50 g of ice at `0^@C` is mixed with 50 g of water at `80^@C` . The final temperature of the mixture is (latent heat of fusion of ice `=80 cal //g , s_(w) = 1 cal //g ^@C)`

Options

A. `0^@C`
B. `40^@C`
C. `60^@C`
D. less than `0^@C`

Answer

To solve the problem, we need to calculate the final temperature of the mixture when 50 g of ice at \(0^\circ C\) is mixed with 50 g of water at \(80^\circ C\).

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of ice, \(m_i = 50 \, \text{g}\)
- Initial temperature of ice, \(T_i = 0^\circ C\)
- Mass of water, \(m_w = 50 \, \text{g}\)
- Initial temperature of water, \(T_w = 80^\circ C\)
- Latent heat of fusion of ice, \(L_f = 80 \, \text{cal/g}\)
- Specific heat of water, \(s_w = 1 \, \text{cal/g} \cdot ^\circ C\)

2. **Calculate the Heat Lost by Water:**
The heat lost by the water as it cools down to the final temperature \(T\) can be expressed as:
\[ Q_w = m_w \cdot s_w \cdot (T_w - T) = 50 \cdot 1 \cdot (80 - T) \]
\[ Q_w = 50 \cdot (80 - T) = 4000 - 50T \, \text{cal} \]

3. **Calculate the Heat Gained by Ice:**
The heat gained by the ice consists of two parts: the heat required to melt the ice and the heat required to raise the temperature of the melted ice (water) to the final temperature \(T\):
\[ Q_i = m_i \cdot L_f + m_i \cdot s_w \cdot (T - T_i) \]
\[ Q_i = 50 \cdot 80 + 50 \cdot 1 \cdot (T - 0) \]
\[ Q_i = 4000 + 50T \, \text{cal} \]

4. **Set the Heat Lost Equal to Heat Gained:**
Since the heat lost by the water equals the heat gained by the ice, we can set the two equations equal to each other:
\[ Q_w = Q_i \]
\[ 4000 - 50T = 4000 + 50T \]

5. **Solve for \(T\):**
Rearranging the equation gives:
\[ 4000 - 4000 = 50T + 50T \]
\[ 0 = 100T \]
\[ T = 0^\circ C \]

6. **Conclusion:**
The final temperature of the mixture is \(0^\circ C\). At this temperature, there will still be some ice remaining since the heat lost by the water is not sufficient to completely melt the ice.