60 g of ice at `0^(@)C` is added to 200 g of water initially at `70^(@)C` in a calorimeter of unknown water equivalent W. If the final temperature of the mixture is `40^(@)C`, then the value of W is [Take latent heat of fusion of ice `L_(f) = 80 cal g^(-1)` and specific heat capacity of water `s = 1 cal g^(-1).^(@)C^(-1)`]

Options

A. 70 g
B. 80 g
C. 40 g
D. 20 g

Answer

To solve the problem, we will use the principle of conservation of energy, which states that the heat lost by the water will be equal to the heat gained by the ice and the calorimeter.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of ice, \( m_{ice} = 60 \, g \)
- Initial temperature of ice, \( T_{ice} = 0^\circ C \)
- Mass of water, \( m_{water} = 200 \, g \)
- Initial temperature of water, \( T_{water} = 70^\circ C \)
- Final temperature of the mixture, \( T_{final} = 40^\circ C \)
- Latent heat of fusion of ice, \( L_f = 80 \, cal/g \)
- Specific heat capacity of water, \( s = 1 \, cal/g^\circ C \)

2. **Calculate the heat gained by the ice:**
- The ice first absorbs heat to melt into water at \( 0^\circ C \):
\[ Q_{melt} = m_{ice} \times L_f = 60 \, g \times 80 \, cal/g = 4800 \, cal \]
- After melting, the water at \( 0^\circ C \) will absorb heat to reach \( 40^\circ C \):
\[ Q_{heat\_up} = m_{ice} \times s \times \Delta T = 60 \, g \times 1 \, cal/g^\circ C \times (40 - 0)^\circ C = 2400 \, cal \]
- Total heat gained by the ice:
\[ Q_{ice} = Q_{melt} + Q_{heat\_up} = 4800 \, cal + 2400 \, cal = 7200 \, cal \]

3. **Calculate the heat lost by the water:**
- The water loses heat as it cools from \( 70^\circ C \) to \( 40^\circ C \):
\[ Q_{water} = (m_{water} + W) \times s \times \Delta T = (200 \, g + W) \times 1 \, cal/g^\circ C \times (70 - 40)^\circ C \]
\[ Q_{water} = (200 + W) \times 30 \, cal = 6000 + 30W \, cal \]

4. **Set the heat gained by the ice equal to the heat lost by the water:**
\[ Q_{ice} = Q_{water} \]
\[ 7200 \, cal = 6000 + 30W \]

5. **Solve for \( W \):**
\[ 7200 - 6000 = 30W \]
\[ 1200 = 30W \]
\[ W = \frac{1200}{30} = 40 \, g \]

### Final Answer:
The value of \( W \) is \( 40 \, g \).