Question

The power of a refrigerator that can make 15 kg of ice at 0 °C from water at 30 °C in one hour is

• A. 6600 W
• B. 1925 W
• C. 2200 W
• D. 4620 W

Hint:

This is a multi-step calculation. Break it down into sensible heat and latent heat. Always convert time to seconds to get the power in Watts (J/s). Using standard approximate values like \(c = 4200\) J/kg°C and \(L_f = 3.36 \times 10^5\) J/kg is common in competitive exams and usually leads to one of the given options.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept: The refrigerator extracts heat to cool water from \(30^\circ\text{C}\) to \(0^\circ\text{C}\) and then to freeze it into ice at \(0^\circ\text{C}\). Power is the rate of heat extraction per unit time.
Step 2: Key Formulas: Total Heat Extracted (\(Q\)): \[ Q = Q_{\text{cooling}} + Q_{\text{freezing}} \] \[ Q = mc\Delta T + mL_f \] Where: - \( m = 15 \) kg (mass of water) - \( c = 1000 \text{ cal/kg}^\circ\text{C} = 1 \text{ kcal/kg}^\circ\text{C} \) (specific heat of water) - \( L_f = 80 \text{ kcal/kg} \) (latent heat of fusion) - \( 1 \text{ kcal} = 4200 \text{ J} \) Power (\(P\)): \[ P = \frac{Q}{t} \] Where \( t \) is time in seconds.
Step 3: Detailed Calculation: 1. Calculate Heat \( Q \) in kcal: - Heat to cool water (\( 30^\circ\text{C} \to 0^\circ\text{C} \)): \[ Q_1 = 15 \times 1 \times (30 - 0) = 450 \text{ kcal} \] - Heat to freeze water at \( 0^\circ\text{C} \): \[ Q_2 = 15 \times 80 = 1200 \text{ kcal} \] - Total Heat: \[ Q = 450 + 1200 = 1650 \text{ kcal} \] 2. Convert to Joules: \[ Q = 1650 \times 4200 \text{ J} \] \[ Q = 6,930,000 \text{ J} \] 3. Calculate Power: Time \( t = 1 \text{ hour} = 3600 \text{ s} \). \[ P = \frac{6,930,000}{3600} \] \[ P = \frac{69300}{36} \] \[ P = 1925 \text{ W} \]
Step 4: Final Answer: The power required is 1925 W.