# What makes a process reversible?

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

• 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 – By Symmetry Jan 29 '18 at 11:13
• yes you are right, i meant to write Tl/Th. – user1234 Jan 29 '18 at 11:25

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.