# Re-Visiting Carnot’s Theorem

The proof by contradiction used in physics textbooks is one of those arguments that appear surprising, then self-evident, then deceptive in its simplicity. You – or maybe only: I – cannot resist turning it over and over in your head again, viewing it from different angles.

tl;dr: I just wanted to introduce the time-honored tradition of ASCII text art images to illustrate Carnot’s Theorem, but this post got out of hand when I mulled about how to  refute an erroneous counter-argument. As there are still research papers being written about Carnot’s efficiency I feel vindicated for writing a really long post though.

Carnot‘s arguments prove that there is a maximum efficiency of a thermodynamic heat engine – a machine that turns heat into mechanical energy. He gives the maximum value by evaluating one specific, idealized process, and then proves that a machine with higher efficiency would give rise to a paradox. The engine uses part of the heat available in a large, hot reservoir of heat and turns it into mechanical work and waste heat – the latter dumped to a colder ‘environment’ in a 4-step process. (Note that while our modern reformulation of the proof by contradiction refers to the Second Law of Thermodynamics, Carnot’s initial version was based on the caloric theory.)

The efficiency of such an engine η – mechanical energy per cycle over input heat energy – only depends on the two temperatures (More details and references here):

$\eta_\text{carnot} = \frac {T_1-T_2}{T_1}$

These are absolute temperatures in Kelvin; this universal efficiency can be used to define what we mean by absolute temperature.

I am going to use ‘nice’ numbers. To make ηcarnot equal to 1/2, the hot temperature
T1 = 273° = 546 K, and the colder ‘environment’ has T2 = 0°C = 273 K.

If this machine is run in reverse, it uses mechanical input energy to ‘pump’ energy from the cold environment to the hot reservoir – it is a heat pump using the ambient reservoir as a heat source. The Coefficient of Performance (COP, ε) of the heat pump is heat output over mechanical input, the inverse of the efficiency of the corresponding engine. εcarnot is 2 for the temperatures given above.

If we combine two such perfect machines – an engine and a heat pump, both connected to the hot space and to the cold environment, their effects cancel out: The mechanical energy released by the engine drives the heat pump which ‘pumps back’ the same amount of energy.

In the ASCII images energies are translated to arrows, and the number of parallel arrows indicates the amount of energy per cycle (or power). For each device, the number or arrows flowing in and out is the same; energy is always conserved. I am viewing this from the heat pump’s perspective, so I call the cold environment the source, and the hot environment room.

Neither of the heat reservoirs are heated or cooled in this ideal case as the same amount of energy flows from and to each of the heat reservoirs:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
| | | |                         | | | |
v v v v                         ^ ^ ^ ^
| | | |                         | | | |
|------------|                 |---------------|
|   Engine   |->->->->->->->->-|   Heat pump   |
|  Eta = 1/2 |->->->->->->->->-| COP=2 Eta=1/2 |
|------------|                 |---------------|
| |                             | |
v v                             ^ ^
| |                             | |
|----------------------------------------------------------|
|        Cold source at temperature T_2 = 0°C = 273 K      |
|----------------------------------------------------------|

If either of the two machines works less than perfectly and in tandem with a perfect machine, anything is still fine:

If the engine is far less than perfect and has an efficiency of only 1/4 – while the heat pump still works perfectly – more of the engine’s heat energy input is now converted to waste heat and diverted to the environment:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
| | | |                           | |
v v v v                           ^ ^
| | | |                           | |
|------------|                 |---------------|
|   Engine   |->->->->->->->->-|   Heat pump   |
|  Eta = 1/4 |                 | COP=2 Eta=1/2 |
|------------|                 |---------------|
| | |                             |
v v v                             ^
| | |                             |
|----------------------------------------------------------|
|        Cold source at temperature T_2 = 0°C = 273 K      |
|----------------------------------------------------------|

Now two net units of energy flow from the hot room to the environment (summing up the arrows to and from the devices):

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
| |
v v
| |
|------------------|
|   Combination:   |
| Eta=1/4 COP=1/2  |
|------------------|
| |
v v
| |
|----------------------------------------------------------|
|        Cold source at temperature T_2 = 0°C = 273 K      |
|----------------------------------------------------------|

Using a real-live heat pump with a COP of 3/2 (< 2) together with a perfect engine …

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
| | | |                             | | |
v v v v                             ^ ^ ^
| | | |                             | | |
|------------|                 |-----------------|
|   Engine   |->->->->->->->->-|    Heat pump    |
|  Eta = 1/2 |->->->->->->->->-|     COP=3/2     |
|------------|                 |-----------------|
| |                                 |
v v                                 ^
| |                                 |
|----------------------------------------------------------|
|        Cold source at temperature T_2 = 0°C = 273 K      |
|----------------------------------------------------------|

… causes again a non-paradoxical net flow of one unit of energy from the room to the environment.

