I was working on a "simple" heat transfer question and have now confused myself!

This question refers to the case where an object is brought into contact with a thermal reservoir, causing the object to be heated.

There is an example of this in this textbook, problem 4.38.

I can solve the problem that is posed in the text book. However, I wondered what would happen if I proposed that the object was heated to a temperature above that of the thermal reservoir.

The second law states that this should be impossible, however I wanted to confirm this by proving that the entropy change of the universe for such a process would be negative.

I did the calculation and the following is my result:


$T_{Res} = 363 K$

$m = 1 kg$

$C_{P} = 4.187 kJ / (kg.K)$

$T_{initial} = 273 K$

$T_{final} = 373 K$ (deliberately picked to be above reservoir temperature)


$Q = m * C_{P} * (T_{final} - T_{initial})$

$Q = 1 kg * 4.187 kJ / (kg.K) * (373K - 273K)$

$Q = 418.7 kJ$

$\Delta S_{Universe} = -Q/T + m * C_{P} * ln ( (T_{final} / T_{initial})$

$\Delta S_{Universe} = -418.7 kJ / 363 K + 1 kg * 4.187 kJ / (kg.K) * ln (373K / 273K)$

$\Delta S_{Universe} = 0.153 kJ / kg$

This is very confusing to me, as I thought that any process with a $\Delta S_{Universe}$ of greater than zero was possible (as long as it obeyed the 1st law, which this does). However it is clearly an impossible process, as a 363 K reservoir cannot possibly heat an object to 373 K.

Have I misunderstood the significance of positive entropy change? Or, is there something wrong with my (and the textbook's) calculation?


I have done more research into this issue and have made some progress toward the answer. However it is still not complete. My first thought was that perhaps the inaccuracy stemmed from the fact that a true thermal reservoir does not exist.

So I did the calculation again using another body of water, but with a mass one million times greater than that of the body being heated. I did this for a range of final temperatures and the result is as follows:

Entropy Changes for Various Final Temperatures

If you look at the table you will notice that the maximum increase in entropy occurs when the body is heated to the initial temperature of the reservoir (in my case the larger body of water).

So this proves that the equilibrium temperature of the body is that of the reservoir, which makes sense. We can calculate the equilibrium temperature by taking the derivative of the entropy change of the universe with respect to the temperature of the smaller body. For two bodies that exchange heat and change temperature the entropy change of the universe is given by the following:

$\Delta S_{Universe} = m_{Res} * C_{P} * ln ( (T_{ResFinal} / T_{ResInitial}) + m_{Body} * C_{P} * ln ( (T_{final} / T_{initial})$

To get the entropy of the universe as a function of $T_{final}$ we can use the first law:

-$m_{Res} * C_{P} * (T_{ResFinal} - T_{ResInitial}) = m_{Body} * C_{P} * (T_{final} - T_{initial})$

Solving for $T_{ResFinal} $:

$T_{ResFinal} = T_{ResInitial} - \displaystyle\frac{m_{Body} * C_{P} * (T_{final} - T_{initial})}{m_{Res} * C_{P}} $

Substituting into our expression for the entropy change of the universe:

$\Delta S_{Universe} = m_{Res} * C_{P} * ln ( (T_{ResInitial} - \displaystyle\frac{m_{Body} * C_{P} * ((T_{final} - T_{initial})}{m_{Res} * C_{P}}) / T_{ResInitial}) + m_{Body} * C_{P} * ln ( (T_{final} / T_{initial})$

Expanding and cancelling like terms:

$\Delta S_{Universe} = m_{Res} * C_{P} * ln ( 1 - \displaystyle\frac{m_{Body} * (T_{final} - T_{initial})}{m_{Res}*T_{ResInitial}}) + m_{Body} * C_{P} * ln ( (T_{final} / T_{initial})$

Separating terms containing T_{final}:

$\Delta S_{Universe} = m_{Res} * C_{P} * ln ( 1 + \displaystyle\frac{m_{Body} * T_{initial}}{m_{Res}*T_{ResInitial}} - \displaystyle\frac{m_{Body}*T_{final}}{m_{Res}*T_{ResInitial}}) + m_{Body} * C_{P} * ln ( (T_{final} / T_{initial})$

Taking the derivative with respect to $T_{final}$:

$ \displaystyle\frac{d\Delta S_{Universe}}{dT_{final}} = m_{Res} * C_{P} * \displaystyle\frac{-m_{Body}/(m_{Res}*T_{ResInitial})}{(1+m_{Body}*T_{initial}/(m_{Res}*T_{ResInitial}))-T_{final}*m_{Body}/(m_{Res}*T_{ResInitial})}+ \displaystyle\frac{m_{Body} * C_{P}}{T_{final}}$

At equilibrium, the entropy of the universe has reached a maximum, which is why no further changes occur. Therefore $d\Delta S_{Universe}$ = 0, giving:

$ m_{Res} * C_{P} * \displaystyle\frac{m_{Body}/(m_{Res}*T_{ResInitial})}{(1+m_{Body}*T_{initial}/(m_{Res}*T_{ResInitial}))-T_{final}*m_{Body}/(m_{Res}*T_{ResInitial})} = \displaystyle\frac{m_{Body} * C_{P}}{T_{final}}$

Solving for $T_{final}$:

$ T_{final} = \displaystyle\frac{1+m_{Body} *T_{initial}/(m_{Res}*T_{ResInitial})}{m_{Res}* C_{P}*m_{Body}/(m_{Res}*T_{ResInitial})}/(\displaystyle\frac{1}{(m_{Res}*C_{P}}+\displaystyle\frac{1}{(m_{Body}*C_{P}})$

The above expression asymptotically approaches $T_{ResInitial}$ as $m_{Res}$ approaches infinity.

