Thermal Physics: Practice Problem & Solution
Carnot engine, having an efficiency of $\eta = 1/10$. 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: (2015)
Solution Explained:
To solve this problem, we apply the core principles of Thermal Physics. Understanding the underlying formula is key to arriving at the correct answer below:
Using $\beta = \frac{1-\eta}{\eta} = 9$, the heat absorbed at lower temperature is $Q_2 = \beta W = 9 \times 10\text{ J} = 90\text{ J}$.
Leave a Reply