In a Carnot’s cycle, the working substance absorbs heat Q1 at temperature T1 and rejects heat Q2 at temperature T2 . The change of entropy during die Carnot’s cycle is

Question

In a Carnot’s cycle, the working substance absorbs heat Q1 at temperature T1 and rejects heat Q2 at temperature T2 . The change of entropy during die Carnot’s cycle is (A)  (Q1/T1)  (B)  (Q2/T2)  (C)  (Q1/T1)+(Q2/T2)  (D) Zero

Tardigrade Admin · Accepted Answer

In a Carnot's cycle the working substance absorbs heat Q1 at temperature T1 and rejects heat Q2 at temperature T2. So, the change of entropy during the Carnot's cycle =(Q1/T1)-(Q2/T2)=0

Q. 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

$\frac{Q _{1}}{T _{1}}$

$\frac{Q _{2}}{T _{2}}$

$\frac{Q _{1}}{T _{1}} + \frac{Q _{2}}{T _{2}}$

Zero

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
$= \frac{Q _{1}}{T _{1}} - \frac{Q _{2}}{T _{2}} = 0$

Q. In a Carnot’s cycle, the working substance absorbs heat Q1​ at temperature T1​ and rejects heat Q2​ at temperature T2​ . The change of entropy during die Carnot’s cycle is

T1​Q1​​

T2​Q2​​

T1​Q1​​+T2​Q2​​