A refrigerator converts `100` g of water at `25^(@)C` into ice at `-10^(@)C` in one hour and `50` minutes. The quantity of heat removed per minute is (specific heat of ice = `0.5 "cal"//g^(@)C`, latent heat of fusion = `80 "cal"//g`)

Options

A. `50"cal"`
B. `100 ` cal
C. `200` cal
D. `75` cal

Explanation

To solve the problem, we need to calculate the total heat removed when converting 100 g of water at \(25^\circ C\) into ice at \(-10^\circ C\) and then find the quantity of heat removed per minute.

### Step-by-Step Solution:

1. **Calculate the heat removed to cool the water from \(25^\circ C\) to \(0^\circ C\)**:
- Formula: \( Q_1 = m \cdot c \cdot \Delta T \)
- Where:
- \( m = 100 \, \text{g} \)
- \( c = 1 \, \text{cal/g}^\circ C \) (specific heat of water)
- \( \Delta T = 25 - 0 = 25^\circ C \)
- Calculation:
\[ Q_1 = 100 \, \text{g} \cdot 1 \, \text{cal/g}^\circ C \cdot 25^\circ C = 2500 \, \text{cal} \]

2. **Calculate the heat removed to freeze the water at \(0^\circ C\) to ice at \(0^\circ C\)**:
- Formula: \( Q_2 = m \cdot L_f \)
- Where:
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion)
- Calculation:
\[ Q_2 = 100 \, \text{g} \cdot 80 \, \text{cal/g} = 8000 \, \text{cal} \]

3. **Calculate the heat removed to cool the ice from \(0^\circ C\) to \(-10^\circ C\)**:
- Formula: \( Q_3 = m \cdot c_{ice} \cdot \Delta T \)
- Where:
- \( c_{ice} = 0.5 \, \text{cal/g}^\circ C \) (specific heat of ice)
- \( \Delta T = 0 - (-10) = 10^\circ C \)
- Calculation:
\[ Q_3 = 100 \, \text{g} \cdot 0.5 \, \text{cal/g}^\circ C \cdot 10^\circ C = 500 \, \text{cal} \]

4. **Total heat removed (Q)**:
- Combine all the heat quantities:
\[ Q = Q_1 + Q_2 + Q_3 = 2500 \, \text{cal} + 8000 \, \text{cal} + 500 \, \text{cal} = 11000 \, \text{cal} \]

5. **Calculate the total time taken in minutes**:
- Total time = 1 hour and 50 minutes = \(60 + 50 = 110 \, \text{minutes}\)

6. **Calculate the heat removed per minute**:
- Formula: \( \text{Heat removed per minute} = \frac{Q}{\text{Total time}} \)
- Calculation:
\[ \text{Heat removed per minute} = \frac{11000 \, \text{cal}}{110 \, \text{minutes}} = 100 \, \text{cal/min} \]

### Final Answer:
The quantity of heat removed per minute is \(100 \, \text{cal/min}\).