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 g
B. 24 g
C. 36.6 g
D. 30 g

Explanation

To solve the problem, we need to determine how much steam will condense into water when it is passed into 22 g of water at 20°C, and then calculate the total mass of water present when the temperature reaches 90°C.

### Step-by-Step Solution:

1. **Identify the known values:**
- Mass of water (m_water) = 22 g
- Initial temperature of water (T_initial) = 20°C
- Final temperature of water (T_final) = 90°C
- Latent heat of steam (L) = 540 cal/g
- Specific heat of water (c_water) = 1 cal/g°C (approximately)

2. **Calculate the heat required to raise the temperature of the water from 20°C to 90°C:**
\[ Q_{water} = m_{water} \cdot c_{water} \cdot (T_{final} - T_{initial}) \]
\[ Q_{water} = 22 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (90°C - 20°C) \]
\[ Q_{water} = 22 \cdot 1 \cdot 70 = 1540 \, \text{cal} \]

3. **Determine the mass of steam that will condense to provide this heat:**
The heat released by the steam when it condenses can be calculated using the formula:
\[ Q_{steam} = m_{steam} \cdot L \]
Setting \(Q_{steam} = Q_{water}\):
\[ 1540 \, \text{cal} = m_{steam} \cdot 540 \, \text{cal/g} \]
Solving for \(m_{steam}\):
\[ m_{steam} = \frac{1540}{540} \approx 2.85 \, \text{g} \]

4. **Calculate the total mass of water present:**
The total mass of water when the steam condenses is the initial mass of water plus the mass of the condensed steam:
\[ m_{total} = m_{water} + m_{steam} \]
\[ m_{total} = 22 \, \text{g} + 2.85 \, \text{g} \approx 24.85 \, \text{g} \]

### Final Answer:
The mass of water that will be present when the water acquires a temperature of 90°C is approximately **24.85 g**.