# Why must the total change in entropy be 0 for a heat cycle to be as efficient as possible?

If $\Delta S_{total}$ were not zero, then heat flow would take place at different temperatures.

It is easy to see why this is true by examining the opposite: that when heat flow takes place at the same temperature, $\Delta S$ = 0. Consider two equal and opposite heat flows: $dQ_{1}$ and $dQ_{2}$ where $dQ_{1} = -dQ_{2}$. If these flows take place at the same temperature, $T$, then $dQ_{1}/T + dQ_{2}/T$ = 0.

Why is this important, however. That heat flow take place at the same temperature, if a process is to be comparatively efficient?

• The idea is that any time you have transfer of energy via heating between two objects at different temperatures, you can put a little heat engine between those two object and divert some of that energy flow into work. So: if during the operation of a heat engine, the system exchanges energy via heat with a reservoir at different a temperature, we are not extracting the maximum work. To extract the maximum work, we put a little Carnot engine between those two objects. Commented Nov 19, 2015 at 5:11
• Hi march, if I understand your comment correctly, I now have the following: carnot engines are characteristically known to have the greatest efficiency. They also have 0 $\Delta S$. Because heat engines can only have $\Delta S = 0$ when they are carnot ones, such engines are also the most efficient. Then I must ask the question: is there any relationship between a carnot engine having $\Delta S = 0$ and being the most efficient engine?
– Muno
Commented Nov 19, 2015 at 5:26
• They are directly related. Any time a thermally isolated system has $\Delta S > 0$, that means that we could have taken advantage of the process to do work: $\Delta S$ is caused by non-quasi-static processes within the working system of an engine and also caused by heat transfer across finite temperature differences. See here, perhaps. Commented Nov 19, 2015 at 5:31
• You wrote (from the link you gave) "In order for the the Clausius inequality to be strict, there must be, in one of the other processes, more heat flow out compared to the reversible version of the cycle−making δQ/Tres more negative for the process−of less heat flow in compared to the reversible version−making δQ/Tres smaller for the process. This is consistent with the engine being less efficient, because either there's more waste, or there's the same amount of waste but less available energy." This solves my question. Thanks.
– Muno
Commented Nov 20, 2015 at 16:16

Making the total entropy tend to zero, makes the efficiency tend to the Carnot efficiency, which is indeed the maximum efficiency for a cyclic process , according to Carnot's theorem $$\frac{dQ_1}{T_1} + \frac{dQ_2}{T_2}= \Delta S_\text{total}\ge 0 \\ \implies \frac{dQ_2}{T_2} + \frac{dQ_1}{T_2}\ge - \frac{dQ_1}{T_1} + \frac{dQ_1}{T_2}\\ \implies \frac{dw}{T_2}\ge -dQ_1\left(\frac{1}{T_2}- \frac{1}{T_1}\right)\\ \implies \frac{dw}{T_1}\ge - \left(1- \frac{T_2}{T_1}\right)\\ \implies \frac{dw}{T_1}\le 1- \frac{T_2}{T_1}\\ \therefore \frac{dw}{dQ_1} = \eta = \textrm{efficiency}\le 1- \frac{T_2}{T_1} \\ \implies \eta \le \eta_\text{Carnot}\;.$$