Steam at `100^(@)C` is passed into `20 g` of water at `10^(@)C` when water acquire a temperature of `80^(@)C`, the mass of water present will be
[Take specific heat of water `= 1 cal g^(-1).^(@) C^(-1)` and latent heat of steam ` = 540 cal g^(-1)`]

Options

A. `24 g`
B. `31.5 g`
C. `42.5 g`
D. `22.5 g`

Explanation

To solve the problem, we need to calculate the mass of steam that condenses into water when it is passed into the existing water. We will use the principle of conservation of energy, where the heat gained by the water is equal to the heat lost by the steam.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Mass of water (m_w) = 20 g
- Initial temperature of water (T_initial_w) = 10°C
- Final temperature of water (T_final) = 80°C
- Specific heat of water (c_w) = 1 cal/g°C
- Latent heat of steam (L) = 540 cal/g
- Initial temperature of steam (T_initial_s) = 100°C

2. **Calculate Heat Gained by Water:**
The heat gained by the water (Q_gained) can be calculated using the formula:
\[ Q_{\text{gained}} = m_w \cdot c_w \cdot (T_{\text{final}} - T_{\text{initial}_w}) \]
Substituting the values:
\[ Q_{\text{gained}} = 20 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (80°C - 10°C) \]
\[ Q_{\text{gained}} = 20 \cdot 1 \cdot 70 = 1400 \, \text{cal} \]

3. **Calculate Heat Lost by Steam:**
The heat lost by the steam (Q_lost) consists of two parts: the heat released when steam condenses into water and the heat released when the water formed from steam cools down to the final temperature.
- Let m_s be the mass of steam condensed.
- The heat lost by steam when it condenses:
\[ Q_{\text{lost, condensation}} = m_s \cdot L \]
- The heat lost when the water from steam cools from 100°C to 80°C:
\[ Q_{\text{lost, cooling}} = m_s \cdot c_w \cdot (T_{\text{initial}_s} - T_{\text{final}}) \]
Combining both:
\[ Q_{\text{lost}} = m_s \cdot L + m_s \cdot c_w \cdot (100°C - 80°C) \]
\[ Q_{\text{lost}} = m_s \cdot 540 + m_s \cdot 1 \cdot 20 \]
\[ Q_{\text{lost}} = m_s \cdot (540 + 20) = m_s \cdot 560 \]

4. **Set Heat Gained Equal to Heat Lost:**
According to the principle of conservation of energy:
\[ Q_{\text{gained}} = Q_{\text{lost}} \]
Thus,
\[ 1400 = m_s \cdot 560 \]

5. **Solve for m_s:**
\[ m_s = \frac{1400}{560} = 2.5 \, \text{g} \]

6. **Calculate Total Mass of Water:**
The total mass of water after steam condenses:
\[ \text{Total mass of water} = \text{mass of existing water} + \text{mass of steam condensed} \]
\[ \text{Total mass of water} = 20 \, \text{g} + 2.5 \, \text{g} = 22.5 \, \text{g} \]

### Final Answer:
The mass of water present will be **22.5 g**.