We have two different set ups:

  1. A 200K brick in a 300K environment
  2. A 300K brick in a 200K environment

From which one of these can you extract more work? Assume equal mass and heat capacity.

Using these equations:

Efficiency - $$\eta=\frac{T_H−T_L}{T_H}$$


$$W = ΔQ => W≤Q_H\times\eta_C$$

I regard both as cases as infinitesimal Carnot engines.

a) $T_L$ is varying, and after integration I get m*c*16.67 (m and c are the mass and heat capacity)

$$ \int_{200}^{300} {\frac{300 - T}{300}}{dT} = 100 - 250/3 = 16.67 $$

b) $T_H$ is varying, after integration the result is m*c*19.

$$ \int_{300}^{200} {\frac{T - 200}{T}}{dT} = -100 + 200 * \log(3/2) = 18.9$$ Therefore, I conclude that I can get more work from b; the problem is that the correct answer is marked a.

Does anyone have any idea what I am doing wrong? (Or perhaps can you assure me I'm correct?)

  • $\begingroup$ Please show how you do your integration - I don't get the same numbers that you are getting... $\endgroup$
    – Floris
    Commented Aug 23, 2014 at 18:41
  • $\begingroup$ Sorry, guess I was kind of lazy with copying it. a) $ \int_{200}^{300} {\frac{300 - T}{300}}{dT} = 100 - 250/3 = 16.67$ b) $\int_{300}^{200} {\frac{T - 200}{T}}{dT} = -100 + 200 * \log(3/2) = 19$ $\endgroup$
    – golanor
    Commented Aug 23, 2014 at 18:57
  • $\begingroup$ Welcome to Physics SE! Your comment actualy helps to understand your problem :) Please click the edit Button and put it in your Question. $\endgroup$ Commented Aug 23, 2014 at 19:07

2 Answers 2


As a rule, when the initial temperature difference is the same, the system that rejects heat at the lowest temperature will be the more efficient. That is because this involves the least amount of entropy.

So without doing the math, I would say that case (b) will give rise to the greater amount of extracted work - which is also your conclusion.

And yes, your math looks about right. In case (a), the integral is 16.7; in (b) it is 18.9 . This confirms that (b) allows the extraction of more work from the brick.

I think the answer given was wrong.

  • $\begingroup$ Do you have some intuitive way to know which of the processes involve a lower amount of entropy? $\endgroup$
    – golanor
    Commented Aug 23, 2014 at 19:28
  • $\begingroup$ +1 I just added an answer, looking at it in terms of exergy, hope you don't mind. $\endgroup$
    – Ellie
    Commented Aug 24, 2014 at 1:31
  • $\begingroup$ @phonon - I don't mind at all. I actually think there is a problem with my answer... I will think this through more carefully and update when I figured out what I don't like. $\endgroup$
    – Floris
    Commented Aug 24, 2014 at 1:34
  • $\begingroup$ @Floris I don't see a problem, maybe the formulation of "rejects heat at the lowest temperature" could be improved, e.g. by clarifying that "lowest" here corresponds to lowest temperature of the surrounding. You've followed it up with "least amount of entropy" which is correct again, as in the opposite case, a process with high entropy generation, would greatly reduce the possible obtainable work. $\endgroup$
    – Ellie
    Commented Aug 24, 2014 at 1:49

Floris's answer is of course very nice. I will just show very quickly how you can solve these scenarios in terms of exergy.

Definition: In the case of a system coupled to a reservoir(environment), exergy is interpreted as the potential of a system to do work as it achieves equilibrium with the reservoir. Sometimes also referred to as "maximum shaft work" that can be done.

We know that the upper limit of conversion of heat $Q$ into work $W$ is given Carnot's efficiency, so exergy is usually very simply defined as: $$B=Q\left(1-\frac{T_{\rm reservoir}}{T_{\rm system}}\right) $$

Now when you substitute the expression of $Q$ taken from the above, in total entropy, you will see that as the exergy of a system decreases as the entropy increases, i.e. any entropy generation, bringing the system towards equilibrium, will reduce the amount of energy available to be converted to work.

Back to your problem: let's see which case gives the maximum power generated during the process of reaching equilibrium, if $S$ the total entropy, then we solve for $dS=0$, in terms of $B$ it is equivalent to: $$\frac{dB}{dT}=-\frac{T_r /T_s^2}{1-T_r/T_s}B$$ ($r$ for reservoir and $s$ for system) Now knowing the maximum power generated corresponds to the case $dB/dT<0$, which in your case corresponds to case $b$ where $T_s>T_r$, as was also confirmed earlier by Floris.

In terms of intuition, there are many ways to see why case $b$ may lead to the maximum exergy during the equilibration process of a system with reservoir, one would be:

From the second law of thermodynamics we know that for a system to do work, there has to be some heat evacuated to the environment, as heat cannot be 100% converted to work. Why case $b$? well if the environment is at a lower temperature, then it is easier for the system to do work and to dispose heat to the cooler surrounding, whereas in case $a$, we have an exergy crisis, where the reservoir being hotter than the system, the system has to spend energy just to be able to evacuate heat, without which a process cannot yield any work, which just amounts to saying there will be less energy to convert to useful work.

Bear in mind the example of the refrigerator, where the role of the compressor is to compress the fluid, until its temperature surpasses that of the kitchen, so that heat transfer can take place and the fluid cools off again...


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