Question

In a Carnot’s cycle, the working substance absorbs heat $ Q_{1} $ at temperature $ T_{1} $ and rejects heat $ Q_{2} $ at temperature $ T_{2} $ . The change of entropy during die Carnot’s cycle is

• A. $ \frac{Q_{1}}{T_{1}} $
• B. $ \frac{Q_{2}}{T_{2}} $
• C. $ \frac{Q_{1}}{T_{1}}+\frac{Q_{2}}{T_{2}} $
• D. Zero

Answer 1

Answer: (D)

In a Carnot's cycle the working substance absorbs heat $Q_{1}$ at temperature $T_{1}$ and rejects heat $Q_{2}$ at temperature $T_{2}$. So, the change of entropy during the Carnot's cycle
$=\frac{Q_{1}}{T_{1}}-\frac{Q_{2}}{T_{2}}=0$