## Yet another cute physics problem

I’ve spent all day either teaching or writing draft grant applications and am consequently a bit knackered, so in lieu of one of my usual rambling dissertations here is another example from the file marked Cute Physics Problems, this time from thermodynamics. It’s quite straightforward. Or is it? Most people I’ve asked this question in private have got it wrong, so let’s see if the blogosphere is smarter:

*Three identical bodies of constant heat capacity are at temperatures of 300, 300 and 100 K. If no work is done on the system and no heat transferred to it from outside, what is the highest temperature to which any one of the bodies can be raised by the operation of heat engine(s)?*

November 22, 2011 at 5:18 pm

Sorry about the glitch; a few votes got lost. Please feel free to vote again if yours went astray.

November 22, 2011 at 5:34 pm

Let the answer be X. Then the other two bodies will finish at identical lower temperatures, denoted Y. From energy conservation,

X + 2Y = 300 + 300 + 100 = 700.

Clearly Y > 100, which is the starting temperature of the coldest body. From the equation above it follows that X < 500. Therefore, of the answers given, the only one possible is 400K.

More interesting is to work out how to do it…

November 22, 2011 at 6:57 pm

There, I’ve put Anton’s comment back….

November 22, 2011 at 8:55 pm

I kind of used a common sense approach. Basically, you can use all the energy from the 100 degree item to move the heat from one of the 300 degree items to the other. Or you can use one of the three hundred degree items to move the 100 degrees to the other 300 degree item. Either way, you get 400.

November 22, 2011 at 9:01 pm

So you think you can remove

allthe heat from the coldest object? Or have I misunderstood your reasoning?November 22, 2011 at 5:39 pm

Anton, I (temporarily) removed your comment as I’m enjoying seeing how many people give the wrong answer. I’ll replace it in due course, and also give a solution…

November 22, 2011 at 6:11 pm

If I were you I would change the range of options so as to provide two possibles between 300 and 500. Then my 3-line trick wouldn’t work any more.

November 22, 2011 at 6:36 pm

it’s very annoying. I added new options but it refuses to display them on the poll itself, only on the answers. Ah, right. It’s done it now.

November 22, 2011 at 6:18 pm

This is a great problem for teaching thermodynamics because there’s such a straightforward parallel with mechanics problems where you conserve energy and momentum… (or is that the clue Anton gave?!)

November 22, 2011 at 6:44 pm

Yes indeed, there’s something other than energy to think about.

November 22, 2011 at 7:29 pm

This is no. 172 in Pippard’s “Cavendish Problems in Classical Physics” and it has an asterisk by it, signifying that it is one of the tougher questions in the book.

The highest temperature of the hottest body is reached when the heat engines work adiabatically; since entropy S is proportional to log T then the sum over the bodies of (log T) is conserved, so that

2log300 + log100 = log X + 2logY

Combine this with the energy conservation equation

2*300 + 100 = 700 = X + 2Y

and you get a cubic equation for X, whose only sensible solution is 400K.

That gets you the answer in the absence of hints based on the alternatives presented in the multiple choice. But it still doesn’t tell you HOW…

November 22, 2011 at 8:31 pm

Does this relate to Carnot heat engines?

Since no work can be done on the system, the only way to move energy between bodies is to use the temperature difference between two of the bodies (the 100k as our cold, one 300k as our hot).

The efficiency of a Carnot engine is 1-Cold/Hot = 2/3 in our case.

So we get 2/3 of the energy out, letting us boost the second body by an energy sufficient to increase its temperature by 100K, giving 400K as the correct answer.

And since Carnot’s engines are the most efficient, that’s the maxiumum it can be.

I suppose it must all come down to the fact that to extract usable energy requires entropy to be raised; we can raise the entropy by sharing the heat between two of the bodies, and the most efficient way possible is to let them tend to their mean. The energy we get out of that entropy raising (or if you prefer, the entropy we lower elsewhere in the system) is enough to raise the body by 100K.

Fun stuff to think about, Peter.

November 22, 2011 at 8:32 pm

I think the wording of this one could be a bit better, as answers here tend to depend heavily on whatever definition of heat engine, one applies.

