I was reading about Carnot cycle in "Thermodynamics by Cengel & Boles", now it is stated that Carnot cycle is a totally reversible cycle, i.e it is an internally as well as externally reversible cycle (in all of the processes involved) and it is very well explained how it is like that(pages attached) but what I don't understand is why then we still get an entropy change in the T-S diagram of Carnot cycle, that is if it is completely reversible then shouldn't entropy change of the system must be zero?
The four processes explained as reversible.]1[It says in the first line in section 9-2 that Carnot is composed of four totally reversible processes ]2
$$W_{total}=nR\left ( T_h\ln \frac{V_2}{V_1}-T_c\ln\frac{V_3}{V_4} \right )$$
Also for adiabatic processes we have :
$$T_h V_2^{\gamma-1 }=T_c V_3^{\gamma-1}\tag{2}$$
$$T_hV_1^{\gamma-1}=T_cV_4^{\gamma-1}\tag{3}$$
Substituting for $V_2$ and $V_4$ from $(2)$ and $(3)$ in the first equation, we can find the ratio $V_3\over V_1$. (Find the final result yourself!)
Best Answer
Ah, but if you look at your book states it is a $T$-$s$ diagram - and I wonder if $s$ is used instead of $S$ because $s$ represents the entropy of the gas going around the Carnot cycle and not the entire entropy of the universe. The entropy of the gas considered changes in the Carnot cycle because it heats up and cools down and expands and contracts, but overall the entropy change of the universe will be zero as the processes are reversible.
So because heat is going into and out of the system (the gas considered in the Carnot cycle) and because the system does work and expands and contracts the entropy of the gas in the system changes and the entropy of the surrounding system changes so that overall the entropy change of the universe is equal to zero.