A steam at `100^@C` is passed into `1 kg` of water contained in a calorimeter of water equivalent `0.2 kg` at `9^@C` till the temperature of the calorimeter and water in it is increased to `90^@C`. Find the mass of steam condensed in `kg(S_(w) = 1 cal//g^@C, & L_("steam") = 540 cal//g)`.

Options

A. 1
B. 0.18
C. 0.81
D. 0.135

Explanation

To solve the problem, we need to find the mass of steam condensed when steam at 100°C is passed into a calorimeter containing 1 kg of water at 9°C, which eventually reaches a temperature of 90°C.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of water, \( M_w = 1 \, \text{kg} = 1000 \, \text{g} \)
- Water equivalent of the calorimeter, \( M_c = 0.2 \, \text{kg} = 200 \, \text{g} \)
- Initial temperature of water and calorimeter, \( T_i = 9°C \)
- Final temperature of the system, \( T_f = 90°C \)
- Specific heat of water, \( S_w = 1 \, \text{cal/g°C} \)
- Latent heat of steam, \( L = 540 \, \text{cal/g} \)

2. **Calculate the Total Heat Gained by Water and Calorimeter:**
The total mass of the water and calorimeter system is:
\[ M_{total} = M_w + M_c = 1000 \, \text{g} + 200 \, \text{g} = 1200 \, \text{g} \]
The heat gained by the water and calorimeter when the temperature rises from 9°C to 90°C is calculated as:
\[ Q_{gain} = M_{total} \cdot S_w \cdot (T_f - T_i) = 1200 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (90 - 9) \, °C \]
\[ Q_{gain} = 1200 \cdot 1 \cdot 81 = 97200 \, \text{cal} \]

3. **Calculate the Heat Lost by the Steam:**
Let the mass of steam condensed be \( M_s \) (in grams). The steam first condenses into water at 100°C and then cools down to 90°C. The heat lost by the steam is given by:
\[ Q_{loss} = M_s \cdot L + M_s \cdot S_w \cdot (100 - 90) \]
\[ Q_{loss} = M_s \cdot 540 + M_s \cdot 1 \cdot 10 = M_s \cdot (540 + 10) = M_s \cdot 550 \]

4. **Set Heat Gained Equal to Heat Lost:**
Since the heat gained by the water and calorimeter is equal to the heat lost by the steam, we can set up the equation:
\[ Q_{gain} = Q_{loss} \]
\[ 97200 = M_s \cdot 550 \]

5. **Solve for the Mass of the Steam:**
Rearranging the equation to find \( M_s \):
\[ M_s = \frac{97200}{550} \approx 176.73 \, \text{g} \]

6. **Convert to Kilograms:**
To convert grams to kilograms:
\[ M_s = \frac{176.73}{1000} \approx 0.17673 \, \text{kg} \approx 0.18 \, \text{kg} \]

### Final Answer:
The mass of steam condensed is approximately **0.18 kg**.

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