Steam at `100^@C` is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at `15^@C`till the temperature of the calorimeter and its contents rises to `80^@C` . The mass of the steam condensed in kilogram is

Options

A. `0.130`
B. `0.065`
C. `0.260`
D. `0.135`

Explanation

To solve the problem, we will use the principle of calorimetry, which states that the heat lost by the steam will be equal to the heat gained by the water and the calorimeter.

### Step-by-Step Solution

1. **Identify Given Data:**
- Mass of water, \( m_w = 1.1 \, \text{kg} \)
- Water equivalent of calorimeter, \( m_c = 0.02 \, \text{kg} \)
- Initial temperature of water and calorimeter, \( t_1 = 15^\circ C \)
- Final temperature, \( t_2 = 80^\circ C \)
- Temperature of steam, \( t_s = 100^\circ C \)
- Latent heat of vaporization of steam, \( L = 536 \, \text{kcal/kg} \)

2. **Calculate the Heat Required by Water and Calorimeter:**
- The heat required by the water can be calculated using the formula:
\[ Q_w = m_w \cdot c_w \cdot (t_2 - t_1) \]
where \( c_w = 1 \, \text{kcal/kg}^\circ C \) (specific heat capacity of water).
- The heat required by the calorimeter is:
\[ Q_c = m_c \cdot c_c \cdot (t_2 - t_1) \]
where \( c_c = 1 \, \text{kcal/kg}^\circ C \) (specific heat capacity of the calorimeter).

- Now, substituting the values:
\[ Q_w = 1.1 \cdot 1 \cdot (80 - 15) = 1.1 \cdot 1 \cdot 65 = 71.5 \, \text{kcal} \]
\[ Q_c = 0.02 \cdot 1 \cdot (80 - 15) = 0.02 \cdot 1 \cdot 65 = 1.3 \, \text{kcal} \]

- Total heat required:
\[ Q_{total} = Q_w + Q_c = 71.5 + 1.3 = 72.8 \, \text{kcal} \]

3. **Calculate the Heat Given by the Steam:**
- The heat given by the steam consists of two parts:
- The heat released when steam condenses to water:
\[ Q_{condensation} = m \cdot L \]
- The heat released by the condensed water as it cools from \( t_s \) to \( t_2 \):
\[ Q_{cooling} = m \cdot c_w \cdot (t_s - t_2) \]

- Therefore, the total heat given by the steam is:
\[ Q_{steam} = m \cdot L + m \cdot c_w \cdot (t_s - t_2) \]
\[ Q_{steam} = m \cdot (L + c_w \cdot (t_s - t_2)) \]

4. **Set Heat Given Equal to Heat Required:**
- According to the principle of conservation of energy:
\[ Q_{steam} = Q_{total} \]
\[ m \cdot (L + c_w \cdot (t_s - t_2)) = 72.8 \]

- Substituting the values:
\[ m \cdot (536 + 1 \cdot (100 - 80)) = 72.8 \]
\[ m \cdot (536 + 20) = 72.8 \]
\[ m \cdot 556 = 72.8 \]

5. **Solve for \( m \):**
\[ m = \frac{72.8}{556} \approx 0.130 \, \text{kg} \]

### Final Answer:
The mass of the steam condensed is approximately \( 0.130 \, \text{kg} \).

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