Thermodynamic temperature

In Feynman's lectures an argument for efficiency of engines goes as: If an engine absorbs heat $Q_1$ at temperature $T_1$ and heat $Q_2$ at temperature $T_2$, and both times give the same heat $Q$s at thermodynamic temperature $1\,\mathrm{K}$ then $\frac{Q_1}{T_1}=\frac{Q_2}{T_2}$. Does this imply that no matter from which temperature the body absorbs heat the heat it gives to the standard temperature will always be constant? Because then only it makes sense for the relationship to hold for any two temperatures $T_1$ and $T_2$.

• Could you include a citation to the specific part in the Feynman lectures you're talking about? – Arturo don Juan Oct 16 '17 at 3:23
• It's Book one, lecture in thermodynamics where he talks about the Carnot engine – user166680 Oct 16 '17 at 6:14

I am not familiar with this part of Feynman's lecture, but based on the information you provided I think his point was regarding entropy. Specifically:

$$S = \frac{Q}{T}$$

Where $S$ is entropy [J/K], $Q$ is heat [J], and $T$ is temperature [K] (stricly speaking it should be $\Delta S=\Delta Q/T$ for the change in the amount of entropy and thermal energy due to it being transferred, but I'm trying to be consistent with your notation). This tells you how much entropy is transferred by $Q$ joules of heat at temperature $T$. If in both cases (1 and 2) the heat engine rejected the same amount of heat, $Q_{s}$, to the cold bath at the same temperature, $T_{s}=1K$, then that means for both cases the heat engine rejected the same amount of entropy, $Q_{s}/T_{s}$, into the cold bath. At steady state operation a heat engine accumulates no entropy (by definition of "steady state" it means no quantity--entropy included--changes over time). This means that the heat engine must dispose of all entropy absorbed or generated by rejecting it into a cold bath (this is why all heat engines require a cold reservoir). Therefore, if the heat engine rejected the same amount of entropy in both cases, it must have absorbed the same amount of entropy from the hot bath for both cases (no entropy generation assuming Carnot heat engines). So, the entropy absorbed was the same for case 1 and case 2:

$$S_{1}=S_{2}$$

And so, it follows that

$$\frac{Q_{1}}{T_{1}}=\frac{Q_{2}}{T_{2}}$$

It is entirely possible, however, for the heat engine to absorb a different amount of entropy in a different cycle (either from absorbing a different $Q$, or from a different $T$, or both). In this case, the heat it gives to the standard temperature (the cold bath) would be different.