10 g of ice of `0^(@)C` is mixed with `100 g` of water at `50^(@)C` in a calorimeter. The final temperature of the mixture is [Specific heat of water `= 1 cal g^(-1).^(@)C^(-1)`, latent of fusion of ice `= 80 cal g^(-1)`]

Options

A. `31.2^(@)C`
B. `32.8^(@)C`
C. `36.7^(@)C`
D. `38.2^(@)C`

Explanation

To solve the problem of mixing 10 g of ice at 0°C with 100 g of water at 50°C, we will use the principle of conservation of energy, specifically the concept of calorimetry, which states that the heat lost by the hot water will be equal to the heat gained by the ice.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of ice, \( m_{ice} = 10 \, \text{g} \)
- Initial temperature of ice, \( T_{ice} = 0 \, \text{°C} \)
- Mass of water, \( m_{water} = 100 \, \text{g} \)
- Initial temperature of water, \( T_{water} = 50 \, \text{°C} \)
- Specific heat of water, \( c_{water} = 1 \, \text{cal/g°C} \)
- Latent heat of fusion of ice, \( L_{fusion} = 80 \, \text{cal/g} \)

2. **Calculate the Heat Gained by the Ice:**
- The ice will first melt, which requires heat:
\[ Q_{melt} = m_{ice} \times L_{fusion} = 10 \, \text{g} \times 80 \, \text{cal/g} = 800 \, \text{cal} \]
- After melting, the water from the melted ice will increase its temperature from 0°C to the final temperature \( T_f \):
\[ Q_{ice\_temp} = m_{ice} \times c_{water} \times (T_f - T_{ice}) = 10 \, \text{g} \times 1 \, \text{cal/g°C} \times (T_f - 0) = 10 T_f \, \text{cal} \]

3. **Calculate the Heat Lost by the Water:**
- The water will cool down from 50°C to the final temperature \( T_f \):
\[ Q_{water} = m_{water} \times c_{water} \times (T_{water} - T_f) = 100 \, \text{g} \times 1 \, \text{cal/g°C} \times (50 - T_f) = 5000 - 100 T_f \, \text{cal} \]

4. **Set Up the Heat Exchange Equation:**
- According to the principle of calorimetry:
\[ Q_{lost} = Q_{gained} \]
Therefore:
\[ 5000 - 100 T_f = 800 + 10 T_f \]

5. **Solve for \( T_f \):**
- Rearranging the equation:
\[ 5000 - 800 = 100 T_f + 10 T_f \]
\[ 4200 = 110 T_f \]
\[ T_f = \frac{4200}{110} \approx 38.18 \, \text{°C} \]

6. **Final Answer:**
- The final temperature of the mixture is approximately \( 38.18 \, \text{°C} \).