The entropy change in melting `1g` of ice at `0^@C`. (Latent heat of ice is `80 cal g^(-1)`)

Options

A. `23.07 JK^(-1) mol^(-1)`
B. `20.07 JK^(-1) mol^(-1)`
C. `22.07 JK^(-1) mol^(-1)`
D. `21.07 JK^(-1) mol^(-1)`

Explanation

To find the entropy change in melting 1 gram of ice at \(0^\circ C\), we can follow these steps:

### Step 1: Understand the given data
- The latent heat of fusion of ice is given as \(80 \, \text{cal/g}\).
- We need to calculate the entropy change for melting \(1 \, \text{g}\) of ice.

### Step 2: Use the formula for entropy change
The formula for the change in entropy (\(\Delta S\)) during a phase change is given by:
\[ \Delta S = \frac{Q_{\text{rev}}}{T} \]
where:
- \(Q_{\text{rev}}\) is the heat absorbed or released during the process (in joules),
- \(T\) is the absolute temperature in Kelvin.

### Step 3: Convert temperature to Kelvin
The melting point of ice is \(0^\circ C\). To convert this to Kelvin:
\[ T = 0 + 273 = 273 \, \text{K} \]

### Step 4: Calculate the heat absorbed (Q)
Since we are melting \(1 \, \text{g}\) of ice, the heat absorbed (\(Q\)) is:
\[ Q = \text{latent heat} \times \text{mass} = 80 \, \text{cal/g} \times 1 \, \text{g} = 80 \, \text{cal} \]

### Step 5: Convert calories to joules
To convert calories to joules, we use the conversion factor \(1 \, \text{cal} = 4.184 \, \text{J}\):
\[ Q = 80 \, \text{cal} \times 4.184 \, \text{J/cal} = 334.72 \, \text{J} \]

### Step 6: Calculate the entropy change
Now, we can substitute \(Q\) and \(T\) into the entropy change formula:
\[ \Delta S = \frac{Q}{T} = \frac{334.72 \, \text{J}}{273 \, \text{K}} \approx 1.225 \, \text{J/K} \]

### Final Result
Thus, the entropy change in melting \(1 \, \text{g}\) of ice at \(0^\circ C\) is approximately:
\[ \Delta S \approx 1.225 \, \text{J/K} \]