High temperature difference heat transfer becomes low temperature difference heat transfer, and the heat absorption becomes smaller. 
Although h2-h3 does not change, the amount of heat released is reduced.
Why is B incorrect? ? ? ?
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The temperature of the high-temperature heat source is 2000K, and the temperature of the high-temperature working fluid is 1900K. Their entropy change signs are different, and the temperature is different. When calculating, they are calculated according to the constant temperature.
Note: The working medium also has entropy change, but the entropy change of the high-temperature working medium and low temperature The entropy changes of the working fluid will definitely cancel each other out, because the entropy of the cycle becomes 0.
For the closed system, the entropy increase is equal to the entropy flow + entropy production. If the heat is absorbed, the entropy flow must be greater than 0.
Book page 50: under the TS process line The premise of expressing the area as heat is a reversible process,
and the premise of expressing the area under the process line of the pv diagram as a volumetric work is the quasi-balanced process (calculated from the perspective of the working fluid) if it is calculated from the perspective of obtaining work from the outside, it must also be reversible The process is
clearly the entropy change of the high-temperature working fluid. Because it is endothermic, the entropy becomes positive.
Only a temperature is given, which means external reversibility;
the efficiency calculated by temperature and energy is different, indicating
U9~21, 27, 28 , Not done at the end of the period