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Can someone explain this explanation to me:

A completely different argument focuses on the requirement that the device only move heat at one or other of the two temperatures. Then, provided we have previously independently developed the concept of entropy, for all systems, we almost immediately (reversible entails no net entropy change) have the Qs being in the same ratio as the Ts, whence the familiar formula for efficiency.

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Since the system - engine plus sources - is isolated and reversible, we have $\Delta S = 0$. The entropy change of the system after a complete cycle of the engine is just $$\Delta S=\Delta S_{\mathrm{sources}},$$ since $\Delta S_{\mathrm{engine}}=0$ for a cycle. Consider the hot reservoir at temperature $T_1$. To calculate its change of entropy we imagine a reversible process where the source transfer heat $|Q_1|$ at constant temperature $T_1$. Hence, $$\Delta S_{\mathrm{source1}}=-\frac{|Q_1|}{T_1}.$$ For the cold reservoir at temperature $T_2$ we got $$\Delta S_{\mathrm{source2}}=+\frac{|Q_2|}{T_2}.$$ Therefore $$\Delta S=-\frac{|Q_1|}{T_1}+\frac{|Q_2|}{T_2}=0,$$ i.e. $$\frac{|Q_2|}{|Q_1|}=\frac{T_2}{T_1}.$$

On the other hand, the efficiency of a thermal engine is $$\eta=\frac{W}{|Q_1|}=1-\frac{|Q_2|}{|Q_1|},$$ so any reversible engine has $$\eta=1-\frac{T_2}{T_1}.$$

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