Question

A Carnot cycle operates between temperature limits of 900 and 300K The best supplied at the high muncrature is 4 MJ. Calculate the change in entropy during the heat addition and rejection processes the heat rejection and the work output

Answer 1

Answer:Work=
`(8)/(3)MJ`

Step-by-step explanation:

Cycle is operated between 900 and 300 K

heat supplied at high temperature is 4MJ

Let heat supplied be
`Q_1`

and heat rejected is
`Q_2`

`\eta =1-(T_L)/(T_H)=1-(Q_2)/(Q_1)`

`(Q_2)/(Q_1)=(T_L)/(T_H)`

`Q_2=4* (300)/(900)=(4)/(3) MJ`

Change in entropy during heat addition
`\Delta =(Q_1)/(T_H)=(4* 10^3)/(900)=4.44 KJ/K`

Change entropy during heat rejection
`\Delta =(Q_2)/(T_L)=-4.44 KJ/K`

`W=4-(4)/(3)=(8)/(3) MJ`