I just want someone to verify what I think.

I originally thought that for a reversible process to occur, no irreversibilities like friction can occur. Nevertheless, if we consider the Carnot efficiency, the equation is given as $1-\frac{T_h}{T_l}$. clearly, the efficiency is not 100%. this means that some heat was energy was converted to some useless output, let say friction.

I am therefore concluding that what makes something reversible is that the work produced, including the useless friction can all be converted back to the original state without any extra input.

Am I correct? thanks

  • $\begingroup$ What exactly are $T_h$ and $T_l$ meant to be? If they are meant to be the temperature of the high and low temperature reservoirs, then I think you have them the wrong way round, as $T_h > T_l\;\Rightarrow\; \frac{T_h}{T_l} > 1$. where that is what is meant by your notation or not, an explanation would make your question clearer $\endgroup$ – By Symmetry Jan 29 '18 at 11:13
  • $\begingroup$ yes you are right, i meant to write Tl/Th. $\endgroup$ – user1234 Jan 29 '18 at 11:25

Your first two sentences are correct. But then you go astray. A Carnot cycle is indeed less than 100% efficient, yet it is an ideal (reversible) cycle without friction or any other 'dissipative' processes. Its inefficiency is a case of the Second Law of Thermodynamics: no cyclic process can take in heat at one temperature and give out an equal quantity of work; some heat has to be given out at a lower temperature (even if there are no dissipative processes at work). The reason for the Second Law can be understood using the methods of statistical mechanics, but this answer stops here!


No, this is not the way to think about this. In the Carnot cycle some heat is, necessarily deposited in the low temperature reservoir, rather than being used to do work, but this is not due to friction and does not represent some sort of irreversibly. Your last sentence is true in the sense that it essentially says "a process is reversible if you can reverse it" but I don't think it is a good way to think about it.

In thermodynamics transferring energy as heat and transferring energy as work are not the same thing. A heat engine takes in energy as heat from a reservoir at temperature $T_1$, does a certain amount of work, deposits a certain mount of heat in a reservoir at temperature $T_2$ and returns to its original state. Kelvin's statement of the second law is that not all of the energy you put in can be converted into work, i.e. some heat must be deposited in the second reservoir. In other words the heat being "wasted" is necessary in order to satisfy the second law. It is preventing a net reduction in entropy over the cycle.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.