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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

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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

User Andrew Morris
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