Question

Questions 28, 29, and 30 refer to the following information: A Carnot engine absorbs 900 J of heat each cycle and provides 350 J of work. QUESTION 28 What is its efficiency? (a) 257% efficient (b) 39% efficient (c) 61% efficient (d) 3% efficient QUESTION 29 How much heat is rejected each cycle? (a) 1450 J (b) 200 J (c) 900 J (d) 550 J QUESTION 30 If the engine rejects heat at 283 K, what's its maximum temperature? (a) 463K (b) 165K (c) 105 K (d) 4,630 K

please tell me your answer, all the step to the answer and why, i got the answer

Answer 1

28 ) Efficiency =( Work done / Heat in ) * 100

   = ( 350/900)*100

   = 39 %

29) Heat rejected = Heat in - Work done

   = 900 J - 350 J

   = 550 J

30) heat in / T max = heat rejected / T min

Tmax = ( 283*900/550) = 463 K

Answer 2

Answer 3

efficiency = work /heat = 39%

heat rejected each cycle = 550 J = 900 -350 J

maximum temp = T,

1 - (Tc/Th ) = 0.39

Th = 463 K

Answer 4

Q28) efficiency = work done / energy absorbed = 350 900 = 0.39 = 39 %(b)

Q29) heat rejected = heat absorbed - work done = 900 - 350 = 550 J (d)

Q 30)

0.39 = 1 - T/TH

T/TH = 0.61

TH = T/0.61 = 463 K

Answer 5

28)since, efficiency=(net work done by the gas)/(net heat absorbed by the gas)

=350/900

=0.388

=38.8% ~ 39% Option (b)

29) net energy in = net energy out

   heat absorbed= work done + heat released

   900 = 350 + heat released

   heat released = 550J Option (d)

30)since, efficiency = 1 - (T_L/T_H)

   0.388 = 1 - (283/T_H)

   283/T_H = 0.612

T_H = 283/0.612 =462.8K ~ 463K Option (a)