In the previous question, if the specific latent heat of vaporization of water at `0^@C` is `eta` times the specific latent heat of freezing of water at `0^@C`, the fraction of water that will ultimately freeze is

Options

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

Explanation

To solve the problem, we need to find the fraction of water that will ultimately freeze when the specific latent heat of vaporization of water at \(0^\circ C\) is \(\eta\) times the specific latent heat of freezing of water at \(0^\circ C\).

### Step-by-Step Solution:

1. **Define Variables**:
- Let \(L_F\) be the specific latent heat of freezing of water at \(0^\circ C\).
- Let \(L_V\) be the specific latent heat of vaporization of water at \(0^\circ C\).
- According to the problem, we have \(L_V = \eta L_F\).

2. **Conservation of Energy**:
- We know that the heat lost by the water that freezes must equal the heat gained by the water that vaporizes.
- Let the mass of water that vaporizes be \(x\).
- The heat gained by the vaporized water is given by:
\[ \text{Heat gained} = x \cdot L_V = x \cdot \eta L_F \]
- Let the total mass of water be \(m\). Therefore, the mass of water that freezes is \(m - x\).
- The heat lost by the water that freezes is given by:
\[ \text{Heat lost} = (m - x) \cdot L_F \]

3. **Setting Up the Equation**:
- By the conservation of energy, we can equate the heat gained and the heat lost:
\[ x \cdot \eta L_F = (m - x) \cdot L_F \]

4. **Simplifying the Equation**:
- We can cancel \(L_F\) from both sides (assuming \(L_F \neq 0\)):
\[ x \cdot \eta = m - x \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ x \cdot \eta + x = m \]
- Factoring out \(x\):
\[ x(1 + \eta) = m \]

6. **Solving for \(x\)**:
- Thus, we find:
\[ x = \frac{m}{1 + \eta} \]

7. **Finding the Mass that Freezes**:
- The mass of water that freezes is:
\[ m - x = m - \frac{m}{1 + \eta} \]
- Finding a common denominator:
\[ m - x = \frac{m(1 + \eta) - m}{1 + \eta} = \frac{m\eta}{1 + \eta} \]

8. **Calculating the Fraction of Water that Freezes**:
- The fraction of water that ultimately freezes is:
\[ \text{Fraction} = \frac{\text{Mass that freezes}}{\text{Total mass}} = \frac{\frac{m\eta}{1 + \eta}}{m} = \frac{\eta}{1 + \eta} \]

### Final Answer:
The fraction of water that will ultimately freeze is:
\[ \frac{\eta}{1 + \eta} \]