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 $10\,J$, the amount of energy absorbed from the reservoir at lower temperature is :

• A. 90 J
• B. 99 J
• C. 100 J
• D. 1 J

Answer 1

The Correct Option is A

To solve this problem, let's start by understanding the efficiency of a Carnot engine and how it applies when the engine is used as a refrigerator.

1. Efficiency of a Carnot engine as a heat engine is given by:
   Efficiency = \frac{W}{Q_H},
   where W is the work done and Q_H is the heat absorbed from the hot reservoir.
2. Given, the efficiency of the Carnot engine is \frac{1}{10}. Therefore, we can write: \frac{1}{10} = \frac{W}{Q_H}.
3. Rearranging the formula gives us Q_H = 10W.
4. When the Carnot engine is used as a refrigerator, it uses work done on the system to absorb heat from the cold reservoir. The work done, W, is given as 10\,J.
5. For a Carnot refrigerator, the heat absorbed from the reservoir at lower temperature, Q_L, can be described by the equation: Q_L = Q_H - W.
6. Substituting for Q_H using the value of W:
   Q_H = 10 \times 10 = 100\,J
7. Therefore, the heat absorbed from the cold reservoir is: Q_L = 100 - 10 = 90\,J.

Thus, the amount of energy absorbed from the reservoir at the lower temperature is 90 J.

Hence, the correct answer is: 90 J