In the most extreme case  a poor heat pump (not worth this name) with a COP of 1 just translates mechanical energy into heat energy 1:1. This is a resistive heating element, a heating rod, and net heat fortunately flows from hot to cold without paradoxes:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
| |                                |
v v                                ^
| |                                |
|------------|                 |-----------------|
|   Engine   |->->->->->->->->-|   'Heat pump'   |
|  Eta = 1/2 |                 |     COP = 1     |
|------------|                 |-----------------|
|
v
|
|----------------------------------------------------------|
|        Cold source at temperature T_2 = 0°C = 273 K      |
|----------------------------------------------------------|

The textbook paradox is encountered, when an ideal heat pump is combined with an allegedly better-than-possible engine, e.g. one with an efficiency:

ηengine = 2/3 (> ηcarnot = 1/2)

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
| | |                           | | | |
v v v                           ^ ^ ^ ^
| | |                           | | | |
|------------|                 |---------------|
|   Engine   |->->->->->->->->-|   Heat pump   |
|  Eta = 2/3 |->->->->->->->->-| COP=2 Eta=1/2 |
|------------|                 |---------------|
|                               | |
v                               ^ ^
|                               | |
|----------------------------------------------------------|
|        Cold source at temperature T_2 = 0°C = 273 K      |
|----------------------------------------------------------|

The net effect / heat flow is then:

|----------------------------------------------------------|
|        Hot room at temperature T_1 = 273°C = 546 K       |
|----------------------------------------------------------|
|
^
|
|------------------|
|   Combination:   |
| Eta=3/2; COP=1/2 |
|------------------|
|
^
|
|----------------------------------------------------------|
|       Cold source at temperature T_2 = 0°C = 273 K       |
|----------------------------------------------------------|

One unit of heat would flow from the environment to the room, from the colder to the warmer body without any other change being made to the system. The combination of these machines would violate the Second Law of Thermodynamics; it is a Perpetuum Mobile of the Second Kind.

If the heat pump has a higher COP than the inverse of the perfect engine’s efficiency, a similar paradox arises, and again one unit of heat flows in the forbidden direction:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
| |                             | | |
v v                             ^ ^ ^
| |                             | | |
|------------|                 |---------------|
|   Engine   |->->->->->->->->-|   Heat pump   |
|  Eta = 1/2 |                 |    COP = 3    |
|------------|                 |---------------|
|                               | |
v                               ^ ^
|                               | |
|----------------------------------------------------------|
|        Cold source at temperature T_2 = 0°C = 273 K      |
|----------------------------------------------------------|

A weird question: Can’t we circumvent the paradoxes if we pair the impossible superior devices with poorer ones (of the reverse type)?

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
| | |                             | |
v v v                             ^ ^
| | |                             | |
|------------|                 |---------------|
|   Engine   |->->->->->->->->-|   Heat pump   |
|  Eta = 2/3 |->->->->->->->->-|    COP = 1    |
|------------|                 |---------------|
|
v
|
|----------------------------------------------------------|
|        Cold source at temperature T_2 = 0°C = 273 K      |
|----------------------------------------------------------

Indeed: If the COP of the heat pump (= 1) is smaller than the inverse of the (impossible) engine’s efficiency (3/2), there will be no apparent violation of the Second Law – one unit of net heat flows from hot to cold.

An engine with low efficiency 1/4 would ‘fix’ the second paradox involving the better-than-perfect heat pump:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
| | | |                          | | |
v v v v                          ^ ^ ^
| | | |                          | | |
|------------|                 |---------------|
|   Engine   |->->->->->->->->-|   Heat pump   |
|  Eta = 1/4 |                 |     COP=3     |
|------------|                 |---------------|
| | |                            | |
v v v                            ^ ^
| | |                            | |
|----------------------------------------------------------|
|        Cold source at temperature T_2 = 0°C = 273 K      |
|----------------------------------------------------------|

