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:
Data:
$T_{Res} = 363 K$
$m = 1 kg$
$C_{P} = 4.187 kJ / (kg.K)$
$T_{initial} = 273 K$
$T_{final} = 373 K$ (deliberately picked to be above reservoir temperature)
Calculation:
$Q = m * C_{P} * (T_{final} – T_{initial})$
$Q = 1 kg * 4.187 kJ / (kg.K) * (373K – 273K)$
$Q = 418.7 kJ$
$\Delta S_{Universe} = -Q/T + m * C_{P} * ln ( (T_{final} / T_{initial})$
$\Delta S_{Universe} = -418.7 kJ / 363 K + 1 kg * 4.187 kJ / (kg.K) * ln (373K / 273K)$
$\Delta S_{Universe} = 0.153 kJ / kg$
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?
UPDATED WITH FURTHER PROGRESS
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:
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.
Best Answer
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 $\Delta S_{universe}=0$.