A calorimeter contains `0.2 kg` of water at `30^(@)C, 0.1 kg` of water at `60^(@)C` is added to it, the mixture is well stirred and the resulting temperature is found to be `35^(@)C`. The thermal capacity of the calorimeter is:

Options

A. `6300 J//K`
B. `1260 J//K`
C. `4200 J//K`
D. None of these

Explanation

To find the thermal capacity of the calorimeter, we will use the principle of conservation of energy, which states that the heat gained by the calorimeter and the water at a lower temperature must equal the heat lost by the water at a higher temperature.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of water at 30°C, \( m_1 = 0.2 \, \text{kg} \)
- Initial temperature of water at 30°C, \( T_1 = 30 \, \text{°C} \)
- Mass of water at 60°C, \( m_2 = 0.1 \, \text{kg} \)
- Initial temperature of water at 60°C, \( T_2 = 60 \, \text{°C} \)
- Final temperature of the mixture, \( T_f = 35 \, \text{°C} \)
- Specific heat capacity of water, \( c = 4200 \, \text{J/(kg·K)} \)

2. **Calculate the Heat Gained by the Calorimeter:**
- Let the thermal capacity of the calorimeter be \( x \).
- The heat gained by the calorimeter when the temperature rises from 30°C to 35°C is:
\[ Q_{\text{calorimeter}} = x \cdot (T_f - T_1) = x \cdot (35 - 30) = 5x \, \text{J} \]

3. **Calculate the Heat Gained by the Water at 30°C:**
- The heat gained by the water at 30°C is:
\[ Q_{1} = m_1 \cdot c \cdot (T_f - T_1) = 0.2 \cdot 4200 \cdot (35 - 30) = 0.2 \cdot 4200 \cdot 5 = 4200 \, \text{J} \]

4. **Calculate the Heat Lost by the Water at 60°C:**
- The heat lost by the water at 60°C is:
\[ Q_{2} = m_2 \cdot c \cdot (T_2 - T_f) = 0.1 \cdot 4200 \cdot (60 - 35) = 0.1 \cdot 4200 \cdot 25 = 10500 \, \text{J} \]

5. **Apply the Principle of Conservation of Energy:**
- According to the conservation of energy:
\[ \text{Heat gained} = \text{Heat lost} \]
- Therefore, we have:
\[ Q_{\text{calorimeter}} + Q_{1} = Q_{2} \]
- Substituting the values:
\[ 5x + 4200 = 10500 \]

6. **Solve for \( x \):**
- Rearranging the equation:
\[ 5x = 10500 - 4200 \]
\[ 5x = 6300 \]
\[ x = \frac{6300}{5} = 1260 \, \text{J/K} \]

### Final Answer:
The thermal capacity of the calorimeter is \( 1260 \, \text{J/K} \).