A mole of steam is condensed at `100^(@)` C, the water is cooled to `0^(@)`C and frozen to ice . What is the difference in entropies of the steam and ice? The heat of vaporization and fusion are `540 cal" "gm^(-1)` and `80 cal" "gm ^(-1)` respectively . Use the average heat capacity of liquild water as `1 cal" "gm^(-1)` `degree^(-1)`.
a. Entropy change during the condensation of steam is `-26.06 cal//^(@)C`
b. Entropy change during cooling of water from `100^(@)C "to" 0^(@)C "is" - 5.62 cal//^(@)C`
c. Entropy change during freezing of water at `0^(@)C "is" -5.27 cal//^(@)C`
d. Total entropy changwe is `-36.95 cal//^(@)C`

Options

A. Entropy change during the condensation of steam is `-26.06 cal//^(@)C`
B. Entropy change during cooling of water from `100^(@)C "to" 0^(@)C "is" - 5.62 cal//^(@)C`
C. Entropy change during freezing of water at `0^(@)C "is" -5.27 cal//^(@)C`
D. Total entropy changwe is `-36.95 cal//^(@)C`

Answer

To solve the problem, we will calculate the entropy changes for each step involved in the process of condensing steam, cooling water, and freezing it to ice. We will then sum these changes to find the total entropy change.

### Step 1: Entropy Change During the Condensation of Steam
1. **Identify the heat of vaporization**: The heat of vaporization of water is given as \(540 \, \text{cal/g}\).
2. **Calculate the total heat for 1 mole**: Since 1 mole of water is approximately \(18 \, \text{g}\), the total heat released during condensation is:
\[ Q = 18 \, \text{g} \times 540 \, \text{cal/g} = 9720 \, \text{cal} \]
Since this is a condensation process, the heat is released, so we take it as negative:
\[ Q = -9720 \, \text{cal} \]
3. **Convert temperature to Kelvin**: The temperature at which the condensation occurs is \(100^\circ C\), which is:
\[ T = 100 + 273 = 373 \, \text{K} \]
4. **Calculate the entropy change**: The entropy change (\(\Delta S\)) is given by:
\[ \Delta S = \frac{Q}{T} = \frac{-9720 \, \text{cal}}{373 \, \text{K}} \approx -26.06 \, \text{cal/K} \]

### Step 2: Entropy Change During Cooling of Water from \(100^\circ C\) to \(0^\circ C\)
1. **Identify the heat capacity**: The average heat capacity of liquid water is given as \(1 \, \text{cal/g} \cdot \text{K}\).
2. **Calculate the entropy change**: The entropy change during cooling can be calculated using the formula:
\[ \Delta S = \int_{T_1}^{T_2} \frac{C_p}{T} \, dT \]
For linear cooling from \(100^\circ C\) (373 K) to \(0^\circ C\) (273 K):
\[ \Delta S = C_p \ln\left(\frac{T_2}{T_1}\right) = 18 \, \text{g} \times 1 \, \text{cal/g} \cdot \text{K} \times \ln\left(\frac{273}{373}\right) \]
Calculating this gives:
\[ \Delta S \approx -5.62 \, \text{cal/K} \]

### Step 3: Entropy Change During Freezing of Water at \(0^\circ C\)
1. **Identify the heat of fusion**: The heat of fusion is given as \(80 \, \text{cal/g}\).
2. **Calculate the total heat for 1 mole**: The total heat released during freezing is:
\[ Q = 18 \, \text{g} \times 80 \, \text{cal/g} = 1440 \, \text{cal} \]
Again, since this is a freezing process, we take it as negative:
\[ Q = -1440 \, \text{cal} \]
3. **Convert temperature to Kelvin**: The temperature at which freezing occurs is \(0^\circ C\), which is:
\[ T = 273 \, \text{K} \]
4. **Calculate the entropy change**: The entropy change is:
\[ \Delta S = \frac{Q}{T} = \frac{-1440 \, \text{cal}}{273 \, \text{K}} \approx -5.27 \, \text{cal/K} \]

### Step 4: Total Entropy Change
1. **Sum the entropy changes**:
\[ \Delta S_{\text{total}} = \Delta S_{\text{condensation}} + \Delta S_{\text{cooling}} + \Delta S_{\text{freezing}} \]
\[ \Delta S_{\text{total}} = -26.06 \, \text{cal/K} - 5.62 \, \text{cal/K} - 5.27 \, \text{cal/K} \]
\[ \Delta S_{\text{total}} \approx -36.95 \, \text{cal/K} \]

### Final Answer
The total change in entropy is approximately \(-36.95 \, \text{cal/K}\).