What will be the final temperature when 150 g of ice at `0^@C` is mixed with 300 g of water at `50^@C`. Specific heat of water `=1 cal//g//^@C`. Latent heat of fusion of ice `=80 cal//g`.

Options

A. `7.6^@C`
B. `6.7^@C`
C. `5.8^@C`
D. `8.5^@C`

Explanation

To find the final temperature when 150 g of ice at 0°C is mixed with 300 g of water at 50°C, we will use the principle of calorimetry, which states that the heat lost by the water will be equal to the heat gained by the ice.

### Step-by-Step Solution:

1. **Identify the Heat Gained by Ice:**
- The ice will first absorb heat to melt into water at 0°C. The heat required to melt the ice can be calculated using the formula:
\[ Q_{\text{melt}} = m_{\text{ice}} \times L_f \]
where \( m_{\text{ice}} = 150 \, \text{g} \) and \( L_f = 80 \, \text{cal/g} \).
\[ Q_{\text{melt}} = 150 \, \text{g} \times 80 \, \text{cal/g} = 12000 \, \text{cal} \]

2. **Identify the Heat Gained by the Melted Ice:**
- After melting, the water from the ice will further absorb heat to increase its temperature from 0°C to the final temperature \( T \):
\[ Q_{\text{ice}} = m_{\text{water}} \times c \times \Delta T \]
where \( m_{\text{water}} = 150 \, \text{g} \), \( c = 1 \, \text{cal/g°C} \), and \( \Delta T = T - 0 \):
\[ Q_{\text{ice}} = 150 \, \text{g} \times 1 \, \text{cal/g°C} \times T = 150T \, \text{cal} \]

3. **Identify the Heat Lost by Water:**
- The water at 50°C will lose heat as it cools down to the final temperature \( T \):
\[ Q_{\text{water}} = m_{\text{water}} \times c \times \Delta T \]
where \( m_{\text{water}} = 300 \, \text{g} \), \( c = 1 \, \text{cal/g°C} \), and \( \Delta T = 50 - T \):
\[ Q_{\text{water}} = 300 \, \text{g} \times 1 \, \text{cal/g°C} \times (50 - T) = 300(50 - T) \, \text{cal} \]

4. **Set Up the Equation:**
- According to the principle of calorimetry, the heat lost by the water equals the heat gained by the ice:
\[ Q_{\text{water}} = Q_{\text{melt}} + Q_{\text{ice}} \]
This gives us the equation:
\[ 300(50 - T) = 12000 + 150T \]

5. **Solve for \( T \):**
- Expanding and rearranging the equation:
\[ 15000 - 300T = 12000 + 150T \]
\[ 15000 - 12000 = 300T + 150T \]
\[ 3000 = 450T \]
\[ T = \frac{3000}{450} = \frac{60}{9} \approx 6.67 \, °C \]

### Final Answer:
The final temperature when 150 g of ice at 0°C is mixed with 300 g of water at 50°C is approximately **6.67°C**.