Question 15.7: (a) Calculate the work output of a Carnot engine operating b...

Solution (a)
The Carnot efficiency is given by

E f f_{ C }=1-\frac{T_{ c }}{T_{ h }}.                    (15.53)

Substituting the given temperatures yields

E f f_{C}=1-\frac{100 K }{600 K }=0.833.                (15.54)

Now the work output can be calculated using the definition of efficiency for any heat engine as given by

E f f=\frac{W}{Q_{ h }}.                   (15.55)

Solving for W and substituting known terms gives

W=E f f_{ C } Q_{ h }                      (15.56)

= (0.833)(4000 J) = 3333 J.

Solution (b)
Similarly,

E f f_{ C }^{\prime}=1-\frac{T_{ c }}{ T _{ c }^{\prime}}=-\frac{100 K }{250 K }=0.600                    (15.57)

so that

W=E f f^{\prime}{ }_{C} Q_{h}                        (15.58)

= (0.600)(4000 J) = 2400 J.

Discussion
There is 933 J less work from the same heat transfer in the second process. This result is important. The same heat transfer into two perfect engines produces different work outputs, because the entropy change differs in the two cases. In the second case, entropy is greater and less work is produced. Entropy is associated with the unavailability of energy to do work.