But we cannot combine heat pumps and engines at will, just to circumvent the paradox – one counter-example is sufficient: Any realistic engine combined with any realistic heat pump – plus all combinations of those machines with ‘worse’ ones – have to result in net flow from hot to cold …

The Second Law identifies such ‘sets’ of engines and heat pumps that will all work together nicely. It’s easier to see this when all examples are condensed into one formula:

The heat extracted in total from the hot room – Q1 –  is the difference of heat used by the engine and heat delivered by the heat pump, both of which are defined in relation to the same mechanical work W:

$Q_1 = W\left (\frac{1}{\eta_\text{engine}}-\varepsilon_\text{heatpump}\right)$

This is also automatically equal to Qas another quick calculation shows or by just considering that energy is conserved: Some heat goes into the combination of the two machines, part of it – W – flows internally from the engine to the heat pump. But no part of the input Q1 can be lost, so the output of the combined machine has to match the input. Energy ‘losses’ such as energy due to friction will flow to either of the heat reservoirs: If an engine is less-then-perfect, more heat will be wasted to the environment; and if the heat pump is less-than-perfect a greater part of mechanical energy will be translated to heat only 1:1. You might be even lucky: Some part of heat generated by friction might end up in the hot room.

As Q1 has to be > 0 according to the Second Low, the performance numbers have to related by this inequality:

$\frac{1}{\eta_\text{engine}}\geq\varepsilon_\text{heatpump}$

The equal sign is true if the effects of the two machines just cancel each other.

If we start from a combination of two perfect machines (ηengine = 1/2 = 1/εheatpump) and increase either ηengine or εheatpump, this condition would be violated and heat would flow from cold to hot without efforts.

But also an engine with efficiency = 1 would work happily with the worst heat pump with COP = 1. No paradox would arise at first glance  – as 1/1 >= 1:

|----------------------------------------------------------|
|         Hot room at temperature T_1 = 273°C = 546 K      |
|----------------------------------------------------------|
|                                |
v                                ^
|                                |
|------------|                 |-----------------|
|   Engine   |->->->->->->->->-|   'Heat pump'   |
|   Eta = 1  |                 |      COP=1      |
|------------|                 |-----------------|

|----------------------------------------------------------|
|        Cold source at temperature T_2 = 0°C = 273 K      |
|----------------------------------------------------------|

What’s wrong here?

Because of conservation of energy ε is always greater equal 1; so the set of valid combinations of machines all consistent with each other is defined by:

$\frac{1}{\eta_\text{engine}}\geq\varepsilon_\text{heatpump}\geq1$

… for all efficiencies η and COPs / ε of machines in a valid set. The combination η = ε = 1 is still not ruled out immediately.

But if the alleged best engine (in a ‘set’) would have an efficiency of 1, then the alleged best heat pump would have a Coefficient of Performance of only 1 – and this is actually the only heat pump possible as ε has to be both lower equal and greater equal than 1. It cannot get better without creating paradoxes!

If one real-live heat pump is found that is just slightly better than a heating rod – say
ε = 1,1 – then performance numbers for the set of consisent, non-paradoxical machines need to fulfill:

$\eta_\text{engine}\leq\eta_\text{best engine}$

and

$\varepsilon_\text{heatpump}\leq\varepsilon_\text{best heatpump}$

… in addition to the inequality relating η and ε.

If ε = 1,1 is a candidate for the best heat pump, a set of valid machines would comprise:

• All heat pumps with ε between 1 and 1,1 (as per limits on ε)
• All engines with η between 0 and 0,9 (as per inequality following the Second Law plus limit on η).

Consistent sets of machines are thus given by a stronger condition – by adding a limit for both efficiency and COP ‘in between’:

$\frac{1}{\eta_\text{engine}}\geq\text{Some Number}\geq\varepsilon_\text{heatpump}\geq1$

Carnot has designed a hypothetical ideal heat pump that could have a COP of εcarnot = 1/ηcarnot. It is a limiting case of a reversible machine, but feasible on principle. εcarnot  is thus a valid upper limit for heat pumps, a candidate for Some Number. In order to make this inequality true for all sets of machines (ideal ones plus all worse ones) then 1/ηcarnot = εcarnot also constitutes a limit for engines:

$\frac{1}{\eta_\text{engine}}\geq\frac{1}{\eta_\text{carnot}}\geq\varepsilon_\text{heatpump}\geq1$

So in order to rule out all paradoxes, Some Number in Between has to be provided for each set of machines. But what defines a set? As machines of totally different making have to work with each other without violating this equality, this number can only be a function of the only parameters characterizing the system – the two temperatures

Carnot’s efficiency is only a function of the temperatures. His hypothetical process is reversible, the machine can work either as a heat pump or an engine. If we could come up with a better process for a reversible heat pump (ε > εcarnot), the machine run in reverse would be an engine with η less than ηcarnot, whereas a ‘better’ engine would lower the upper bound for heat pumps.

