Rankers Physics

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)
$99\text{ J}$
$90\text{ J}$
$1\text{ J}$
$100\text{ J}$

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 $Q_2 = \beta W = 9 \times 10\text{ J} = 90\text{ J}$.

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