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. A) `(ML)/(mctheta)`
B. B) `(mctheta)/(ML)`
C. C) `(Mctheta)/(L)`
D. D) `(mctheta)/(L)`

Answer

To solve the problem, we need to analyze the heat transfer involved when a small mass of water at a certain temperature is poured onto a larger mass of ice at its melting point.

### Step-by-Step Solution:

1. **Identify the Heat Gained by Ice:**
The heat gained by the ice when it melts can be calculated using the formula:
\[ Q_{\text{gain}} = m_1 \cdot L \]
where \( m_1 \) is the mass of ice melted and \( L \) is the specific latent heat of fusion of ice.

2. **Identify the Heat Lost by Water:**
The heat lost by the water as it cools down to 0°C can be calculated using the formula:
\[ Q_{\text{loss}} = m \cdot c \cdot (θ - 0) \]
where \( m \) is the mass of the water, \( c \) is the specific heat capacity of water, and \( θ \) is the initial temperature of the water.

3. **Set Heat Gain Equal to Heat Loss:**
Since the heat gained by the ice is equal to the heat lost by the water, we can set the two equations equal to each other:
\[ m_1 \cdot L = m \cdot c \cdot θ \]

4. **Solve for the Mass of Ice Melted (m1):**
Rearranging the equation to solve for \( m_1 \):
\[ m_1 = \frac{m \cdot c \cdot θ}{L} \]

5. **Final Result:**
Thus, the mass of ice melted \( m_1 \) is given by:
\[ m_1 = \frac{m \cdot c \cdot θ}{L} \]