Question

1.A heat engine exhausts 9000 J of heat while performing 2300 J of useful work.

What is the efficiency of this engine?

Express your answer using two significant figures.

2.A nuclear power plant operates at 68 % of its maximum theoretical (Carnot) efficiency between temperatures of 690 ∘C and 390 ∘C.

If the plant produces electric energy at the rate of 1.2 GW , how much exhaust heat is discharged per hour?

Express your answer using two significant figures.

Answer 1

1)

efficiency eta = W/(Qc+W) = 2300/(2300+9000) = 0.203 or 20.3 %

2) maximum efficiency is eta _max= 1-(Tc/Th) = 1-((273+390)/(273+690)) = 0.688
68 % means 0.68*0.688 = 0.468 = eta

but eta = power output /power input

power input = power output /eta = 1.2 GW/0.468 = 2.56 GW

But power exhaust = power input - power ouput = 2.56-1.2 = 1.36 GW

in one second plant discharges 1.36*10^9 J

in one hour (3600 S) the plant discharges 3600*1.36*10^9 = 4896*10^9 J = 4.896*10^12 J