# 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.

First: Feynman is just being a little incautious with the sign of W_32. He states that it's sign is negative, but then uses W_13 - W_32. Since W_13 is regarded as positive, that would result in an even larger result when you subtract them, but actually you want a smaller result-- he is saying that the work you get from the machine 1-->3-->2 is less than the work you'd get from just 1-->3. So what he has done is explicitly inserted the minus sign and then treated W_32 like a positive number. The more formally correct thing to do is say W_13 + W_32 = W_12, which is just the correct algebra of the subscripts. So there's nothing wrong in his argument, but the equations cannot be taken literally as written, you have to fix the signs manually. This is a common problem in physics, sign language can be ambiguous-- the classic example is when someone says "I lost -5 dollars", it's very unclear if they gained 5 dollars, or if they are simply doubly emphasizing the fact that it was a loss and not a gain, essentially by stressing that their change in money was -5 dollars, and that constitutes a "loss." We are usually more careful, but Feynman lapsed a little there.

Second: Actually, you have 7 equations there. It looks like there are 9 unknowns, but actually there are only 7, because you don't actually care what Q_1 is, so you could call it 1 and measure all other energies relative to that. So immediately you are down to 8 unknowns. Also, you don't care what the efficiencies actually are, you just want a relationship between them. That relationship will ultimately involve ratios between the efficiences, rather than the efficiences themselves. So that's again one less variable, and now you are down to 7 equation in 7 unknowns, and you can solve it. The actual efficiencies will depend on the actual T, but again you don't really care what the Ts are, you only care about the ratios between them.

• 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. Jan 15, 2017 at 19:58
• Good point, that's not an independent constraint. But it's clear you could not solve for the efficiencies of this system, without being given the Ts. I think what is really happening is that you can take Q_1, T_1, and T_3 as given, and then vary T_2. Looking at the cycle both ways around constrains Q_2 and Q_3, given T_2. In short, you end up with two equations for Q_2 and Q_3, which solves to Q_2 = Q_1*T_2/T_1. Alternatively, treat Q_1 and the efficiencies of the 13 and 32 cycles as given, and solve for the efficiency of the 12 cycle as a function of Q_2. Jan 15, 2017 at 20:11
• What I'm saying is, the logical argument gives you a constraint on the efficiencies, it doesn't tell you the efficiencies. You have to add in the fact that the efficiencies are expressions like 1 - T_1/T_3, for example, and then you can relate the constraint on the efficiencies to the result about Q/T. Again you don't care what T_1 actually is, or what Q_1 actually is, so you can take them both to be 1, and then you are showing Q_2/T_2 = 1, in those units, and you can eliminate T_2 in terms of efficiencies that now look like 1 - T_2, so T_2 is one minus the efficiency of the 12 engine. Jan 15, 2017 at 20:22
• Thus the final result you use your system of equations to show is that Q_2 equals one minus the efficiency of the 12 engine, given any efficiencies for the 13 and 32 engines. Since it works for any of those efficiencies, you don't need to solve for them, they should just cancel out. Jan 15, 2017 at 20:24
• By the way, one thing to point out is that the logical argument Feynman gives has nothing to do with T, nor even any idea what T is. So you could do the whole thing where all you end up with is Q_2/Q_1 equals one minus the efficiency of the 12 engine, without ever mentioning T at all, indeed this is the only thing that logical argument can actually derive. To get this in terms of Q/T, you have to say what T is, and that usually brings in entropy. Without that, T is only another way to talk about the efficiency, via efficiency = 1 - T_3/T_1 kinds of expressions. Jan 15, 2017 at 20:30

Well according to the next paragraph I think that $Q_3$ is known, $Q_1$ is unknown

We can make the argument very clear in the following way: We have just seen that we can always relate the heat absorbed at $T_1$ to the heat delivered at $T_2$ by finding the heat delivered at some other temperature $T_3$. Therefore we can get all the engines’ properties if we introduce a standard temperature, analyzing everything with that standard temperature. In other words, if we knew the efficiency of an engine running between a certain temperature T and a certain arbitrary standard temperature, then we could work out the efficiency for any other difference in temperature. Because we assume we are using only reversible engines, we can work from the initial temperature down to the standard temperature and back up to the final temperature again. We shall define the standard temperature arbitrarily as one degree. We shall also adopt a special symbol for the heat which is delivered at this standard temperature: we shall call it $Q_S$. In other words, when a reversible engine absorbs the heat Q at temperature T, it will deliver, at the unit temperature, a heat $Q_S$. If one engine, absorbing heat $Q_1$ at $T_1$, delivers the heat $Q_S$ at one degree, and if an engine absorbing heat $Q_2$ at temperature $T_2$ will also deliver the same heat $Q_S$ at one degree, then it follows that an engine which absorbs heat $Q_1$ at the temperature $T_1$ will deliver heat $Q_2$ if it runs between $T_1$ and $T_2$, as we have already proved by considering engines running between three temperatures. So all we really have to do is to find how much heat $Q_1$ we need to put in at the temperature $T_1$ in order to deliver a certain amount of heat $Q_S$ at the unit temperature. If we discover that, we have everything. The heat Q, of course, is a function of the temperature T. It is easy to see that the heat must increase as the temperature increases, for we know that it takes work to run an engine backwards and deliver heat at a higher temperature.It is also easy to see that the heat $Q_1$ must be proportional to $Q_S$. So the great law is something like this: for a given amount of heat $Q_S$ delivered at one degree from an engine running at temperature T degrees, the heat Q absorbed must be that amount $Q_S$ times some increasing function of the temperature:
$Q = Q_Sf(T).$

Also I have found the relation between efficiencies:
$\frac{W_{12}+W_{32}}{Q_1}=\frac{W_{13}}{Q_1}$
$\eta_{\text{m.3}} + \frac{W_{32}}{Q_1}=\eta_{\text{m.1}}$
$\eta_{\text{m.3}} + \eta_{\text{m.2}}\frac{Q_2}{Q_1}=\eta_{\text{m.1}}$
$\eta_{\text{m.3}} + \eta_{\text{m.2}}(1-\eta_{\text{m.3}})=\eta_{\text{m.1}}$
$\eta_{\text{m.3}} =\frac{\eta_{\text{m.1}} - \eta_{\text{m.2}}}{1-\eta_{\text{m.2}}}$
But I don't know whether it is that what Feynman meant... I am still thinking that we should not use the temperature...