50 g ice at `0^(@)C` in kept in an insulating vessel and 50 g watat at `100^(@)C` is mixed in it. Then the final temperature of the mixture is (neglect the heat loss)

Options

A. (a)`10^(@)C`
B. (b)`0^(@)C lt Tm lt 20^(@)C`
C. (c)`20^(@)C`
D. (d)above `20^(@)C`

Answer

To solve the problem of mixing 50 g of ice at 0°C with 50 g of water at 100°C, we will use the principle of conservation of energy, where the heat gained by the ice and the melted water equals the heat lost by the hot water.

### Step-by-Step Solution:

1. **Identify the Masses and Temperatures:**
- Mass of ice, \( m_i = 50 \, \text{g} \)
- Initial temperature of ice, \( T_i = 0 \, \text{°C} \)
- Mass of water, \( m_w = 50 \, \text{g} \)
- Initial temperature of water, \( T_w = 100 \, \text{°C} \)

2. **Specific Heat and Latent Heat:**
- Specific heat of water, \( C_w = 1 \, \text{cal/g°C} \)
- Latent heat of fusion of ice, \( L = 80 \, \text{cal/g} \)

3. **Set Up the Heat Gain and Loss Equation:**
- Let \( T \) be the final temperature of the mixture.
- The heat gained by the ice as it melts and warms up to \( T \) is given by:
\[ Q_{\text{gain}} = m_i \cdot L + m_i \cdot C_w \cdot (T - T_i) \]
- The heat lost by the water as it cools down to \( T \) is given by:
\[ Q_{\text{loss}} = m_w \cdot C_w \cdot (T_w - T) \]

4. **Substituting Values:**
- Substitute the known values into the equations:
\[ Q_{\text{gain}} = 50 \cdot 80 + 50 \cdot 1 \cdot (T - 0) \]
\[ Q_{\text{loss}} = 50 \cdot 1 \cdot (100 - T) \]

5. **Equating Heat Gain and Loss:**
- Set the heat gained equal to the heat lost:
\[ 50 \cdot 80 + 50 \cdot (T) = 50 \cdot (100 - T) \]

6. **Simplifying the Equation:**
- Simplify the equation:
\[ 4000 + 50T = 5000 - 50T \]
- Combine like terms:
\[ 100T = 5000 - 4000 \]
\[ 100T = 1000 \]
\[ T = 10 \, \text{°C} \]

7. **Final Result:**
- The final temperature of the mixture is \( T = 10 \, \text{°C} \).

### Conclusion:
The final temperature of the mixture is **10°C**.