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:
Power is the rate at which energy is transferred. In this case, it's the rate at which the refrigerator removes heat from the water. The total heat removed consists of two parts: the heat removed to cool the water from 30 °C to 0 °C (sensible heat), and the heat removed to convert the water at 0 °C into ice at 0 °C (latent heat).
Step 2: Key Formula or Approach:
1. Heat removed to cool water: \(Q_1 = mc\Delta T\), where \(c\) is the specific heat capacity of water.
2. Heat removed to freeze water: \(Q_2 = mL_f\), where \(L_f\) is the latent heat of fusion of ice.
3. Total heat removed: \(Q_{total} = Q_1 + Q_2\).
4. Power: \(P = \frac{Q_{total}}{\text{time}}\).
Standard values to be used: \(c_{water} = 4200 \text{ J/kg}^\circ C\), \(L_{f,ice} = 336000 \text{ J/kg}\).
Step 3: Detailed Explanation:
Given values:
- Mass of water, \(m = 15\) kg.
- Initial temperature = 30 °C.
- Final temperature = 0 °C.
- Time, \(t = 1\) hour = \(3600\) seconds.
Calculate the heat removed to cool the water (\(Q_1\)):
\[ Q_1 = mc\Delta T = 15 \text{ kg} \times 4200 \text{ J/kg}^\circ C \times (30 - 0)^\circ C \] \[ Q_1 = 15 \times 4200 \times 30 = 1,890,000 \text{ J} \] Calculate the heat removed to freeze the water (\(Q_2\)):
\[ Q_2 = mL_f = 15 \text{ kg} \times 336000 \text{ J/kg} \] \[ Q_2 = 5,040,000 \text{ J} \] Calculate the total heat removed (\(Q_{total}\)):
\[ Q_{total} = Q_1 + Q_2 = 1,890,000 + 5,040,000 = 6,930,000 \text{ J} \] Calculate the power of the refrigerator (\(P\)):
\[ P = \frac{Q_{total}}{t} = \frac{6,930,000 \text{ J}}{3600 \text{ s}} = \frac{69300}{36} = 1925 \text{ W} \] Step 4: Final Answer:
The power of the refrigerator is 1925 W. Therefore, option (B) is correct.