Thermal Physics - Carnot Cycle

An engine absorbs heat at a temperature of 727^oC and exhaust heat at a temperature of 527^oC. If the engine operates at maximum possible efficiency, for 2000 joules of heat input the mount of work the engine performs is most nearly 

A. 400 J
B. 1450 J
C. 1600 J
D. 2000 J
E. 2760 J

(GR9277 #16)

Solution:

T_H = 727^oC = 727 + 273 = 1000 K
T_C = 527^oC = 527 + 273 = 800 K

Q_H = 2000 J
W = ?

Efficiency of heat engine cycle: η = W/Q_H

Carnot efficiency: η = (T_H − T_C) /T_H = 1 − T_C/T_H

W/Q_H = 1 − T_C/T_H
W = Q_H (1 − T_C/T_H) =  2000  (1 − 800/1000) = 400 J

Answer: A 

Thermal Physics - Carnot Cycle

In the cycle shown above, KL and NM represent isotherms, while KN and LM represent reversible adiabats. A system is carried through the Carnot cycle KLMN, taking in heat Q₂ from the hot reservoir T₂ and releasing heat Q₁ to the cold reservoir T₁. All of the following statements are true, EXCEPT:

A. Q₁/T₁ = Q₂/T₂
B. The entropy of the hot reservoir decreases
C. The entropy of the system increases
D. The work done is equal to the net heat absorbed, Q₂ − Q₁
E. The efficiency of the cycle is independent of the working substance.

(GR8677 #95)

Solution:

Carnot engine efficiency:

where W = Q₂ − Q₁ → work done
(D) TRUE

The efficiency η does not depend on the working substance
(E) TRUE

W = Q₂ − Q₁
η = (Q₂ − Q₁)/Q₂ = 1 − (Q₁/Q₂) = 1 − (T₁/T₂)
Q₁/Q₂ = T₁/T₂
Q₁/T₁ = Q₂/T₂
(A) TRUE

The hot reservoir has decreasing entropy because it gets cooler as the cycle proceeds.
(B) TRUE

In order to approach the Carnot efficiency, the processes involved in the heat engine cycle must be reversible and involve no change in entropy.
(C) FALSE

Answer: C