In summer, Kamal decides to take a bath at noon. Temperature of the water in overhead tank is `45^(@)C` so he decides to mix ince from refrigerator. The temperature of the ice in refrigerator. The temperature of the ice in refrigerator is `-10^(@)C`. How much ice will be needed to prepare 20 kg of water at `25^(@)C` for bath? (Assume no loss to surrounding latent heat of fusion of ice `=80cal//gm`, specific heat of ice `=0.5cal//gm^(@)C`

Options

A. `400//115kg`
B. `40//13kg`
C. `40//11kg`
D. `400//135`kg

Explanation

To solve the problem of how much ice is needed to prepare 20 kg of water at 25°C from water at 45°C and ice at -10°C, we will use the principles of heat transfer and conservation of energy.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let the mass of ice needed be \( m \) grams.
- The mass of water from the overhead tank is \( 20 - m \) grams (since the total mass of the mixture is 20 kg).
- The initial temperature of the water \( T_w = 45^\circ C \).
- The final temperature of the mixture \( T_f = 25^\circ C \).
- The initial temperature of the ice \( T_i = -10^\circ C \).

2. **Calculate the Heat Required to Warm the Ice to 0°C**:
- The specific heat of ice \( c_{ice} = 0.5 \, \text{cal/g°C} \).
- The heat required to raise the temperature of ice from -10°C to 0°C:
\[ Q_1 = m \cdot c_{ice} \cdot \Delta T = m \cdot 0.5 \cdot (0 - (-10)) = 5m \, \text{cal} \]

3. **Calculate the Heat Required to Melt the Ice**:
- The latent heat of fusion of ice \( L_f = 80 \, \text{cal/g} \).
- The heat required to melt the ice at 0°C:
\[ Q_2 = m \cdot L_f = m \cdot 80 = 80m \, \text{cal} \]

4. **Calculate the Heat Required to Raise the Temperature of Melted Ice to 25°C**:
- The specific heat of water \( c_{water} = 1 \, \text{cal/g°C} \).
- The heat required to raise the temperature of the melted ice (now water) from 0°C to 25°C:
\[ Q_3 = m \cdot c_{water} \cdot \Delta T = m \cdot 1 \cdot (25 - 0) = 25m \, \text{cal} \]

5. **Total Heat Required for the Ice**:
- The total heat required to convert the ice at -10°C to water at 25°C:
\[ Q_{ice} = Q_1 + Q_2 + Q_3 = 5m + 80m + 25m = 110m \, \text{cal} \]

6. **Calculate the Heat Released by the Water**:
- The heat released by the water cooling from 45°C to 25°C:
\[ Q_{water} = (20 - m) \cdot c_{water} \cdot \Delta T = (20 - m) \cdot 1 \cdot (25 - 45) = (20 - m) \cdot (-20) = -20(20 - m) = -400 + 20m \, \text{cal} \]

7. **Set the Heat Gained by Ice Equal to the Heat Lost by Water**:
- According to the principle of conservation of energy:
\[ Q_{ice} = -Q_{water} \]
\[ 110m = 400 - 20m \]

8. **Solve for m**:
- Rearranging the equation:
\[ 110m + 20m = 400 \]
\[ 130m = 400 \]
\[ m = \frac{400}{130} \approx 3.08 \, \text{kg} \]

### Final Answer:
The mass of ice needed is approximately **3.08 kg**.