Anyway, according to the Wikipedia definition, the engine sits between to reservoirs and does work to a third one. How exactly the energy is split between the cold reservoir and the third one depends on the efficiency of the engine, which is limited by Carnot’s theorem.

Say it is at maximal efficiency, and that somehow the third substance turns the mechanical energy created by the system into preserved heat (which is probably somewhat fishy and is just my take on what might have been meant by the question). In the beginning, the maximal efficiency will be 2/3. This will of course change as heat is transferred, but to begin with, let’s assume that it won’t. That is, the warm reservoir will transfer heat with a ratio of 2/3 to the third one, and 1/3 to the cold one, until they reach the same temperature. This will happen when it hits 150K, where the third substance reaches the temperature of 400K (as someone else has guessed). Now, recall that I overshot by assuming constant efficiency. Thinking a bit more carefully about how this changes, one could probably come up with an exact solution by integrating, but assuming that the right answer is part of the poll, it must be 350K, as we now know that it is strictly less than 400K and strictly greater than 300K. Recall also that I claimed I didn’t understand the question, so this is just ramblings with numbers.

Okay, so for the hell of it, doing this precisely, the temperature of the colder substance as a function of the hotter is

y'(x) = -y(x)/x

The unique solution to this with initial condition y(100) = 300 is

y(x) = 30000/x

which has a unique fixed point in at sqrt(30000). Thus the highest possible temperature is going to be

700 – 2sqrt(30000)

which is approximately 353.6. Now, this number isn’t on the list, so maybe my interpretation of the question is completely off (pretty likely), or the author didn’t want to put in this weird answer.

November 22, 2011 at 8:45 pm

I believe the general definition of a heat engine is very standard in physics.

November 22, 2011 at 8:53 pm

Implausible as it might be, can’t you just do the whole thing in one Carnot cycle (keeping efficiency at the maximum throughout)?

Actually, isn’t such a cycle the only reversible process, as otherwise you are not conserving entropy?

November 22, 2011 at 8:54 pm

Here’s my favourite solution:

Let A, B and C be the final temperatures of the three bodies in units of 100K.

Conservation of energy (heat) gives us A + B + C = 3 + 3 + 1

Since, as Anton says, the maximum will be reached by an adiabatic process, entropy in = entropy out so log (A) + log(B) + log (C) = log(3) + log(3) +log (1), or ABC = 3 x 3 x 1

So we need to satisfy these two equations and maximise A.

Assuming that this solution will have B = C, we can combine the two

equations and eliminate B to give:

A + 6A^{-1/2} = 7

which has an exact solution at A = 4, so the correct answer is 400K.

November 22, 2011 at 9:21 pm

There’s another solution of 900K, with the other two bodies having been sucked of heat down to an impressive -100K each…

November 22, 2011 at 9:31 pm

That’s colder than my office!

November 22, 2011 at 9:38 pm

Surely you mean hotter, Peter!

March 27, 2017 at 12:18 am

The energy does not conserve for this answer and the equation for entropy is wrong. Since the heat capacity, c, is constant Q=c(T_f – T_0) and the entropy is S=C(T_f^2 – T_0^2). The sum of entropy has to be larger or equal than zero and the energy must conserve. Therefore, the answer is 500k in order to respect these two conditions.

November 22, 2011 at 9:12 pm

Just dump one of the 300K’s into the 100K. They both reach an average temperature of 200K, and you get 100K of work out in the process, which you can use to elevate the temperature of the other body 100 degrees, ending at 400K.

November 23, 2011 at 6:05 am

Plz delete this

November 22, 2011 at 10:56 pm

Am I being obtuse, or isn’t the difficulty that while conservation of energy allows the (400/200/200) solution because it contains the same amount of energy as the starting (300/300/300), no heat engine can pump the required heat from one 300K body to the other because the coefficient of performance required violates the 2nd Law – so you can’t get from one state to the other?

Too late in the evening to do the integration…

November 23, 2011 at 3:01 pm

300+300+100 = 700 (as in the problem statement, and conserved). But

400+200+200 = 800

300 +300 +300 = 900

Doing the integration will get the right answer (supposing these were typos) but it’s the hard way. It’s worth adding that the end solution is

400, 150, 150

so that the coldest body is warmed by only 50 degress, yet the one initially at 300 that is ‘sacrificed’ for the other is cooled by 150 degrees.

