Stream at `100^(@)C` is passed into 20 g of water at `10^(@)C`. When water acquires 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. 31.5g
C. 42.5g
D. 22.5 g

Explanation

To solve the problem, we will use the principle of conservation of energy, which states that the heat lost by the steam will be equal to the heat gained by the water.

### Step-by-Step Solution:

1. **Identify the given data:**
- Initial temperature of steam, \( T_s = 100^\circ C \)
- Mass of water, \( m_w = 20 \, g \)
- Initial temperature of water, \( T_w = 10^\circ C \)
- Final temperature of water, \( T_f = 80^\circ C \)
- Specific heat of water, \( c_w = 1 \, \text{cal g}^{-1} \, ^\circ C^{-1} \)
- Latent heat of steam, \( L = 540 \, \text{cal g}^{-1} \)

2. **Calculate the heat gained by the water:**
\[ Q_{gain} = m_w \cdot c_w \cdot (T_f - T_w) \]
Substituting the values:
\[ Q_{gain} = 20 \, g \cdot 1 \, \text{cal g}^{-1} \, ^\circ C^{-1} \cdot (80^\circ C - 10^\circ C) \]
\[ Q_{gain} = 20 \cdot 1 \cdot 70 = 1400 \, \text{cal} \]

3. **Set up the equation for heat lost by steam:**
The heat lost by the steam when it condenses to water at \( 100^\circ C \) is given by:
\[ Q_{loss} = m' \cdot L \]
where \( m' \) is the mass of steam that condenses.

4. **Equate the heat gained by water to the heat lost by steam:**
\[ Q_{gain} = Q_{loss} \]
\[ 1400 \, \text{cal} = m' \cdot 540 \, \text{cal g}^{-1} \]

5. **Solve for \( m' \):**
\[ m' = \frac{1400}{540} \approx 2.59 \, g \]

6. **Calculate the total mass of water present:**
The total mass of water after the steam has condensed is:
\[ m_{total} = m_w + m' = 20 \, g + 2.59 \, g \approx 22.59 \, g \]

### Final Answer:
The mass of water present when the water acquires a temperature of \( 80^\circ C \) is approximately \( 22.59 \, g \).