Where does the energy flow differ between a forward and reverse process?

Question

Mathematically ($W=\int PdV$) and by the First Law, I understand that $1 \rightarrow 4 \rightarrow 3 \rightarrow 2 \rightarrow 1$ and $1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 1$ are not symmetric; the first requires work, whilst the latter gives it to the environment. What I am having trouble with is bellyfeeling why. A physical example: imagine a machine that works thusly:

(1-2) A piston loaded with 3 bricks in a hot reservoir isothermally expands when 1 brick is removed.

(2-3) Then, removed from contact from any reservoir, it adiabatically expands when the 2nd brick is removed, cooling to the temperature of the cold reservoir, where

(3-4) the 2nd brick is replaced, isothermal compression takes place, then

(4-1) away from reservoir the 3rd is added, the temperature rising to the intial hot temperature.

Given that the two possible cycles are asymmetric, at what point (1,2,3,4) does the internal energy of the gas or the potential energy of the bricks differ for the two different cycles?

Answer 1

In all of them. In 1-2, 2-3, 3-4, as well as 4-1, the increase of the internal energy has the opposite sign than it has for 2-1, 3-2, 4-3, and 1-4, respectively, because they're the same processes run backwards. Not surprisingly, if one sums the four terms in the 1-2-3-4-1 sequence, he gets the opposite sign than in the opposite 1-4-3-2-1 sequence, too.

The adjective "asymmetric" is misleading or possibly wrong. There's a perfect symmetry, the time reversal symmetry, between the two directions of the reversible cycle. But one shouldn't forget that the time reversal symmetry exchanges the initial and final state, so it flips the sign of $\Delta E = E_f-E_i$.

A different, more nontrivial question you could have asked it why in 1-2-3-4-1 (or, equivalently 1-4-3-2-1), the total work done/absorbed over the cycle is nonzero. But it's just the case, $\oint p\,dV$ or the are of the curve inside gives the work per cycle.