To find the thermal capacity of the calorimeter, we can 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 Variables:**
- Mass of cold water, \( m_1 = 0.2 \, \text{kg} = 200 \, \text{g} \)
- Initial temperature of cold water, \( T_1 = 30^\circ \text{C} \)
- Mass of hot water, \( m_2 = 0.1 \, \text{kg} = 100 \, \text{g} \)
- Initial temperature of hot water, \( T_2 = 60^\circ \text{C} \)
- Final temperature of the mixture, \( T_f = 35^\circ \text{C} \)
2. **Calculate Heat Gained by Cold Water:**
\[
Q_{\text{gain}} = m_1 \cdot c \cdot (T_f - T_1)
\]
Where \( c \) (specific heat of water) is \( 1 \, \text{cal/g}^\circ \text{C} \).
\[
Q_{\text{gain}} = 200 \cdot 1 \cdot (35 - 30) = 200 \cdot 1 \cdot 5 = 1000 \, \text{cal}
\]
3. **Calculate Heat Lost by Hot Water:**
\[
Q_{\text{loss}} = m_2 \cdot c \cdot (T_2 - T_f)
\]
\[
Q_{\text{loss}} = 100 \cdot 1 \cdot (60 - 35) = 100 \cdot 1 \cdot 25 = 2500 \, \text{cal}
\]
4. **Set Up the Heat Balance Equation:**
The heat lost by the hot water is equal to the heat gained by the cold water plus the heat gained by the calorimeter:
\[
Q_{\text{loss}} = Q_{\text{gain}} + C \cdot (T_f - T_1)
\]
Where \( C \) is the thermal capacity of the calorimeter.
5. **Substituting the Values:**
\[
2500 = 1000 + C \cdot (35 - 30)
\]
\[
2500 = 1000 + 5C
\]
6. **Solve for C:**
\[
2500 - 1000 = 5C
\]
\[
1500 = 5C
\]
\[
C = \frac{1500}{5} = 300 \, \text{cal/K}
\]
7. **Convert to Joules:**
Since \( 1 \, \text{cal} = 4.2 \, \text{J} \):
\[
C = 300 \, \text{cal/K} \times 4.2 \, \text{J/cal} = 1260 \, \text{J/K}
\]
### Final Answer:
The thermal capacity of the calorimeter is \( 1260 \, \text{J/K} \).
