Steam at `100^@C` is allowed to pass into a vessel containing 10 g of ice and 100 g of water at `0^@C`, until all the ice is melted and the temperature is raised to `40^@C`. Neglecting water equivalent of the vessel and the loss due to radiation etc. The approximate amount of steam condensed is

Options

A. 9 g
B. 18 g
C. 15 g
D. 21 g

Explanation

To solve the problem, we need to calculate the amount of steam that condenses when it comes into contact with the ice and water mixture. We will use the principle of conservation of energy, where the heat lost by the steam equals the heat gained by the ice and water.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of ice, \( m_{ice} = 10 \, \text{g} \)
- Mass of water, \( m_{water} = 100 \, \text{g} \)
- Initial temperature of ice and water, \( T_{initial} = 0^\circ C \)
- Final temperature of the mixture, \( T_{final} = 40^\circ C \)
- Latent heat of fusion of ice, \( L_f = 80 \, \text{cal/g} \)
- Specific heat of water, \( c_{water} = 1 \, \text{cal/g}^\circ C \)
- Latent heat of vaporization of steam, \( L_v = 540 \, \text{cal/g} \)

2. **Calculate the Heat Required to Melt the Ice:**
\[ Q_{melt} = m_{ice} \times L_f = 10 \, \text{g} \times 80 \, \text{cal/g} = 800 \, \text{cal} \]

3. **Calculate the Heat Required to Raise the Temperature of the Melted Ice (now water) from 0°C to 40°C:**
- After melting, the total mass of water becomes \( m_{total} = m_{ice} + m_{water} = 10 \, \text{g} + 100 \, \text{g} = 110 \, \text{g} \).
\[ Q_{heat} = m_{total} \times c_{water} \times (T_{final} - T_{initial}) = 110 \, \text{g} \times 1 \, \text{cal/g}^\circ C \times (40 - 0)^\circ C = 4400 \, \text{cal} \]

4. **Total Heat Gained by the Ice and Water:**
\[ Q_{total} = Q_{melt} + Q_{heat} = 800 \, \text{cal} + 4400 \, \text{cal} = 5200 \, \text{cal} \]

5. **Set Up the Heat Lost by the Steam:**
- Let \( m \) be the mass of the steam that condenses.
- The heat lost by the steam consists of two parts: the heat released when steam condenses and the heat released when the resulting water cools from 100°C to 40°C.
\[ Q_{steam} = m \times L_v + m \times c_{water} \times (100 - 40) = m \times 540 \, \text{cal/g} + m \times 1 \, \text{cal/g}^\circ C \times 60^\circ C = m \times (540 + 60) = m \times 600 \, \text{cal/g} \]

6. **Equate Heat Gained and Heat Lost:**
\[ Q_{total} = Q_{steam} \]
\[ 5200 \, \text{cal} = m \times 600 \, \text{cal/g} \]

7. **Solve for \( m \):**
\[ m = \frac{5200 \, \text{cal}}{600 \, \text{cal/g}} \approx 8.67 \, \text{g} \]
Rounding off, we find that approximately \( 9 \, \text{g} \) of steam is condensed.

### Final Answer:
The approximate amount of steam condensed is **9 g**.