4 kg of ice at `-15^(@)C` are added to 5 kg of water at `15^(@)C` . The temperature of the mixture equals

Options

A. `-15^(@)C`
B. `0^(@)C`
C. ` 5^(@)C`
D. `15^(@)C`

Explanation

To solve the problem of finding the final temperature of the mixture when 4 kg of ice at -15°C is added to 5 kg of water at 15°C, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Mass of ice, \( m_{ice} = 4 \, \text{kg} \)
- Initial temperature of ice, \( T_{ice} = -15°C \)
- Mass of water, \( m_{water} = 5 \, \text{kg} \)
- Initial temperature of water, \( T_{water} = 15°C \)

2. **Calculate the Heat Required to Raise the Temperature of Ice to 0°C**:
- The specific heat of ice, \( c_{ice} = 2.1 \, \text{J/g°C} \) or \( 2100 \, \text{J/kg°C} \).
- The temperature change for ice, \( \Delta T_{ice} = 0 - (-15) = 15°C \).
- Heat absorbed by ice to reach 0°C:
\[ Q_{ice} = m_{ice} \cdot c_{ice} \cdot \Delta T_{ice} = 4 \, \text{kg} \cdot 2100 \, \text{J/kg°C} \cdot 15°C = 126000 \, \text{J} \]

3. **Calculate the Heat Released by Water When it Cools to 0°C**:
- The specific heat of water, \( c_{water} = 4.18 \, \text{J/g°C} \) or \( 4180 \, \text{J/kg°C} \).
- The temperature change for water, \( \Delta T_{water} = 15 - 0 = 15°C \).
- Heat released by water when it cools to 0°C:
\[ Q_{water} = m_{water} \cdot c_{water} \cdot \Delta T_{water} = 5 \, \text{kg} \cdot 4180 \, \text{J/kg°C} \cdot 15°C = 313500 \, \text{J} \]

4. **Compare Heat Absorbed by Ice and Heat Released by Water**:
- Heat required to raise ice to 0°C: \( Q_{ice} = 126000 \, \text{J} \)
- Heat released by water: \( Q_{water} = 313500 \, \text{J} \)

5. **Determine If Ice Melts**:
- Since \( Q_{water} > Q_{ice} \), the ice will not only reach 0°C but will also start to melt.
- The latent heat of fusion of ice is \( L_{fusion} = 334000 \, \text{J/kg} \).

6. **Calculate the Amount of Ice that Melts**:
- The remaining heat after raising the ice to 0°C:
\[ Q_{remaining} = Q_{water} - Q_{ice} = 313500 \, \text{J} - 126000 \, \text{J} = 187500 \, \text{J} \]
- The mass of ice that can melt with the remaining heat:
\[ m_{melted} = \frac{Q_{remaining}}{L_{fusion}} = \frac{187500 \, \text{J}}{334000 \, \text{J/kg}} \approx 0.56 \, \text{kg} \]

7. **Final State of the Mixture**:
- After melting, there will still be some ice left since the initial mass of ice was 4 kg and only about 0.56 kg melted.
- Therefore, the final temperature of the mixture will be 0°C, as there will be both water and ice present at this temperature.

### Conclusion:
The final temperature of the mixture is **0°C**.