The entropy change can be calculated by using the expression, `DeltaS=(q_("rev"))/(T)`. When water freezes in a glass beaker, choose the correct statement amongst the following:

Options

A. `DeltaS_("system")` decreases but `DeltaS_("surroundings")` remains the same.
B. `DeltaS_("system")` increases but `DeltaS_("surroundings")` decreases.
C. `DeltaS_("system")` decrease but `DeltaS_("surroundings")` increases.
D. `DeltaS_("system")` decreases and `DeltaS_("surroundings")` also decreases.

Explanation

To solve the question regarding the entropy change when water freezes in a glass beaker, we can follow these steps:

### Step 1: Understand the Concept of Entropy
Entropy (S) is a measure of the disorder or randomness of a system. The change in entropy (ΔS) can be calculated using the formula:
\[ \Delta S = \frac{q_{\text{rev}}}{T} \]
where \( q_{\text{rev}} \) is the heat exchanged reversibly and \( T \) is the temperature in Kelvin.

### Step 2: Analyze the Process of Freezing
When water freezes, it transitions from a liquid state to a solid state. This process involves the release of heat (exothermic process) as the water molecules lose energy and arrange themselves into a more ordered solid structure (ice).

### Step 3: Determine the Change in Entropy for the System
- For the system (water freezing):
- The entropy of the system decreases because the liquid water (more disorder) is converting into solid ice (less disorder).
- Therefore, \( \Delta S_{\text{system}} < 0 \) (negative change).

### Step 4: Determine the Change in Entropy for the Surroundings
- For the surroundings:
- As the water freezes and releases heat, the surroundings gain this heat, leading to an increase in entropy.
- Therefore, \( \Delta S_{\text{surroundings}} > 0 \) (positive change).

### Step 5: Apply the Second Law of Thermodynamics
According to the second law of thermodynamics, the total entropy change of the universe (system + surroundings) must be greater than or equal to zero:
\[ \Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} \]
Since \( \Delta S_{\text{system}} < 0 \) and \( \Delta S_{\text{surroundings}} > 0 \), it is possible that \( \Delta S_{\text{total}} \) remains constant or increases, depending on the magnitudes of the changes.

### Step 6: Choose the Correct Statement
From the analysis:
- \( \Delta S_{\text{system}} \) decreases (negative).
- \( \Delta S_{\text{surroundings}} \) increases (positive).

Thus, the correct statement is that the entropy of the system decreases while the entropy of the surroundings increases.

### Final Answer
The correct option is: **ΔS system decreases but ΔS surrounding increases.**

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