Question

A refrigerator works between $4^{\circ}C$ and $30^{\circ}C$. It is required to remove $600\, calories$ of heat every second in order to keep the temperature of the refrigerated space constant. The power required is : $(Take\, 1 \,cal\, =\, 4.2 \,Joules)$

• A. 23.65 W
• B. 236.5 W
• C. 2365 W
• D. 2.365 W

Answer 1

The Correct Option is B

To solve this problem, we need to determine the power required for the refrigerator to remove 600 calories of heat per second. We'll use the concept of the coefficient of performance (COP) for a refrigerator and basic energy conversion.

Step 1: Understand and Define COP for a Refrigerator

The coefficient of performance (COP) of a refrigerator is given by:

\text{COP} = \frac{Q_L}{W}

Where:

• Q_L is the heat removed from the cold reservoir (in Joules).
• W is the work done (in Joules).

The COP can also be expressed in terms of the temperatures of the hot (T_H) and cold (T_L) reservoirs (in Kelvin):

\text{COP} = \frac{T_L}{T_H - T_L}

Step 2: Convert Temperatures to Kelvin

The refrigerator operates between 4^{\circ}C and 30^{\circ}C. Convert these to Kelvin:

• T_L = 4 + 273 = 277\, \text{K}
• T_H = 30 + 273 = 303\, \text{K}

Step 3: Calculate COP

Using the formula for COP:

\text{COP} = \frac{277}{303 - 277} = \frac{277}{26}

\text{COP} \approx 10.65

Step 4: Convert Heat Removed to Joules

The refrigerator removes 600 calories of heat every second. Convert this to Joules:

600 \, \text{cal} \times 4.2 \, \text{J/cal} = 2520 \, \text{J}

Step 5: Calculate Work Done

Using the COP formula W = \frac{Q_L}{\text{COP}}:

W = \frac{2520 \, \text{J}}{10.65} \approx 236.51 \, \text{W}

Conclusion:

The power required for the refrigerator is approximately 236.5 W. Therefore, the correct answer is:

236.5 \, \text{W}