Topics in Physics Thermo Question

In our 7th period class today Alex asked how can a Carnot cycle be the most efficient, if going from point C back to the adiabatic line from point A via an isobaric process requires less work into the gas than going from C back to the adiabatic line using an isothermal process. For convenience lets call the point where we reach the adiabatic from A as point D. We can reach this point D from point C with an isothermal process or an isobaric process.

The explanation is subtle. The efficiency for a cycle must be considered in conjunction with the temperature limits it operates within. In other words, when comparing types of cycles, one must compare them with the same heat input temperatures and same heat rejection temperatures. In order to compress isobarically from C to D one must be reducing the temperature. Therefore Alex's suggested cycle either would require a lower heat rejection temperature (the low temperature point would be at D only rather than the whole line from C to D as in teh isothermal case ) than the Carnot cycle, which would allow it to get cooler going from C to D and thus have better efficiency, but only because the heat rejection temperature is lower, or it would have to have a point C at a higher temperature than the point C for the Carnot cycle, so that the temperature at the isobaric D would match the temperature of the isothermal point D) which would give it a lower efficiency.