If water kept in an insultated vessel at `- 10^(@)C` suddenly freezes, the entropy change of the system

Options

A. decreases
B. increases
C. is zero
D. equals to that of the surroundings.

Explanation

To solve the problem of finding the entropy change of the system when water at `-10°C` suddenly freezes in an insulated vessel, we can follow these steps:

### Step 1: Understand the System
We have water at `-10°C` in an insulated vessel. An insulated vessel means that no heat can enter or leave the system. Therefore, the system is isolated from its surroundings.

**Hint:** Remember that in an insulated system, the heat exchange (Q) with the surroundings is zero.

### Step 2: Identify the Process
The water is freezing, which means it is undergoing a phase change from liquid to solid. The freezing process releases latent heat, but since the vessel is insulated, this heat cannot escape or enter.

**Hint:** Consider the implications of a phase change on the internal energy of the system.

### Step 3: Calculate the Change in Heat (Q)
Since the vessel is insulated, the heat exchange (Q) is zero. This means that the total heat change in the system during the freezing process is also zero.

**Hint:** Recall that for an insulated system, the heat transfer (Q) is zero, which affects the entropy calculation.

### Step 4: Calculate the Change in Entropy (ΔS)
The change in entropy (ΔS) of a system is given by the formula:
\[ \Delta S = \frac{Q}{T} \]
Where:
- \(Q\) is the heat exchanged (which is zero in this case)
- \(T\) is the absolute temperature in Kelvin

Since \(Q = 0\), we have:
\[ \Delta S = \frac{0}{T} = 0 \]

**Hint:** Remember that entropy change is directly related to heat transfer and temperature.

### Step 5: Conclusion
The change in entropy of the system when the water at `-10°C` freezes in an insulated vessel is zero.

**Final Answer:**
\[ \Delta S = 0 \]