Heat Engine & Refrigerator: Practice Problem & Solution
A carnot engine having an efficiency of $1/10$ 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: (2017-Delhi)
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Heat Engine & Refrigerator. Understanding the underlying formula is key to arriving at the correct answer below:
Coefficient of performance $\beta = \frac{1-eta}{\eta} = 9$. Heat absorbed from lower reservoir $Q_2 = \beta \times W = 9 \times 10\text{ J} = 90\text{ J}$.
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