Question

In a cold storage, ice melts at the rate of 2 kg per hour when the external temperature is 20°C. The minimum power output of the motor used to drive the refrigerator which just prevents the ice from melting is (latent heat of fusion of ice = 80 cal g\(^{-1}\))

• A. \( 28.5 \text{ W} \)
• B. \( 13.6 \text{ W} \)
• C. \( 9.75 \text{ W} \)
• D. \( 16.4 \text{ W} \)

Hint:

For problems involving refrigeration, always use the latent heat formula \( Q = mL \) to calculate heat energy and apply the coefficient of performance (COP) formula to find the minimum power required.

Answer 1

The Correct Option is B

Given: \[ \text{Latent heat of fusion of ice} = 80 \, \text{cal/g} \] \[ 1 \, \text{cal} = 4.184 \, \text{J} \] So, \[ L = 80 \times 4.184 = 334.72 \, \text{J/g} \]

Step 2: Convert Mass of Ice

\[ m = 2 \, \text{kg} = 2000 \, \text{g} \]

Step 3: Calculate Total Heat Required

\[ Q = m \cdot L = 2000 \times 334.72 = 669440 \, \text{J} \]

Step 4: Convert Time to Seconds

\[ t = 1 \, \text{hour} = 3600 \, \text{seconds} \]

Step 5: Calculate Power

\[ P = \frac{Q}{t} = \frac{669440}{3600} \approx 185.96 \, \text{W} \]

Conclusion:

The minimum power required by the motor is: \[ \boxed{185.96 \, \text{W}} \] If multiple-choice options are given and this value isn't among them, then either an approximation or a simplification (e.g., using \(1 \, \text{cal} = 4.2 \, \text{J}\)) may be expected.

Answer 2

Step 1: Understanding the Given Data 

• Mass of ice melting per hour: \( m = 2 \, \text{kg} = 2000 \, \text{g} \)
• Latent heat of fusion of ice: \( L = 80 \, \text{cal g}^{-1} \)
• Conversion: \( 1 \, \text{cal} = 4.2 \, \text{J} \)

Step 2: Calculating the Total Heat Required per Hour

\[ Q = m \cdot L = 2000 \cdot 80 = 160000 \, \text{cal} \] \[ Q = 160000 \cdot 4.2 = 672000 \, \text{J} \]

Step 3: Calculating the Minimum Power Required

Time \( t = 1 \, \text{hour} = 3600 \, \text{s} \) \[ P = \frac{Q}{t} = \frac{672000}{3600} = 186.67 \, \text{W} \]

Coefficient of Performance (COP) of an Ideal Refrigerator

• Cold reservoir temperature: \( T_C = 0^\circ C = 273 \, \text{K} \)
• Hot reservoir temperature: \( T_H = 20^\circ C = 293 \, \text{K} \)

\[ COP = \frac{T_C}{T_H - T_C} = \frac{273}{293 - 273} = \frac{273}{20} = 13.65 \]

Step 4: Determining the Work Done (Minimum Power)

\[ W = \frac{Q}{COP} = \frac{186.67}{13.65} \approx 13.67 \, \text{W} \]

Conclusion:

The minimum power required by the refrigerator is: \[ \boxed{13.6 \, \text{W}} \]