A piece of ice of mass `100 g` and at temperature `0^@ C` is put in `200 g` of water of `25^@ C`. How much ice will melt as the temperature of the water reaches `0^@ C` ? (specific heat capacity of water `=4200 J kg^(-1) K^(-1)` and latent heat of fusion of ice `= 3.4 xx 10^(5) J Kg^(-1)`).

Options

A. 128 g
B. 185.4 g
C. 92.8g
D. 61.8g

Explanation

To solve the problem, we need to determine how much ice will melt when it is put into water at a higher temperature until the water reaches 0°C. We can use the principle of conservation of energy, where the heat lost by the water will be equal to the heat gained by the ice.

### Step-by-step Solution:

1. **Identify the Given Values:**
- Mass of ice, \( m_{\text{ice}} = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Mass of water, \( m_{\text{water}} = 200 \, \text{g} = 0.2 \, \text{kg} \)
- Initial temperature of water, \( T_{\text{initial}} = 25^\circ C \)
- Final temperature of the system, \( T_{\text{final}} = 0^\circ C \)
- Specific heat capacity of water, \( c = 4200 \, \text{J/kg/K} \)
- Latent heat of fusion of ice, \( L_f = 3.4 \times 10^5 \, \text{J/kg} \)

2. **Calculate the Heat Lost by the Water:**
The heat lost by the water as it cools down from 25°C to 0°C can be calculated using the formula:
\[ Q_{\text{water}} = m_{\text{water}} \cdot c \cdot \Delta T \]
Where \( \Delta T = T_{\text{initial}} - T_{\text{final}} = 25 - 0 = 25 \, \text{K} \).
\[ Q_{\text{water}} = 0.2 \, \text{kg} \cdot 4200 \, \text{J/kg/K} \cdot 25 \, \text{K} \]
\[ Q_{\text{water}} = 0.2 \cdot 4200 \cdot 25 = 21000 \, \text{J} \]

3. **Relate Heat Lost by Water to Heat Gained by Ice:**
The heat gained by the ice to melt can be expressed as:
\[ Q_{\text{ice}} = m' \cdot L_f \]
Where \( m' \) is the mass of the melted ice.

4. **Set Heat Lost Equal to Heat Gained:**
Since the heat lost by the water will be equal to the heat gained by the ice:
\[ Q_{\text{water}} = Q_{\text{ice}} \]
\[ 21000 \, \text{J} = m' \cdot 3.4 \times 10^5 \, \text{J/kg} \]

5. **Solve for the Mass of Melted Ice:**
Rearranging the equation gives:
\[ m' = \frac{21000 \, \text{J}}{3.4 \times 10^5 \, \text{J/kg}} \]
\[ m' = \frac{21000}{340000} \approx 0.0618 \, \text{kg} = 61.8 \, \text{g} \]

### Conclusion:
The mass of ice that will melt as the temperature of the water reaches 0°C is approximately **61.8 grams**.