# Feynman Lectures. Derivation of efficiency law from logic

Question is about Feynman Lectures. Lecture 44. Ch.44-4 The efficiency of an ideal engine (feynmanlectures.caltech.edu/I_44.html).

In this chapter Feynman derives the formula $$\frac{Q_1}{T_1} = \frac{Q_2}{T_2}$$ ¹⁾ by two ways: first, by using the properties of the ideal gas, and second, by logic. But trying to do it by the 2nd way (from the logic), he did not do it. Here is a quote:

Now we shall see how this universal law could also be obtained by logical argument, without knowing the properties of any specific substances, as follows. Suppose that we have three engines and three temperatures, let us say $$T_1$$, $$T_2$$, and $$T_3$$. Let one engine absorb heat $$Q_1$$ from the temperature $$T_1$$ and do a certain amount of work $$W_{13}$$, and let it deliver heat $$Q_3$$ to the temperature $$T_3$$ (Fig. 44–8). Let another engine run backwards between $$T_2$$ and $$T_3$$. Suppose that we let the second engine be of such a size that it will absorb the same heat $$Q_3$$, and deliver the heat $$Q_2$$. We will have to put a certain amount of work, $$W_{32}$$, into it—negative because the engine is running backwards. When the first machine goes through a cycle, it absorbs heat $$Q_1$$ and delivers $$Q_3$$ at the temperature $$T_3$$; then the second machine takes the same heat $$Q_3$$ out of the reservoir at the temperature $$T_3$$ and delivers it into the reservoir at temperature $$T_2$$. Therefore the net result of the two machines in tandem is to take the heat $$Q_1$$ from $$T_1$$ and deliver $$Q_2$$ at $$T_2$$. The two machines are thus equivalent to a third one, which absorbs $$Q_1$$ at $$T_1$$, does work $$W_{12}$$, and delivers heat $$Q_2$$ at $$T_2$$, because $$W_{12}=W_{13}−W_{32}$$, as one can immediately show from the first law, as follows:
$$W_{13} - W_{32} = (Q_1 - Q_3) - (Q_2 - Q_3) = Q_1 - Q_2 = W_{12}.$$
We can now obtain the laws which relate the efficiencies of the engines because there clearly must be some kind of relationship between the efficiencies of engines running between the temperatures $$T_1$$ and $$T_3$$, and between $$T_2$$ and $$T_3$$, and between $$T_1$$ and $$T_2$$.

First, I don't understand why W32 is negative and why is calculated as Q2−Q3 and not Q3−Q2.
Mathematics must be correct in all cases, even if we don't know is the work done on or by the machine 2.

Second. How to calculate efficiencies? I try next:
$$\eta _{\text{machine 1}}=\tfrac{W_{13}}{Q_1}$$

$$\eta _{\text{machine 2}}=\tfrac{W_{32}}{Q_2}$$

$$\eta _{\text{machine 3}}=\tfrac{W_{12}}{Q_1}$$
$$W_{12} = Q_1-Q_2$$
$$W_{13} = Q_1-Q_3$$
$$W_{32} = Q_2-Q_3$$
$$(W_{13} - W_{32}= W_{12})$$
So we have 6 equations and 9 unknowns. The system is unsolvable. Besides the system does not contain temperature, and we cannot use $$PV\sim T$$, because it's the ideal gas equation (and in 2nd way we shouldn't use properties of any substance).

¹⁾ Here $$Q_1$$ is the heat absorbed from the heat pad at temperature $$T_1$$, $$Q_2$$ is the heat delivered to the heat pad at temperature $$T_2$$.

Feynman derives the formula $$W_{13} - W_{32}= W_{12}$$ from the assumption that the engine 3 will absorb heat $$Q_1$$ and deliver $$Q_2$$. How he knows that? Could it be that the engine 3 absorbs $$Q_1$$, but delivers unknown $$Q_?$$. Hence it follows that 3rd engine's work is $$Q_1 - Q_?$$, and so equation $$W_{13} - W_{32}= W_{12}$$ is wrong.
• So we have next:<br> $\tfrac{\eta _{\text{machine 1}}}{\eta _{\text{machine 2}}}=\tfrac{W_{13}}{1}\tfrac{Q_2}{W_{32}}$<br> $\tfrac{\eta _{\text{machine 2}}}{\eta _{\text{machine 3}}}=\tfrac{W_{32}}{Q_2}\tfrac{1}{W_{12}}$<br> $W_{12} = 1-Q_2$<br> $W_{13} = 1-Q_3$<br> $W_{32} = Q_2-Q_3$<br> The equation $(W_{13} - W_{32}= W_{12})$ is deduced from the previous, so it is only an identity. Commented Jan 15, 2017 at 19:58