# Why does the temperature of the working fluid have to be close to the temperature of the cold reservoir during a Carnot cycle?

I understand why the temperature of the hot reservoir has to be minimally higher than the temperature of the hot working fluid during the isothermal expansion phase of the Carnot cycle (to limit new entropy being produced in the working fluid that we have to get rid of). But during the isothermal compression phase why do we need the cold reservoir to be only minimally cooler than the cold working fluid? The working fluid looses the same entropy independent of the temperature of the cold reservoir, so why do we need to minimize the new entropy created in the cold reservoir? What is wrong with just letting the cold reservoir get more and more entropy as long as the working fluid returns to the same state as before the cycle?

• If there is resistance to heat transfer within the working fluid, the volume average temperature of the working fluid will differ from that of the hot and cold reservoirs. So which temperature do we use, and, with an irreversible process and non-uniform temperature taking place within the working fluid, what temperature do we use to calculate the work? Nov 9, 2023 at 20:46

The work $$W_0$$ produced in the Carnot cycle is the result of transporting a certain amount of entropy, say, $$S_0$$ from a higher temperature $$T_h$$ to a lower temperature $$T_{\ell}$$. Here $$W_0=S_0(T_h-T_{\ell})$$ and this is done by isothermally absorbing $$S_0$$ at $$T_h$$, then adiabatically expanding after which isothermally rejecting the same amount of entropy $$S_0$$ at a lower temperature $$T_{\ell}$$, followed by an adiabatic compression stage to close the cycle. If you increase the temperature at which the entropy $$S_0$$ is rejected to the environment or lower it at which it is absorbed or both, then less thermal work is done than otherwise would be available.