`m_(1)` gram of ice at `-10^(@)C` and `m_(2)` gram of water at `50^(@)C` are mixed in insulated container. If in equilibrium state we get only water at `0^(@)C` then latent heat of ice is :

Options

A. `(50M_(2))/(M_(1))-5`
B. `(5M_(1))/(M_(2))-50`
C. `(50M_(2))/(M_(1))`
D. `(5M_(2))/(M_(1))-5`

Answer

To solve the problem, we need to consider the heat exchange between the ice and the water. The heat gained by the ice must equal the heat lost by the water in an insulated system.

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- Mass of ice, \( m_1 \) grams at \( -10^\circ C \)
- Mass of water, \( m_2 \) grams at \( 50^\circ C \)
- Final equilibrium temperature is \( 0^\circ C \)

2. **Calculate the heat gained by the ice**:
- The ice first warms up from \( -10^\circ C \) to \( 0^\circ C \).
- The specific heat of ice is \( c_{ice} = 0.5 \, \text{cal/g}^\circ C \).
- The heat gained by the ice to raise its temperature:
\[ Q_{ice} = m_1 \cdot c_{ice} \cdot \Delta T_{ice} = m_1 \cdot 0.5 \cdot (0 - (-10)) = m_1 \cdot 0.5 \cdot 10 = 5m_1 \, \text{cal} \]

3. **Calculate the heat required to melt the ice**:
- Let \( L_f \) be the latent heat of fusion of ice.
- The heat required to melt the ice at \( 0^\circ C \):
\[ Q_{melt} = m_1 \cdot L_f \]

4. **Total heat gained by the ice**:
- The total heat gained by the ice is the sum of the heat to warm it and the heat to melt it:
\[ Q_{total \, ice} = Q_{ice} + Q_{melt} = 5m_1 + m_1 \cdot L_f \]

5. **Calculate the heat lost by the water**:
- The water cools from \( 50^\circ C \) to \( 0^\circ C \).
- The specific heat of water is \( c_{water} = 1 \, \text{cal/g}^\circ C \).
- The heat lost by the water:
\[ Q_{water} = m_2 \cdot c_{water} \cdot \Delta T_{water} = m_2 \cdot 1 \cdot (0 - 50) = -50m_2 \, \text{cal} \]

6. **Set the heat gained by ice equal to the heat lost by water**:
- In an insulated system, the heat gained by the ice equals the heat lost by the water:
\[ 5m_1 + m_1 \cdot L_f = 50m_2 \]

7. **Rearranging the equation**:
- Rearranging gives:
\[ m_1 \cdot L_f = 50m_2 - 5m_1 \]

8. **Solve for the latent heat of fusion \( L_f \)**:
- Dividing both sides by \( m_1 \):
\[ L_f = \frac{50m_2 - 5m_1}{m_1} \]
- This simplifies to:
\[ L_f = \frac{50m_2}{m_1} - 5 \]

### Final Expression:
The latent heat of fusion of ice is given by:
\[ L_f = \frac{50m_2}{m_1} - 5 \]