Steam is passes into 22 g of water at `20^@C`. The mass of water that will be present when the water acquires a temperature of `90^@C` (Latent heat of steam is `540 cal//g`) is

Options

A. 24.8 gm
B. 24 gm
C. 36.6 gm
D. 30 gm

Answer

To solve the problem, we need to determine the mass of water present when steam is passed into 22 g of water at 20°C, and the final temperature of the mixture is 90°C. We will use the principle of conservation of energy, where 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 (m1) = 22 g
- Initial temperature of water (T1) = 20°C
- Final temperature of water (T_final) = 90°C
- Latent heat of steam (L) = 540 cal/g
- Specific heat of water (c) = 1 cal/g°C

2. **Calculate the Heat Required to Raise the Temperature of Water:**
The heat required (Q1) to raise the temperature of the water from 20°C to 90°C can be calculated using the formula:
\[ Q_1 = m_1 \cdot c \cdot (T_{final} - T_1) \]
Substituting the values:
\[ Q_1 = 22 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (90°C - 20°C) = 22 \cdot 1 \cdot 70 = 1540 \, \text{cal} \]

3. **Set Up the Heat Balance Equation:**
The heat gained by the water (Q1) is equal to the heat lost by the steam (Q2) plus the heat required to raise the temperature of the condensed steam to 90°C (Q3):
\[ Q_1 = Q_2 + Q_3 \]
Where:
- \( Q_2 = m \cdot L \) (heat released by steam condensing)
- \( Q_3 = m \cdot c \cdot (100°C - 90°C) \) (heat required to raise the temperature of condensed steam from 100°C to 90°C)

4. **Substituting the Values:**
From the above, we have:
\[ 1540 = m \cdot 540 + m \cdot 1 \cdot 10 \]
Simplifying this gives:
\[ 1540 = 540m + 10m = 550m \]

5. **Solve for m:**
Rearranging the equation to find m:
\[ m = \frac{1540}{550} \approx 2.8 \, \text{g} \]

6. **Calculate the Total Mass of Water:**
The total mass of water when the system reaches 90°C is the sum of the original water and the condensed steam:
\[ \text{Total mass} = m + m_1 = 2.8 \, \text{g} + 22 \, \text{g} = 24.8 \, \text{g} \]

### Final Answer:
The mass of water present when the water acquires a temperature of 90°C is **24.8 g**.