A thermally insulated vessel contains `150g` of water at `0^(@)C`. Then the air from the vessel is pumped out adiabatically. A fraction of water turms into ice and the rest evaporates at `0^(@)C` itself. The mass of evaporated water will be closest to :
(Latent heat of vaporization of water `=2.10xx10^(6)jkg^(-1)` and Latent heat of Fusion of water `=3.36xx10^(5)jkg^(-1)`)

Options

A. `150g`
B. `20g`
C. `130g`
D. `35g`

Answer

To solve the problem, we need to analyze the energy exchanges that occur when air is pumped out of the thermally insulated vessel containing water at 0°C. The process leads to some of the water turning into ice while the rest evaporates, both at 0°C.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of water (m) = 150 g = 0.150 kg
- Latent heat of fusion of water (L_f) = \(3.36 \times 10^5 \, \text{J/kg}\)
- Latent heat of vaporization of water (L_v) = \(2.10 \times 10^6 \, \text{J/kg}\)

2. **Define Variables:**
- Let \(x\) be the mass of water that turns into ice (in kg).
- The mass of water that evaporates will then be \(0.150 - x\) kg.

3. **Set Up the Energy Balance Equation:**
- The heat lost by the water turning into ice is given by:
\[ Q_{\text{loss}} = x \cdot L_f \]
- The heat gained by the water turning into vapor is given by:
\[ Q_{\text{gain}} = (0.150 - x) \cdot L_v \]
- Since the system is adiabatic, the heat lost by the ice formation equals the heat gained by the vaporization:
\[ x \cdot L_f = (0.150 - x) \cdot L_v \]

4. **Substitute the Values:**
\[ x \cdot (3.36 \times 10^5) = (0.150 - x) \cdot (2.10 \times 10^6) \]

5. **Expand and Rearrange the Equation:**
\[ 3.36 \times 10^5 x = 0.150 \cdot 2.10 \times 10^6 - 2.10 \times 10^6 x \]
\[ 3.36 \times 10^5 x + 2.10 \times 10^6 x = 0.150 \cdot 2.10 \times 10^6 \]
\[ (3.36 \times 10^5 + 2.10 \times 10^6) x = 0.150 \cdot 2.10 \times 10^6 \]

6. **Calculate the Right Side:**
\[ 0.150 \cdot 2.10 \times 10^6 = 315000 \, \text{J} \]

7. **Combine the Coefficients:**
\[ (3.36 \times 10^5 + 2.10 \times 10^6) = 2.4336 \times 10^6 \]
\[ 2.4336 \times 10^6 x = 315000 \]

8. **Solve for \(x\):**
\[ x = \frac{315000}{2.4336 \times 10^6} \approx 0.129 \, \text{kg} \approx 129 \, \text{g} \]

9. **Calculate the Mass of Evaporated Water:**
\[ \text{Mass of evaporated water} = 150 \, \text{g} - x \approx 150 \, \text{g} - 129 \, \text{g} \approx 21 \, \text{g} \]

### Final Answer:
The mass of evaporated water will be closest to **20 g**.