If you have found one truly reversible process, both η and ε associated with it are necessarily the upper bounds of performance of the respective machines, so you cannot push Some Number in one direction or the other, and the efficiencies of all reversible engines have to be equal – and thus equal to ηcarnot. The ‘resistive heater’ with ε = 1 is the iconic irreversible device. It will not turn into a perfect engine with η = 1 when ‘run in reverse’.

The seemingly odd thing is that 1/ηcarnot appears like a lower bound for ε at first glance if you just declare ηcarnot an upper bound for corresponding engines and take the inverse, while in practice and according to common sense it is the maximum value for all heat pumps, including irreversible ones. (As a rule of thumb a typical heat pump for space heating has a COP only 50% of 1/ηcarnot.)

But this ‘contradiction’ is yet another way of stating that there is one universal performance indicator of all reversible machines making use of two heat reservoirs: The COP of a hypothetical ‘superior’ reversible heat pump would be at least 1/ηcarnot  … as good as Carnot’s reversible machine, maybe better. But the same is true for the hypothetical superior engine with an efficiency of at least ηcarnot. So the performance numbers of all reversible machines (all in one set, characterized by the two temperatures) have to be exactly the same.

1. I love that you posted this. I sometimes come close to thinking about following a pure math stream, as it is probably easier to access from where I live now, but then I come in contact with a conversation like this and I feel that excitement that suggests applications of the theory is a lot more fun.

1. elkement says:

Thanks, Michelle, for using the word ‘excitement’ :-) It is indeed what I wanted to convey (on a meta-level, in addition to the facts) – and perhaps it takes a STEM enthusiast to detect it!
What I like about Carnot’s Theorem is that you can learn about one of the ‘deep and unintuitive’ secrets of nature … using only the simplest of math! The same goes for special relativity.

BTW – I had tried to visit your blog recently, and noticed you switched it to private mode. Will you open it again or start something new? :-)

1. I found the introduction to conservation laws in one of my classes like this, too: simpler with math, and really cool.

As for the blog, I want to get something new to replace of the old. I started playing at it in the spring, but ran out of time before getting anything done, so I only switched off to private mode. Just this weekend I decided to reduce my class load, so maybe one of these days… even if it is only a page, right? ;)
I am guessing, by the prolonged periods between posts, that you have been busy?

1. elkement says:

Looking forward to any new web project of yours, ‘small’ as it may be :-)
Yes, I have been busy – but I was also sort of exhausted after the two posts from August (plus the background research for it). I still ‘plan’ for about two posts per month, but maybe I will skip the second September post as I was not able to cut this one down to reasonable length :-) I’ll never learn the art of splitting one long post into several smaller ones …

1. The sign of a busy mind!

2. Reading this forced me to the inescapable conclusion that I should have paid closer attention to Physics in College.

1. elkement says:

I suppose you paid close attention – but it is the type of thing you forget if you don’t really need it ‘on a daily basis’ … or if it had been presented in a breath-taking, Feynman-style way. Carnot’s theorem would deserve stunning presentation: Feynman called its discovery ‘one of the few times in history when engineering made a contribution to fundamental physics’.

But typically in lectures only the paradox is presented – and the weird question about fixing the paradox that I have covered here is not answered. Perhaps it is confusing or superfluous as nobody has that question anyway. But on the other hand, it is that type of question(s) related to everyday physics that might turn amateur scientists into crackpots, when they feel they have found an ‘error’ in the way physics is presented in introductory classes.

3. I absolute love the ASCII art. With our slick graphics these days, we forgot how good we had it before :)

I follow your argument – which must mean it works! Any glimmer of hope I had of over-unity machines has just evaporated. Not bad for a Monday morning :)

1. elkement says:

The ASCII art was first only a way to create ‘temporary’ images for playing with different numbers quickly. I wanted to replace them with ‘better ones’ later – and then I thought the same: What was good enough for more complex diagrams in RFCs is good enough for me :-)

Thanks a lot for reading and following this long-winded, repetitive thread – I do really appreciate it!

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