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 if 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 solve the problem of finding the thermal capacity of the calorimeter, we will use the principle of conservation of energy, which states that the heat lost by the hot water will be equal to the heat gained by the cold water and the calorimeter.

### Step-by-Step Solution:

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

2. **Calculate Heat Lost by Hot Water:**
The heat lost by the hot water can be calculated using the formula:
\[ Q_{\text{lost}} = m_2 \cdot c \cdot (T_2 - T_f) \]
Substituting the values:
\[ Q_{\text{lost}} = 0.1 \cdot 4200 \cdot (60 - 35) = 0.1 \cdot 4200 \cdot 25 \]
\[ Q_{\text{lost}} = 0.1 \cdot 4200 \cdot 25 = 10500 \, \text{J} \]

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

4. **Calculate Heat Gained by Calorimeter:**
Let the thermal capacity of the calorimeter be \( X \). The heat gained by the calorimeter is:
\[ Q_{\text{gained, calorimeter}} = X \cdot (T_f - T_1) = X \cdot (35 - 30) = 5X \]

5. **Set Up the Energy Balance Equation:**
According to the conservation of energy:
\[ Q_{\text{lost}} = Q_{\text{gained, cold}} + Q_{\text{gained, calorimeter}} \]
Substituting the values:
\[ 10500 = 4200 + 5X \]

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

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

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