From this result we can conclude that any change past this temperature would result in a negative change of entropy for the universe. Therefore, it is impossible to heat the second body to a higher temperature than the first.

I guess this answered my own question, however it seems like an extremely long way to prove something that I thought would be very simple!

In the process of figuring this out I have realised there are some concepts I am not as sure about as I thought I was. I will give these some thought and possibly start another question about these.

Thanks everyone for your help and for reading.

  • $\begingroup$ Hey user, welcome to Physics Stack Exchange. We support Latex markups at Physics.SE. Please try using that for your math expressions :) $\endgroup$ Commented Dec 22, 2014 at 15:08
  • $\begingroup$ Thanks for the input, I have now updated the question to have clearer formatting. As for the CP value, it seems pretty accurate according to the following website: people.ucsc.edu/~bkdaniel/WaterProperties.html $\endgroup$
    – Appguy1
    Commented Dec 22, 2014 at 15:26
  • 3
    $\begingroup$ At the end of all this do you actually still have a question? $\endgroup$
    – Floris
    Commented Dec 29, 2014 at 4:07
  • $\begingroup$ Hi Floris thanks for your reply, as I mentioned, no I no longer have a question about why this process resulted in a positive $ \Delta S_{Universe}$. However if you have a simpler way to prove that this process is impossible please share it with us in the following thread: physics.stackexchange.com/q/155049 Thank you! $\endgroup$
    – Appguy1
    Commented Dec 29, 2014 at 9:40

2 Answers 2


Actually, there is a reversible process that will allow you to end up with a larger temperature, and this does not violate the second law. To see this, you can couple your system to Carnot engine, which will extract and "store" work until both the reservoir and the case are at the same temperature. Now use the stored energy to heat the case. Actually your can use your equation and find what the maximum possible temperature could be, by constraining it to be $\Delta S_{universe}=0$.

  • $\begingroup$ Thanks for your answer, I agree with you that we could use a Carnot engine and then heat the cold body until it is hotter than the first. I still find the question that I proposed puzzling though, because I did not include the Carnot engine in the system that I formulated. I guess the point of confusion for me is that I thought that if a process obeyed E_{Gen} = 0 and \DeltaS_{Universe} >= 0, then it was possible. However I guess there must be some other criteria it needs to meet so that it excludes examples like mine above. I will do some research and possibly post another question. $\endgroup$
    – Appguy1
    Commented Dec 23, 2014 at 7:53
  • $\begingroup$ @Appguy1 But that is the point, your process can be done reversibly (it doesn't matter "how" would you do it) , so it should not be excluded. $\endgroup$
    – user65081
    Commented Dec 23, 2014 at 14:41
  • $\begingroup$ It is true that the final state can be reached, but I am trying to analyse the case of doing this in a single step, and want to be able to prove that this is impossible. Also I don't agree that this can be done reversibly, because any time there is heat exchange across a finite temperature difference, the process is irreversible. Reference: web.mit.edu/16.unified/www/FALL/thermodynamics/notes/… $\endgroup$
    – Appguy1
    Commented Dec 24, 2014 at 0:36

Let's break up the process into two steps, which necessarily must happen in sequence--the first step is one in which the reservoir heats the body up from a temperature of 273 K to its own temperature of 363 K, and the second step is one in which the reservoir heats the body from 363 K to 373 K. The first step involves a positive entropy change, so it can happen spontaneously. But then once the body has reached 363 K, is it possible for the body to then spontaneously heat up further to 373 K? Let's repeat your calculation, but with an initial temperature of 363 K:

$Q = m * C_{P} * (T_{final} - T_{initial})$

$Q = 1 kg * 4.187 kJ / (kg.K) * (373K - 363K)$

$Q = 41.87 kJ$

$\Delta S_{Universe} = -Q/T + m * C_{P} * ln ( (T_{final} / T_{initial})$

$\Delta S_{Universe} = -41.87 kJ / 363 K + 1 kg * 4.187 kJ / (kg.K) * ln (373K / 363K)$

$\Delta S_{Universe} = -0.00156 kJ / kg$

So, this second step involves a net decrease in the entropy of the universe. I think the answer is that you can't always decide whether a process is allowed thermodynamically just by looking at the endpoints--whatever the entire time interval is between the endpoints, it needs to be true that the entropy of the universe is increasing or constant during any shorter time interval within the larger time interval.

  • $\begingroup$ Thanks for your reply, I agree with your analysis. I was getting at the same thing with my table and taking the derivative in order to find the maximum. <br/> I think we have now answered the original question, which was that positive entropy change alone is not enough to say that the process obeys the second law. I wonder if there is a more simple way to work this out rather than taking the derivative as I did in my example, as this is a long calculation just for my two body system. I am going to research this and possibly start another thread. <br/> Thanks again and merry Christmas! $\endgroup$
    – Appguy1
    Commented Dec 25, 2014 at 2:35

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