I am considering a thermally isolated system of constant pressure. Where 10kg of air at 1000K (assume $Cp_{air}=0.98$ kJ/kgK, denote this as the hot reservoir by H) is connected to 10kg of water at 300K (assume $Cp_{water}=$4.2 KJ/kgK and denote this as the cold reservoir by C). The heat engine has a maximum thermal efficiency of 50%.

I know I should consider equilibrium in terms of entropy change. i.e. when

$\Delta S_{system} = 0$

I understand how to do this problem when the heat engine is reversible. Since then $\frac{\delta Q_H}{T_H} = \frac{\delta Q_H}{T_H} $ leading to

$\Delta S_{system} = (mC_P)_{air}\ln(\frac{T_f}{T_H}) + (mC_P)_{water}\ln(\frac{T_f}{T_C}) = 0$

$\Rightarrow T_f =$ equilibrium temperature of system = 376.7K.

However I can't work out how to take account of the fact that the thermal efficiency of the heat engine is only 50%. I have done the following so far:

$\eta_{th} = \frac{\delta W_E}{\delta{Q_H}} = 0.5$

$\Rightarrow \delta W_E = 0.5\delta{Q_H}$

Since 1st law $ \Rightarrow dU=\delta{Q}-\delta{W} $ and for a heat engine completing a cycle, $dU=0$ and hence $\delta{Q_C}=-0.5\delta{Q_H}$. But I know this is only for the maximal case, and I know the thermal efficiency will reduce as the two temperatures equalize.

EDIT: The exact problem is P2.10 below: problem


1 Answer 1


I can think of two interpretations of this question. Did you give the exact problem statement?

One interpretation would be that, irrespective of the nature of the cyclic process, the efficiency is 50% over then entire path. So, $$Q_H=m_HC_H(T_{H0}-T_F)$$ $$Q_C=m_CC_C(T_F-T_{C0})$$ and $$\frac{Q_H-Q_C}{Q_H}=0.5$$

The other interpretation would be that $\Delta S$ is equal to zero up to the point that the temperatures are such that the efficiency is 50%, after which the efficiency is held constant at 50% (as in the first interpretation).


Whoops. I think I got the 2nd interpretation backwards. Start with the efficiency of 50%, and then switch to the $Delta S=0$ path after the absolute temperature of the cold reservoir reaches 50% of that of the hot reservoir.


Try this and see if it works. Solve for intermediate temperatures $T_{H}$ and $T_C$ at which $T_C=T_H/2$ with 50% efficiency and with the instantaneous Carnot efficiency is also 50%: $$\frac{m_CC_C(0.5T_H-T_{C0})}{m_HC_H(1000-T_H)}=0.5$$

What values do you get for $T_{H}$ and $T_C$? Then, use these temperatures as the starting point for a second change in which you use your $\Delta S=0$ equation, with these values as the starting temperatures. What do you get for the final temperature?

  • $\begingroup$ Thank you, I have now added a picture of the exact problem from the book. When I applied the 1st law and assumed a constant thermal efficiency, I also arrived at the same equations that you gave for your first interpretation; which yield a final temperature of Tf=373.13K,which is different to the answer in the book of 385.1K. Unfortunately, I'm not quite sure how I would work the temperature out using your second interpretation, as I thought $\Delta S = 0$ was the definition of equilibrium? $\endgroup$
    – user401751
    Jan 12, 2018 at 21:48
  • $\begingroup$ See Edit #2 above $\endgroup$ Jan 12, 2018 at 23:01
  • $\begingroup$ I have done as you suggested and the intermediate values are $T_H = $ 675.68 and $T_C = $ 337.84. Substituting these into equilibrium equation results in 373.14K again! I'm beginning to wonder if the answer in the book is wrong. $\endgroup$
    – user401751
    Jan 12, 2018 at 23:50
  • $\begingroup$ Those are the same temperatures I get for the first part. But, it doesn't seem possible that you would get the same final temperature again, considering the efficiency in the delta S calc from this point on drops below 0.5. I'll try running the 2nd part of the calculation myself and see what I get. $\endgroup$ Jan 13, 2018 at 0:12
  • $\begingroup$ OK. I get this: $$T_F=(675.68)^{0.98/5.18}(337.84)^{4.2/5.18}=385.18$$ $\endgroup$ Jan 13, 2018 at 1:21

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