Calculate change in entropy when `5g` of pure ice melts to form water at `0^(@)C` Given latent heat of ice is `80cal//g` at `0^(@)C`.

Answer

To calculate the change in entropy when 5 g of pure ice melts to form water at 0°C, we can follow these steps:

### Step 1: Identify the given values
- Mass of ice (M) = 5 g
- Latent heat of ice (L) = 80 cal/g
- Temperature (T) = 0°C = 273 K (in Kelvin)

### Step 2: Calculate the heat (Q) absorbed during the melting process
The heat absorbed when ice melts can be calculated using the formula:
\[ Q = M \times L \]
Substituting the values:
\[ Q = 5 \, \text{g} \times 80 \, \text{cal/g} = 400 \, \text{cal} \]

### Step 3: Convert heat from calories to joules
Since the SI unit of energy is joules, we need to convert calories to joules. The conversion factor is:
\[ 1 \, \text{cal} = 4.184 \, \text{J} \]
Thus,
\[ Q = 400 \, \text{cal} \times 4.184 \, \text{J/cal} = 1673.6 \, \text{J} \]

### Step 4: Calculate the change in entropy (ΔS)
The change in entropy can be calculated using the formula:
\[ \Delta S = \frac{Q}{T} \]
Substituting the values:
\[ \Delta S = \frac{1673.6 \, \text{J}}{273 \, \text{K}} \]

### Step 5: Perform the calculation
Now, calculating ΔS:
\[ \Delta S \approx 6.13 \, \text{J/K} \]

### Final Answer
The change in entropy when 5 g of pure ice melts to form water at 0°C is approximately **6.13 J/K**.

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