of movement and can arrange themselves more randomly. In the solid state (ice), water molecules are locked into a rigid crystalline structure with much less randomness.}$
$\text{Since entropy is a measure of randomness or disorder, the entropy of the system decreases when water freezes:}$
$\Delta\text{S}_{\text{system}} < 0$
$\text{2. Surroundings:}$
$\text{Freezing is an exothermic process, which means heat is released from the system to the surroundings. The heat released during freezing can be represented as the enthalpy of fusion (with a negative sign since the process is reverse of melting):}$
$\text{q}_{\text{released}} = -\Delta\text{H}_{\text{fusion}}$
$\text{This released heat is absorbed by the surroundings, causing an increase in the entropy of the surroundings according to the formula:}$
$\Delta\text{S}_{\text{surroundings}} = \frac{\text{q}_{\text{released}}}{\text{T}} = \frac{-\Delta\text{H}_{\text{fusion}}}{\text{T}} > 0$
$\text{Since}$ $-\Delta\text{H}_{\text{fusion}}$ $\text{is positive (heat is released) and T is always positive, the entropy change of the surroundings is positive.}$
$\text{Therefore, when water freezes:}$
$\text{- The entropy of the system (water) decreases}$
$\text{- The entropy of the surroundings increases}$
$\text{This corresponds to option 3: }\Delta \text{ S (system) decreases but } \Delta \text{ S (surroundings) increases.}$