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.

]]>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.

]]>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.

]]>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? ]]>

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

]]>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.

]]>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?

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