Which outcome occurs when a reversible Carnot cycle operates isothermally with zero temperature difference?

💡 Explanation

When a reversible Carnot cycle operates with an infinitesimally small temperature difference during isothermal processes, heat transfer rates become infinitely slow because heat transfer is proportional to temperature difference divided by thermal resistance. Therefore ideal, infinitely slow heat transfer results, rather than zero entropy change, reduced work output, or irreversible heating, which are results of non-ideal conditions.

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