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. `31.5 g`
C. `42.5 g`
D. `22.5 g`

Explanation

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

### Step-by-Step Solution:

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

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

3. **Set Up the Heat Lost by Steam:**
The heat lost by the steam consists of two parts: the heat released during condensation and the heat released as the resulting water cools down. If M is the mass of steam condensed, the heat lost can be expressed as:
\[ Q_{loss} = M \cdot L + M \cdot c_w \cdot (T_s_{initial} - T_s_{final}) \]
Substituting the values:
\[ Q_{loss} = M \cdot 540 + M \cdot 1 \cdot (100°C - 80°C) = M \cdot 540 + M \cdot 20 = M \cdot (540 + 20) = M \cdot 560 \]

4. **Equate Heat Gained and Heat Lost:**
According to the conservation of energy:
\[ Q_{gain} = Q_{loss} \]
Therefore:
\[ 1400 = M \cdot 560 \]

5. **Solve for M:**
Rearranging the equation to find M:
\[ M = \frac{1400}{560} = 2.5 \, \text{g} \]

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

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

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