The heat of fusion is `334.7Jg^(-1)`. The entropy change in `JK^(-1)kg^(-1)` in melting of 1 g of ice at `0^(@)C` is

Options

A. 1.226
B. 0.293
C. 293
D. 1226

Answer

To find the entropy change during the melting of 1 gram of ice at 0°C, we can follow these steps:

### Step 1: Identify the Heat of Fusion
The heat of fusion (Q) for ice is given as 334.7 J/g. This means that to melt 1 gram of ice, 334.7 Joules of energy is required.

### Step 2: Convert the Heat of Fusion to Kilograms
Since we need the entropy change in J/K/kg, we will convert the heat of fusion from J/g to J/kg.

1 gram = 0.001 kg, so:
\[ Q = 334.7 \, \text{J/g} \times 1000 \, \text{g/kg} = 334700 \, \text{J/kg} \]

### Step 3: Determine the Temperature in Kelvin
The temperature at which the melting occurs is given as 0°C. To convert this to Kelvin:
\[ T = 0°C + 273.15 = 273.15 \, K \]

### Step 4: Calculate the Entropy Change
The formula for the change in entropy (ΔS) is given by:
\[ \Delta S = \frac{Q}{T} \]
Substituting the values we have:
\[ \Delta S = \frac{334700 \, \text{J/kg}}{273.15 \, K} \]

### Step 5: Perform the Calculation
Now we can perform the calculation:
\[ \Delta S = \frac{334700}{273.15} \approx 1225.9 \, \text{J/K/kg} \]

### Final Answer
The entropy change in melting of 1 g of ice at 0°C is approximately:
\[ \Delta S \approx 1225.9 \, \text{J/K/kg} \]