If 10 g of ice at `0^(@)C` is mixed with 10 g of water at `40^(@)C`. The final mass of water in mixture is (Latent heat of fusion of ice = 80 cel/g, specific heat of water `=1 cal/g""^(@)C`)

Options

A. 10 g
B. 15 g
C. 18 g
D. 20 g

Answer

To solve the problem, we need to determine the final mass of water in the mixture when 10 g of ice at 0°C is mixed with 10 g of water at 40°C.

### Step-by-Step Solution:

1. **Identify the given data:**
- Mass of ice, \( m_{ice} = 10 \, \text{g} \)
- Mass of water, \( m_{water} = 10 \, \text{g} \)
- Latent heat of fusion of ice, \( L = 80 \, \text{cal/g} \)
- Specific heat of water, \( s = 1 \, \text{cal/g°C} \)
- Initial temperature of ice, \( T_{ice} = 0°C \)
- Initial temperature of water, \( T_{water} = 40°C \)

2. **Calculate the heat released by the water when it cools down from 40°C to 0°C:**
\[ Q_{released} = m_{water} \times s \times \Delta T \]
Where \( \Delta T = T_{water} - T_{final} = 40°C - 0°C = 40°C \).
\[ Q_{released} = 10 \, \text{g} \times 1 \, \text{cal/g°C} \times 40°C = 400 \, \text{cal} \]

3. **Calculate the heat required to convert ice at 0°C to water at 0°C:**
\[ Q_{absorbed} = m_{ice} \times L \]
Let \( m \) be the mass of ice that melts. Then,
\[ Q_{absorbed} = m \times 80 \, \text{cal/g} \]

4. **Set the heat released equal to the heat absorbed:**
\[ Q_{released} = Q_{absorbed} \]
\[ 400 \, \text{cal} = m \times 80 \, \text{cal/g} \]
Solving for \( m \):
\[ m = \frac{400 \, \text{cal}}{80 \, \text{cal/g}} = 5 \, \text{g} \]

5. **Determine the final mass of water in the mixture:**
- The mass of water formed from the melted ice is \( 5 \, \text{g} \).
- The remaining ice is \( 10 \, \text{g} - 5 \, \text{g} = 5 \, \text{g} \) (still in solid form).
- The total mass of water in the mixture is:
\[ \text{Total mass of water} = m_{water} + m_{melted \, ice} = 10 \, \text{g} + 5 \, \text{g} = 15 \, \text{g} \]

### Final Answer:
The final mass of water in the mixture is **15 g**.