(a) Zero
The maximum efficiency (Carnot efficiency) of a heat engine is given by
[tex]\eta=1-\frac{T_C}{T_H}[/tex]
where
[tex]T_C[/tex] is the low-temperature reservoir
[tex]T_H[/tex] is the high-temperature reservoir
For the heat engine in the problem, we have:
[tex]T_C = 300K[/tex]
[tex]T_H = 300K[/tex]
Therefore, the maximum efficiency is
[tex]\eta=1-\frac{300}{300}=0[/tex]
(b) Zero
The efficiency of a heat engine can also be rewritten as
[tex]\eta = \frac{W}{Q_H}[/tex]
where
W is the work performed by the engine
[tex]Q_H[/tex] is the heat absorbed from the high-temperature reservoir
In this problem, we know
[tex]\eta=0[/tex]
Therefore, since the term [tex]Q_H[/tex] cannot be equal to infinity, the numerator of the fraction must be zero as well, which means
W = 0
So the engine cannot perform any work.
