Question

A Carnot engine having an efficiency of 20 % is used as a refrigerator. If the amount of heat absorbed from the reservoir at low temperature is 200 J, then the work done on the refrigerator is:

• A. 150 J
• B. 50 J
• C. 100 J
• D. 75 J

Hint:

For a refrigerator, Work = Heat Absorbed / COP.

Answer 1

The Correct Option is B

Step 1: Link efficiency and temperatures.
For a Carnot machine the efficiency depends only on the two reservoir temperatures: \[ \eta = 1 - \frac{T_2}{T_1} \] Here $\eta = 20\% = 0.2$, so \[ \frac{T_2}{T_1} = 1 - 0.2 = 0.8 \]

Step 2: Run it as a refrigerator.
The same machine working in reverse is a refrigerator. Its coefficient of performance is \[ COP = \frac{T_2}{T_1 - T_2} = \frac{Q_2}{W} \] where $Q_2$ is heat drawn from the cold side and $W$ is the work done on it.

Step 3: Find the COP.
From $\dfrac{T_2}{T_1} = 0.8$ we get $T_1 = \dfrac{T_2}{0.8} = 1.25\,T_2$. So \[ COP = \frac{T_2}{1.25T_2 - T_2} = \frac{T_2}{0.25T_2} = \frac{1}{0.25} = 4 \]

Step 4: Solve for the work.
With $Q_2 = 200\ \text{J}$, \[ W = \frac{Q_2}{COP} = \frac{200}{4} = 50\ \text{J} \]

Step 5: Conclusion.
The work that must be done on the refrigerator is 50 J. \[ \boxed{50\ \text{J}} \]