# Why Carnot efficiency says heat quotient equals temperature quotient?

I am trying to understand the difference between the concepts of Temperature and Heat.

So far I see it as:

• Heat is atomic kinetic energy flowing in or out of something (i.e the thermodynamic system).
• Temperature is a measure of the average amount of atomic kinetic energy that is collected inside the something.

Both are then clearly different concepts. In a triangle there would be of course "atomic kinetic energy", "heat", "temperature", all playing together. But not being the same.

Why, when calculating the efficiency of a Carnot cycle we assume the following?

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

When we say

$\eta=\frac{W_{net}}{Q_{1}}=\frac{Q_{1}-Q_{2}}{Q_{1}}=1-\frac{Q_{2}}{Q_{1}}=1-\frac{T_{2}}{T_{1}}$

• Did you read the wikipedia en.wikipedia.org/wiki/Carnot_cycle article on the Carnot cycle? – Sanya Aug 17 '16 at 10:57
• My mistake was reading it in a different language, which was not as complete as the English version. Thank you. – Álvaro N. Franz Aug 17 '16 at 12:32
• @ÁlvaroN.Franz That's an excellent point to make - it should be a meta post. Sometimes I think we (particularly native English speakers like myself) make rash assumptions of insufficient research whilst forgetting that most (in the sense of a heavy majority of) people on Earth speak a first language other than English. Granted, English is a language of scientific communication, but this site's charter is to be about physics at all levels. – WetSavannaAnimal Aug 18 '16 at 1:52

The 'assumption' comes from a more involved look at the different processes in the Carnot cycle.

First take the Isothermal expansion stroke from a to b at the start of the cycle. As the temperature is not changing we can say that the change in internal energy is zero thus giving the first law of thermodynamics to be W = Q (Work = Heat).

What this means is that the work done by the gas expanding all goes into the heat energy 'taken' away: $$\ W = PdV = Q$$ $$\ Q_{ab} = \int_{a}^{b} PdV$$ $$\ Q_{1} = \int_{a}^{b} NKT_1 dV/V = NKT_1ln(V_b/V_a)$$

Similarly for the adiabatic compression between c and d we get: $$\ Q_{2} = NKT_2ln(V_c/V_d)$$

Looking through the adiabatic strokes we find that the volume ratios $V_b/V_a$ and $V_c/V_d$ are equivalent so then you just equate the natural logarithms and cancel NK and what you are left with is: $$Q_2/Q_1 = T_2/T_1$$

So basically to answer your question it just comes out of doing the maths for each step in the cycle. Worth noting that this only works for the idealised conditions of the Carnot engine and in reality you can't use the temperatures in this way for efficiency! Hope this helps :)

This is actually how temperature is defined in Carnot's conception of thermodynamics. Here's how he argued.

He showed that all reversible heat engines working between the same two reservoirs must have the same efficiency, otherwise one could use the more efficient one as a heat engine, the less efficient one as a heat pump and thus pump heat without input of work continuously from a colder reservoir to a hotter one - in violation of his formulation of the second law of thermodynamics. That's a simple an neat argument and it's known as Carnot's Theorem: that all reversible heat engines working between the same thermal reservoirs must have the same efficiency, regardless of their makeup.

But his true genius was to understand that this observation could be used to define the "hotness" of something. Up till that time, there wasn't really a deep definition of hotness or temperature aside from in terms of measurement instruments (even thought these, even then, were becoming quite sophisticated). Carnot defines temperature thus: we take a "standard reservoir" and call its temperature unity, by definition.

Then if we have a hotter reservoir, we run a reversible heat engine between the two, and if the heat engine expels $Q_2$ units of heat to the standard reservoir for every $Q_1$ units of heat drawn from the hot reservoir (thus outputting $Q_1-Q_2$ units of work, then we say that the hot reservoir's temperature is $Q_1/Q_2$, by definition. Likewise, if we have a colder reservoir and run a heat engine between the standard and colder reservoir and the heat flows are $Q_1$ from the standard and $Q_2$ to the colder, then the temperature is $Q_2/Q_1$ (less than unity), by definition.

With a bit more work, all these ideas imply the existence of a function of state, entropy, but that is another story. See my answer here for further details.

• Thank you very much "multicellular organism", as you call yourself. Excellent. Beautiful explanation. – Álvaro N. Franz Aug 17 '16 at 12:31

In addition to the answers already given, I want to make the following connection. The efficiency of a Carnot engine can be calculated for a system containing an ideal gas in the way shown by @JC1217. This leads to:

$$\eta_C = 1-\frac{T_L}{T_H}$$ and also

$$\frac{Q_L}{Q_H} = \frac{T_L}{T_H}$$

It is harder to do this calculation for real gasses and fluids, for which the integral of $pdV$ can't be evaluated analytically. However, by Carnot's principle, the efficiency of a reversible machine between the same temperatures $T_L$ and $T_H$ is independent of the internal structure (incl. the working fluid) of the machine. Hence, the formula applies to all Carnot engines, not only those using ideal gas.

I think this was a missing point in the other answers.