Question

A carnot engine having an efficiency of $\frac{1}{10}$ as heat engine, is used as a refrigerator. If the work done on the system is $10 \text{ J}$, the amount of energy absorbed from the reservoir at lower temperature is

• A. $100 \text{ J}$
• B. $99 \text{ J}$
• C. $90 \text{ J}$
• D. $80 \text{ J}$

Hint:

The Carnot cycle is reversible. If you know the efficiency as an engine is $\eta = W/Q_1$, remember that for the refrigerator, $Q_1 = Q_2 + W$. This means $Q_2 = Q_1 - W$. Since $\eta = 1/10$, $Q_1$ must be $100 \text{ J}$ for $10 \text{ J}$ of work. Thus, $Q_2 = 100 - 10 = 90 \text{ J}$.

Answer 1

The Correct Option is C

Step 1: Relate Efficiency to COP: The relationship between the efficiency of a Carnot engine and the COP of a Carnot refrigerator is: $$\beta = \frac{1 - \eta}{\eta}$$ Given $\eta = \frac{1}{10}$: $$\beta = \frac{1 - 1/10}{1/10} = \frac{9/10}{1/10} = 9$$

Step 2: Calculate Energy Absorbed ($Q_2$): The coefficient of performance is defined as the ratio of energy absorbed from the cold reservoir ($Q_2$) to the work done ($W$) on the system: $$\beta = \frac{Q_2}{W}$$ Substituting the known values ($\beta = 9$ and $W = 10 \text{ J}$): $$9 = \frac{Q_2}{10}$$ $$Q_2 = 9 \times 10 = 90 \text{ J}$$ Therefore, the refrigerator absorbs $90 \text{ J}$ of energy from the lower temperature reservoir.