When `1 kg` of ice at `0^(@)C` melts to water at `0^(@)C`, the resulting change in its entropy, taking latent heat of ice to be `80 cal//g` is

Options

A. `8 xx 10^(4)` cal/K
B. `80` cal/K
C. 293 cal/K
D. 273 cal/K

Answer

To find the change in entropy when 1 kg of ice at \(0^\circ C\) melts to water at \(0^\circ C\), we can follow these steps:

### Step 1: Determine the mass of ice in grams
Given that the mass of ice is \(1 \text{ kg}\), we convert this to grams:
\[ \text{Mass of ice} = 1 \text{ kg} = 1000 \text{ g} \]

### Step 2: Calculate the heat required for melting
The heat required (\(Q\)) to melt the ice can be calculated using the formula:
\[ Q = m \times L \]
where:
- \(m\) is the mass of ice in grams,
- \(L\) is the latent heat of ice.

Given that the latent heat of ice is \(80 \text{ cal/g}\):
\[ Q = 1000 \text{ g} \times 80 \text{ cal/g} = 80000 \text{ cal} \]

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

### Step 4: Calculate the change in entropy
The change in entropy (\(\Delta S\)) during the melting process can be calculated using the formula:
\[ \Delta S = \frac{Q}{T} \]
Substituting the values we found:
\[ \Delta S = \frac{80000 \text{ cal}}{273 \text{ K}} \approx 293.37 \text{ cal/K} \]

### Final Answer
Thus, the change in entropy when 1 kg of ice melts to water at \(0^\circ C\) is approximately:
\[ \Delta S \approx 293 \text{ cal/K} \]