Water at `0^@C` contained in a closed vessel, is abruptly opened in an evacuated chamber. If the specific latent heats of fusion and vapourization at `0^@C` are in the ratio `lamda:1` the fraction of water eveporated will be

Options

A. `lamda//1`
B. `lamda//(lamda+1)`
C. `(1-lamda)//lamda`
D. `(lamda-1)//(lamda+1)`

Explanation

To solve the problem, we need to determine the fraction of water that evaporates when water at 0°C is suddenly exposed to an evacuated chamber. We are given that the specific latent heats of fusion (Lf) and vaporization (Lv) at 0°C are in the ratio λ:1.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have water at 0°C in a closed vessel.
- When the vessel is opened in an evacuated chamber, some of the water will evaporate.
- We need to find the fraction of water that evaporates.

2. **Setting Up the Ratios**:
- Given that the specific latent heats of fusion and vaporization are in the ratio λ:1, we can express this as:
\[ \frac{L_f}{L_v} = \lambda \]
- This implies:
\[ L_f = \lambda L_v \]

3. **Defining the Fraction of Water Evaporated**:
- Let the fraction of water that evaporates be denoted by \( k \).
- Therefore, the fraction of water that remains as liquid is \( 1 - k \).

4. **Energy Balance**:
- The energy required to evaporate the fraction \( k \) of water is given by:
\[ \text{Energy for evaporation} = k L_v \]
- The energy released by the remaining fraction \( 1 - k \) of water as it freezes (if it were to freeze) is:
\[ \text{Energy released} = (1 - k) L_f \]

5. **Setting Up the Equation**:
- Since the energy gained by the evaporating water must equal the energy lost by the remaining water, we can set up the equation:
\[ (1 - k) L_f = k L_v \]

6. **Substituting for \( L_f \)**:
- Substitute \( L_f \) from the ratio we established earlier:
\[ (1 - k)(\lambda L_v) = k L_v \]

7. **Simplifying the Equation**:
- Dividing both sides by \( L_v \) (assuming \( L_v \neq 0 \)):
\[ (1 - k) \lambda = k \]
- Rearranging gives:
\[ \lambda - \lambda k = k \]
- Combine like terms:
\[ \lambda = k + \lambda k \]
- Factor out \( k \):
\[ \lambda = k(1 + \lambda) \]

8. **Solving for \( k \)**:
- Rearranging for \( k \):
\[ k = \frac{\lambda}{\lambda + 1} \]

### Final Answer:
The fraction of water that evaporates is:
\[ k = \frac{\lambda}{\lambda + 1} \]