Water of mass `m_(2) = 1 kg` is contained in a copper calorimeter of mass `m_(1) = 1 kg`. Their common temperature `t = 10^(@)C`. Now a piece of ice of mass `m_(2) = 2 kg` and temperature is `-11^(@)C` dropped into the calorimeter. Neglecting any heat loss, the final temperature of system is. [specific heat of copper `=0.1 Kcal//kg^(@)C`, specific heat of water `= 1 Kcal//kg^(@)C`, specific heat of ice `= 0.5 Kcal//kg^(@)C`, latent heat of fusion of ice `= 78.7 Kcal//kg`]

Options

A. `0^(@)C`
B. `4^(@)C`
C. `-4^(@)C`
D. `-2^(@)C`

Explanation

To solve the problem, we need to analyze the heat exchange between the water, the copper calorimeter, and the ice. We will calculate the heat lost by the water and the calorimeter as they cool down to 0°C, and the heat gained by the ice as it warms up to 0°C.

### Step-by-Step Solution:

1. **Identify the given data:**
- Mass of water, \( m_2 = 1 \, \text{kg} \)
- Mass of copper calorimeter, \( m_1 = 1 \, \text{kg} \)
- Initial temperature of water and calorimeter, \( t = 10^\circ C \)
- Mass of ice, \( m_3 = 2 \, \text{kg} \)
- Initial temperature of ice, \( t_3 = -11^\circ C \)
- Specific heat of copper, \( c_1 = 0.1 \, \text{Kcal/kg}^\circ C \)
- Specific heat of water, \( c_2 = 1 \, \text{Kcal/kg}^\circ C \)
- Specific heat of ice, \( c_3 = 0.5 \, \text{Kcal/kg}^\circ C \)
- Latent heat of fusion of ice, \( L_f = 78.7 \, \text{Kcal/kg} \)

2. **Calculate the heat lost by the water and calorimeter when they cool down to 0°C:**
- Heat lost by water:
\[ Q_{\text{water}} = m_2 \cdot c_2 \cdot (t - 0) = 1 \cdot 1 \cdot (10 - 0) = 10 \, \text{Kcal} \]
- Heat lost by copper calorimeter:
\[ Q_{\text{calorimeter}} = m_1 \cdot c_1 \cdot (t - 0) = 1 \cdot 0.1 \cdot (10 - 0) = 1 \, \text{Kcal} \]
- Total heat lost by the system:
\[ Q_{\text{lost}} = Q_{\text{water}} + Q_{\text{calorimeter}} = 10 + 1 = 11 \, \text{Kcal} \]

3. **Calculate the heat gained by the ice to reach 0°C:**
- Heat gained by ice to reach 0°C:
\[ Q_{\text{ice}} = m_3 \cdot c_3 \cdot (0 - t_3) = 2 \cdot 0.5 \cdot (0 - (-11)) = 2 \cdot 0.5 \cdot 11 = 11 \, \text{Kcal} \]

4. **Compare the heat lost and gained:**
- Heat lost by the water and calorimeter is 11 Kcal.
- Heat gained by the ice is also 11 Kcal.
- Since the heat lost by the calorimeter and water equals the heat gained by the ice, the ice will reach 0°C, and the system will stabilize at this temperature.

5. **Conclusion:**
- The final temperature of the system is \( 0^\circ C \).