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 heat transfer that occurs when a fraction of water turns into ice and the remaining water evaporates in a thermally insulated vessel. Here’s the step-by-step solution:

### Step 1: Understand the Problem
We have a thermally insulated vessel containing 150 grams of water at 0°C. When air is pumped out adiabatically, some water turns into ice while the rest evaporates. We need to find the mass of the evaporated water.

### Step 2: Define Variables
Let:
- \( m \) = mass of water that evaporates (in grams)
- \( 150 - m \) = mass of water that turns into ice (in grams)

### Step 3: Apply the Principle of Conservation of Energy
In an adiabatic process, the heat lost by the water turning into ice must equal the heat gained by the water that evaporates. Thus, we can write the equation:

\[ \text{Heat lost by water turning into ice} = \text{Heat gained by evaporating water} \]

### Step 4: Write the Heat Transfer Equations
The heat lost when \( 150 - m \) grams of water turns into ice is given by:

\[ Q_{\text{lost}} = (150 - m) \times L_f \]

where \( L_f \) is the latent heat of fusion of water, which is \( 3.36 \times 10^5 \, \text{J/kg} \) or \( 3.36 \times 10^2 \, \text{J/g} \).

The heat gained when \( m \) grams of water evaporates is given by:

\[ Q_{\text{gained}} = m \times L_v \]

where \( L_v \) is the latent heat of vaporization of water, which is \( 2.10 \times 10^6 \, \text{J/kg} \) or \( 2.10 \times 10^3 \, \text{J/g} \).

### Step 5: Set Up the Equation
Setting the heat lost equal to the heat gained, we have:

\[ (150 - m) \times 3.36 \times 10^2 = m \times 2.10 \times 10^3 \]

### Step 6: Simplify the Equation
Expanding the left side:

\[ 150 \times 3.36 \times 10^2 - m \times 3.36 \times 10^2 = m \times 2.10 \times 10^3 \]

Rearranging gives:

\[ 150 \times 3.36 \times 10^2 = m \times (2.10 \times 10^3 + 3.36 \times 10^2) \]

### Step 7: Calculate the Values
Calculating \( 150 \times 3.36 \times 10^2 \):

\[ 150 \times 3.36 \times 10^2 = 50400 \, \text{J} \]

Now, calculate \( 2.10 \times 10^3 + 3.36 \times 10^2 \):

\[ 2.10 \times 10^3 = 2100 \, \text{J/g} \]
\[ 3.36 \times 10^2 = 336 \, \text{J/g} \]
\[ 2100 + 336 = 2436 \, \text{J/g} \]

### Step 8: Solve for \( m \)
Now substituting back into the equation:

\[ 50400 = m \times 2436 \]

Thus,

\[ m = \frac{50400}{2436} \approx 20.69 \, \text{g} \]

### Step 9: Final Answer
The mass of evaporated water is approximately \( 20.69 \) grams, which rounds to \( 20 \) grams.

### Conclusion
The closest mass of evaporated water is **20 grams**.

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