A small quantity mass m, of water at a temperature `theta ("in " ^(@)C)` is poured on to a larger mass M of ice which is at its melting point. If c is the specific heat capacity of water and L the specific heat capacity of water and L the specific latent heat of fusion of ice, then the mass of ice melted is give by

Options

A. `(ML)/(mc theta)`
B. `(mc theta)/(ML)`
C. `(M c theta)/(L)`
D. `(mc theta)/(L)`

Explanation

To solve the problem, we will use the principle of calorimetry, which states that the heat lost by the hot substance (water) is equal to the heat gained by the cold substance (ice).

### Step-by-Step Solution:

1. **Identify the Given Variables:**
- Mass of water = \( m \)
- Initial temperature of water = \( \theta \) (in °C)
- Specific heat capacity of water = \( C \)
- Mass of ice = \( M \) (but we are looking for the mass of ice melted)
- Latent heat of fusion of ice = \( L \)

2. **Write the Heat Lost by Water:**
The heat lost by the water as it cools down from temperature \( \theta \) to 0°C (the melting point of ice) can be expressed as:
\[ Q_{\text{lost}} = m \cdot C \cdot (\theta - 0) = m \cdot C \cdot \theta \]

3. **Write the Heat Gained by Ice:**
The heat gained by the ice as it melts into water at 0°C can be expressed as:
\[ Q_{\text{gained}} = M \cdot L \]
Here, \( M \) is the mass of ice melted, and \( L \) is the latent heat of fusion.

4. **Apply the Principle of Calorimetry:**
According to the principle of calorimetry:
\[ Q_{\text{lost}} = Q_{\text{gained}} \]
Therefore, we can set the two equations equal to each other:
\[ m \cdot C \cdot \theta = M \cdot L \]

5. **Solve for the Mass of Ice Melted (M):**
Rearranging the equation to solve for \( M \):
\[ M = \frac{m \cdot C \cdot \theta}{L} \]

### Final Answer:
The mass of ice melted is given by:
\[ M = \frac{m \cdot C \cdot \theta}{L} \]

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