= Solution
Because the working substance returns to its original <thermodynamic state>, its net <entropy> change is zero. There is no <entropy> change on either <reversible ideal-gas adiabat>; on each <isothermal process> the <entropy> transfer is the exchanged <heat> divided by the bath's absolute <temperature>. Hence
$$
\frac{Q_h}{T_h}-\frac{Q_c}{T_c}=0,\qquad \frac{Q_c}{Q_h}=\frac{T_c}{T_h}.
$$
The <first law of thermodynamics> over a complete cycle gives $W=Q_h-Q_c$. The <thermal efficiency> is therefore
$$
\boxed{\eta_{\rm Carnot}=\frac{W}{Q_h}=1-\frac{Q_c}{Q_h}=1-\frac{T_c}{T_h}=1-\frac{T_1}{T_2}.}
$$
The use of kelvins is essential. This result does not depend on the working substance. For the <ideal gas> illustration, the two <reversible ideal-gas adiabats> imply $V_C/V_B=V_D/V_A=(T_h/T_c)^{1/(\gamma-1)}$, so $\ln(V_B/V_A)=\ln(V_C/V_D)$; the explicit <heat> expressions in (a) give the same ratio.
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