A thermally isolated vessel contains `100`g of water at `0^(@)C` when air above the water is pumped out, some of the water freezes and some evaporates at `0^(@)C` itself. Calculate the mass at `0^(@)C=2.10xx10^(6) j//kg` and latent heat of fusion of ice `=3.36xx10^(5) j//kg`.

Options

A. `86.2 g`
B. `13.8 g`
C. `76.2 g`
D. `65.6g`

Answer

To solve the problem, we will use the principle of calorimetry, which states that the total heat lost by the water that freezes is equal to the total heat gained by the water that evaporates.

### Step-by-Step Solution:

1. **Define Variables**:
- Let \( m_1 \) be the mass of water that freezes (in kg).
- The total mass of water is given as \( 100 \, \text{g} = 0.1 \, \text{kg} \).
- Therefore, the mass of water that evaporates will be \( 0.1 - m_1 \).

2. **Heat Loss by Freezing**:
- The heat lost when \( m_1 \) kg of water freezes can be calculated using the formula:
\[ Q_{\text{loss}} = m_1 \times L_f \]
- Where \( L_f = 3.36 \times 10^5 \, \text{J/kg} \) (latent heat of fusion).

3. **Heat Gain by Evaporation**:
- The heat gained when \( 0.1 - m_1 \) kg of water evaporates can be calculated using the formula:
\[ Q_{\text{gain}} = (0.1 - m_1) \times L_v \]
- Where \( L_v = 2.10 \times 10^6 \, \text{J/kg} \) (latent heat of vaporization).

4. **Set Up the Equation**:
- According to the principle of calorimetry:
\[ Q_{\text{loss}} = Q_{\text{gain}} \]
- This gives us:
\[ m_1 \times L_f = (0.1 - m_1) \times L_v \]

5. **Substitute Values**:
- Substitute \( L_f \) and \( L_v \) into the equation:
\[ m_1 \times (3.36 \times 10^5) = (0.1 - m_1) \times (2.10 \times 10^6) \]

6. **Expand and Rearrange**:
- Expanding the right side:
\[ 3.36 \times 10^5 m_1 = 0.1 \times 2.10 \times 10^6 - 2.10 \times 10^6 m_1 \]
- This simplifies to:
\[ 3.36 \times 10^5 m_1 + 2.10 \times 10^6 m_1 = 0.1 \times 2.10 \times 10^6 \]

7. **Factor Out \( m_1 \)**:
- Factor out \( m_1 \):
\[ m_1 (3.36 \times 10^5 + 2.10 \times 10^6) = 0.1 \times 2.10 \times 10^6 \]

8. **Calculate the Coefficient**:
- Calculate \( 3.36 \times 10^5 + 2.10 \times 10^6 \):
\[ 3.36 \times 10^5 + 2.10 \times 10^6 = 2.43 \times 10^6 \]

9. **Solve for \( m_1 \)**:
- Now, substitute back into the equation:
\[ m_1 (2.43 \times 10^6) = 0.1 \times 2.10 \times 10^6 \]
- Thus:
\[ m_1 = \frac{0.1 \times 2.10 \times 10^6}{2.43 \times 10^6} \]
- Calculate \( m_1 \):
\[ m_1 = \frac{0.21 \times 10^6}{2.43 \times 10^6} \approx 0.0862 \, \text{kg} = 86.2 \, \text{g} \]

### Final Answer:
The mass of the water that freezes is approximately **86.2 g**.