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In thermodynamics, a reversible process is quasi static (because every point in the system should be in equilibrium to study the state properties and work done is maximum), but why should heat transfer occur between infinitesimally small temperature difference(why not between finite temperature difference)?

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  • $\begingroup$ Because a reversible process consists of a continuous sequence of thermodynamic equilibrium states, and a system can not be at thermodynamic equilibrium if there are finite temperature gradients within the system. $\endgroup$ – Chet Miller May 22 '18 at 3:23
  • $\begingroup$ But how can a finite temperature gradient affect the efficiency of the process.Wont the gradient die out after some time? $\endgroup$ – user196272 May 22 '18 at 4:27
  • $\begingroup$ Suppose I have a hot body and an identical cold body, and I place them into direct contact with one another, allowing them to equilibrate spontaneously via transient heat conduction (automatically involving large finite transient temperature gradients). How much work is done by this process? What is the mechanical efficiency of this process? Now I employ a working fluid between the two bodies and use a sequence of mini Carnot cycles to reversibly extract tiny amounts of heat from the hot body and transfer tiny amounts of heat to the cold body in each cycle. Any work now? $\endgroup$ – Chet Miller May 22 '18 at 11:44
  • $\begingroup$ I think no work is done when heat transfer occurs by direct contact. When a Carnot engine is used to transfer heat reversibly, work is done .But if the temperature difference between the cold and the hot bodies is finite, how can we claim that the engine operates reversibly?(From whatever I learnt, I understood that for an engine to operate reversibly, the temperature difference between the source and the sink should be infinitesimally small) $\endgroup$ – user196272 May 22 '18 at 15:07
  • $\begingroup$ So when you have isothermal expansion of an ideal gas, it can't be reversible (even with just a tiny difference in temperature between the reservoir and the gas)? $\endgroup$ – Chet Miller May 22 '18 at 15:34
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This all depends on precisely what the temperature boundary condition is at the surface of the cylinder. We can solve the transient differential heat balance equations within the cylinder as a function of time, and predict the pressure and the work done by the gas (even if the work is close to quasi static). This is a do-able calculation. But we would need to specify the problem precisely, and be sure that we are comparing apples with apples. As far as good thermodynamics books, I highly recommend "Fundamentals of Engineering Thermodynamics," by Moran et al. I particularly like their alternative treatment of the 2nd law of thermodynamics.

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