A calorimeter contains 0.2kg 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

The correct option is 1260 J/K.

To find the thermal capacity of the calorimeter, we can apply the principle of calorimetry, which states that under ideal conditions (no heat exchange with the surroundings), the total heat lost by the hot body is equal to the total heat gained by the cold bodies.

1. Identify the given values:
- Mass of cold water initially in the calorimeter, $m_1=0.2\text{kg}$
- Initial temperature of the calorimeter and cold water, $T_1=30\text{°C}$
- Mass of hot water added, $m_2=0.1\text{kg}$
- Temperature of hot water added, $T_2=60\text{°C}$
- Final equilibrium temperature of the mixture, $T_f=35\text{°C}$
- Specific heat capacity of water, $s_w=4200\text{J/(kg·K)}$
- Let the thermal capacity of the calorimeter be $C_c$.

2. Calculate the heat lost by the hot water:
The hot water cools down from 60°C to 35°C.
$Q_{\text{lost}}=m_2\times s_w\times (T_2-T_f)$
Substituting the values:
$Q_{\text{lost}}=0.1\text{kg}\times 4200\text{J/(kg·K)}\times (60-35)\text{K}$
$Q_{\text{lost}}=420\times 25=10500\text{J}$

3. Calculate the heat gained by the cold water and the calorimeter:
Both the cold water and the calorimeter warm up from 30°C to 35°C.
$Q_{\text{gained}}=(m_1\times s_w+C_c)\times (T_f-T_1)$
Substituting the values:
$Q_{\text{gained}}=(0.2\text{kg}\times 4200\text{J/(kg·K)}+C_c)\times (35-30)\text{K}$
$Q_{\text{gained}}=(840+C_c)\times 5$

4. Apply the principle of conservation of energy:
$Q_{\text{gained}}=Q_{\text{lost}}$
$(840+C_c)\times 5=10500$
Divide both sides by 5:
$840+C_c=2100$
Subtract 840 from both sides:
$C_c=2100-840=1260\text{J/K}$

Therefore, the thermal capacity of the calorimeter is 1260 J/K.