Question

Six Carnot engines operating between different hot and cold reservoirs are described below. The heat energy transferred to the gas during the isothermal expansion phase of each cycle is indicated.

Answer 1

Part -A

The change in the entropy from largest to smallest is 3>2=4 >1 >5>6.

Part-B

Change in entropy over a complete cycle is 1 =2 = 3= 4 = 5 = 6

Step-by-step explanation:

Part -A

The change in entropy for the reversible process that transfers the heat energy 'Q' at temperature 'T' is

`\Delta S = (Q)/(T)`

1)

`\Delta S_(1) = (Q)/(T)`

`= (1000J)/(600K)= 1.67J/K`

2)

`\Delta S_(2) = (Q)/(T)`

`= (1000J)/(500K)= 2J/K`

3)

`\Delta S_(3) = (Q)/(T)`

`= (1500J)/(600K)= 2.5J/K`

4)

`\Delta S_(4) = (Q)/(T)`

`= (1000J)/(500K)= 2J/K`

5)

`\Delta S_(5) = (Q)/(T)`

`= (500J)/(500K)= 1J/K`

6)

`\Delta S_(6) = (Q)/(T)`

`= (500J)/(600K)= 0.83J/K`

Therefore, The change in the entropy from largest to smallest is 3>2=4 >1 >5>6.

Part-B

`\Delta S = (Q)/(T)`

`Q_(system)=work\,\,done`

Initial and final states of complete cycle are equal. after complete the one cycle the system returns to the original state. So, the change in the entropy will be the zero

Therefore, Change in the overall cycle is 1 =2 = 3= 4 =5=6