CHAP. 4: APPLICATION OF THERMODYNAMICS [CONTENTS] 118
Solution
From relation (4.30) we have
η=
T 2 −T 1
T 2
=
600 − 300
600
=
1
2
.
Using relation (4.29) we calculate the work performed by the engine
Wdone=−W=Q 2 η= 100×
1
2
= 50J.
Since the initial and final state are identical, the internal energy change is
∆U= 0.
We substitute this result into (3.3)
0 =Q+W=Q 2 +Q 1 +W
and calculate the supplied heat
Q 1 =−Q 2 −W=− 100 −(−50) =− 50 J.
The Carnot engine supplied−Q 1 = +50 J of heat to the cooler reservoir.
Example
Prove that the Carnot engine efficiency can never equal one.
Proof
It follows from relation (4.30) thatη= 1 only whenT 1 = 0 K orT 2 →∞. According to the Third
Law of thermodynamics (3.1.4), however, temperature 0 K cannot be attained. An infinitely
high temperatureT 2 cannot be attained, either.
