Derive the Carnot Efficiency without using the Carnot Cycle? How can you derive the Carnot efficiency using only properties of reversible cycles?
 A: Use the second law of thermodynamics. 
Consider as the system a generic reversible engine plus the cold and hot sources. This is a closed and reversible system, therefore its entropy change vanishes. You decompose this change in changes due to the engine and due to the sources. The entropy change of the engine after a cycle vanishes, thus
$$\Delta S=\Delta S_{\mathrm{sources}}=0.$$
The hot source loses $|Q_1|$ at constant temperature $T_1$, whereas the cold source gain $|Q_2|$ at temperature $T_2$. Hence,
$$\Delta S=-\frac{|Q_1|}{T_1}+\frac{|Q_2|}{T_2}=0,$$
i.e.
$$\frac{|Q_2|}{|Q_1|}=\frac{T_2}{T_1}.$$
Plug this into the efficiency
$$\eta=\frac{W}{|Q_1|}=1-\frac{|Q_2|}{|Q_1|},$$
and obtain the efficiency of a reversible engine
$$\eta=1-\frac{T_2}{T_1}.$$
Notice that we have not made any assumption about the cycle or even the agent responsible by the engine. That is in the Core of Carnot theorem. Any engine, regardless its nature, working between the same sources have the same efficiency, so it is natural that this efficiency can be calculated without any reference to the Carnot cycle which is a specific cycle followed by a specific agent (ideal gas).
