Thermal Physics: Practice Problem & Solution
An ideal gas heat engine operates in Carnot cycle between $227\text{ }^\circ\text{C}$ and $127\text{ }^\circ\text{C}$. It absorbs $6 \times 10^4\text{ cal}$ of heat at higher temperature. Amount of heat converted to work is: (2005)
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:
Temperatures are $T_1 = 500\text{ K}$ and $T_2 = 400\text{ K}$. Efficiency $\eta = 1 - \frac{400}{500} = 0.2$. Work $W = \eta Q_1 = 0.2 \times 6 \times 10^4 = 1.2 \times 10^4\text{ cal}$.
Leave a Reply