November 23, 2011 at 3:59 pm

Here’s an explicit description of how you could actually achieve this result with (ideal) heat engines.

Call the two 300-K bodies A and B, and the 100-K one C. Hook up an ideal (Carnot) heat engine so that it uses A and C as its hot and cold reservoirs. Use the work produced by this engine to power a Carnot refrigerator / heat pump (which is, after all, just a Carnot engine running backwards). Use this device to pump heat from C to B. This’ll keep working, heating up B, as long as A’s temperature exceeds C’s.

With a tiny bit of indulgence on the part of the problem setter, you can do it another way: hook up your heat engine to use A and B as the work done to pump heat from C to either A (or B, but let’s say A). This won’t work if T_A=T_B, since you can’t get any work out of the heat engine, but the system in this case is in an unstable equilibrium: if you let the temperature fluctuate a tiny amount, so that T_A>T_B by even a tiny amount, then off it goes.

November 24, 2011 at 9:58 am

Cavendish problems: happy days….

I did it your “favourite” way, but wasn’t satisfied, for two reasons. Calling the final temps A,B,C and the initial ones a,b,c, the mathematical setup is ABC=abc (entropy) and A+B+C=a+b+c (energy). Let A be the highest of the three temperatures, in which case you need to choose B & C to maximise A.

That sounds a bit nasty, but things become easier if you see that the solution will be symmetric with B=C=X, so we have to solve AX^2=abc and A+2X=a+b+c. Trouble is, the best way of seeing this is physical (no way to pump more heat to A when B=C), rather than purely mathematical.

But even so, it’s ugly to end up with a cubic where you have to play the old game of trying to fluke a solution. It works in this case with a bit of trial and error, so (a,b,c)=(100,300,300) gives X=150 and hence A=400.

The fact that we ended up in this case with X=(a+b)/2 seems to offer some hope that there might be a simpler argument, and some commenters tried to follow these lines. But It’s hard to see how this can be right. Mathematica will write out the general solution, but it’s not pretty, and even other simple whole-number examples fail to come out neatly: (a,b,c)=(100,200,300) gives X=135 and A=330, for example. Can anyone see how to deal with such cases without solving a cubic?

November 24, 2011 at 10:19 am

What’s wrong with solving cubics? We’ve known how to do that systematically without ‘fluking’ since the work of Ferro, Tartaglia and Cardano in the Renaissance.

November 24, 2011 at 2:41 pm

Anton: I know there’s an exact solution for cubics. I also know the formula is too long and messy to carry in my head, so in practice I may as well be solving it numerically. There is also a practical issue that the Cardano formula gives complex expressions, which can be real and yet not obviously so. I’d rather there was a simpler way.

November 30, 2011 at 5:37 pm

In this case you also know that 300 300 100 is a solution. That gives you one factor of the cubic, and you can divide to get a quadratic. I have to admit that isn’t what I did. I just looked for an on line cubic equation solver.

November 24, 2011 at 3:29 pm

You can solve this one by inspection!

November 24, 2011 at 3:52 pm

John: I did indeed expect that you would be aware cubics could be solved systematically, but I was confused by your comment about fluking it. I agree that the formula is complicated – though not necessarily complex (couldn’t resist that) as at least one root is guaranteed to be real.

Peter: By ‘inspection’ do you mean bracketing the solution by hand work and then taking a punt on 400?

November 24, 2011 at 4:03 pm

I assume that Peter’s “inspection” is my “fluking” – i.e. you try X=200,

then X=100, from which you learn there is root in between. So now you try X=150, but with no reason to expect that the answer will be a whole number (well, OK, you do expect that, because this is how we set up undergraduate problems – but in the real world it would never happen). You can just about do it in your head, especially if you work in multiples of 100K, but it still requires some trial and error, surely?

November 24, 2011 at 4:53 pm

Not really. I just noted that the equation I got involved a square root, and supposed that it might therefore be solved by a perfect square. Since the answer clearly could not be A=1, the next one to try was A=4, which works, as 4+6/2 = 7.

November 25, 2011 at 11:45 am

“no heat transferred to it from outside” – presumably we are not allowed to transfer heat _out_ of the system, e.g., use the rest of the universe as a cold reservoir at 3K?