In a container of negligible mass 140 g of ice initially at `-15^@C` is added to 200 g of water that has a temperature of `40^@C`. If no heat is lost to the surroundings, what is the final temperature of the system and masses of water and ice in mixture ?

Explanation

To solve the problem, we need to analyze the heat exchange between the ice and the water. We will use the principle of conservation of energy, which states that the heat lost by the warm water will be equal to the heat gained by the cold ice.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of ice, \( m_{\text{ice}} = 140 \, \text{g} \)
- Initial temperature of ice, \( T_{\text{ice}} = -15^\circ \text{C} \)
- Mass of water, \( m_{\text{water}} = 200 \, \text{g} \)
- Initial temperature of water, \( T_{\text{water}} = 40^\circ \text{C} \)
- Specific heat of water, \( c_{\text{water}} = 1 \, \text{cal/g}^\circ \text{C} \)
- Specific heat of ice, \( c_{\text{ice}} = 0.5 \, \text{cal/g}^\circ \text{C} \)
- Latent heat of fusion of ice, \( L_f = 80 \, \text{cal/g} \)

2. **Calculate the Heat Required to Raise the Temperature of Ice to 0°C:**
\[ Q_1 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T_{\text{ice}} = 140 \, \text{g} \cdot 0.5 \, \text{cal/g}^\circ \text{C} \cdot (0 - (-15)) = 140 \cdot 0.5 \cdot 15 = 1050 \, \text{cal} \]

3. **Calculate the Heat Required to Melt the Ice:**
\[ Q_2 = m_{\text{melted ice}} \cdot L_f \]
Let \( m_{\text{melted ice}} \) be the mass of ice that melts. The total heat required to raise the temperature of the ice to 0°C and then melt it is:
\[ Q_{\text{ice}} = Q_1 + Q_2 = 1050 + m_{\text{melted ice}} \cdot 80 \]

4. **Calculate the Heat Lost by Water:**
\[ Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} = 200 \, \text{g} \cdot 1 \, \text{cal/g}^\circ \text{C} \cdot (40 - 0) = 200 \cdot 40 = 8000 \, \text{cal} \]

5. **Set Up the Energy Balance Equation:**
Since no heat is lost to the surroundings, we can set the heat gained by the ice equal to the heat lost by the water:
\[ Q_{\text{ice}} = Q_{\text{water}} \]
\[ 1050 + m_{\text{melted ice}} \cdot 80 = 8000 \]

6. **Solve for the Mass of Melted Ice:**
\[ m_{\text{melted ice}} \cdot 80 = 8000 - 1050 = 6950 \]
\[ m_{\text{melted ice}} = \frac{6950}{80} \approx 86.875 \, \text{g} \]

7. **Calculate the Final Masses:**
- Total mass of water after melting:
\[ m_{\text{final water}} = m_{\text{water}} + m_{\text{melted ice}} = 200 + 86.875 \approx 286.875 \, \text{g} \]
- Remaining mass of ice:
\[ m_{\text{remaining ice}} = m_{\text{ice}} - m_{\text{melted ice}} = 140 - 86.875 \approx 53.125 \, \text{g} \]

8. **Final Temperature of the System:**
Since all the ice that melted is at 0°C and the water is also at 0°C after the heat exchange, the final temperature of the system is:
\[ T_{\text{final}} = 0^\circ \text{C} \]

### Summary of Results:
- Final Temperature: \( 0^\circ \text{C} \)
- Mass of Water in the Mixture: \( \approx 286.875 \, \text{g} \)
- Mass of Remaining Ice: \( \approx 53.125 \, \text{g} \)