I'm confused by the following excerpt from my thermo book, which is referring to a heat engine in contact with two reservoirs:

According to the first law of thermodynamics, the net work transfer is equal to the net heat transfer via $W_{net} = Q_H+Q_L$. Unfortunately, the net positive work transfer cannot be made to approach the heat transfer from the high temperature heat reservoir simply by adjusting the negative heat transfer with the low temperature heat reservoir. Once the thermodynamic temperatures of the heat reservoirs have been established, the second law of thermodynamics limits the possible heat transfers between these heat reservoirs and a heat engine.

I'm confused about the bold text; why can't we simply decrease the heat transfer to the cold reservoir, $Q_L$?

The second law for an irreversible cycle in contact with two reservoirs leads to

$-\frac{Q_L}{Q_H} < \frac{T_L}{T_H}$

For a given $T_L$ and $T_H$, this implies that we can decrease the magnitude of $Q_L$ as much we like (in fact, $Q_L=0$ satisfies this equation).

Why can't we decrease $Q_L$ so that $W_{net}$ approaches $Q_H$?

EDIT: My signs were all messed up, and I figured this out. I said the second law leads to

$-\frac{Q_L}{Q_H} < \frac{T_L}{T_H}$

but it's actually

$\frac{Q_L}{Q_H} < -\frac{T_L}{T_H}$ (big difference in signs)

and $Q_L < 0$. So we can't simply increase $Q_L$ (or decrease its magnitude) to be greater than $-{Q_H} \frac{T_L}{T_H}$.

  • $\begingroup$ read the highlighted text again and note the "negative heat transfer " $\endgroup$
    – hyportnex
    Commented Sep 12, 2018 at 16:06
  • $\begingroup$ You can do what you said. You just can’t do it while working in a cycle. If you try, you just won’t be able to get the working fluid back to its initial state. $\endgroup$ Commented Sep 12, 2018 at 16:17
  • $\begingroup$ I discuss this point in a note here (see the "party boat" discussion). Cooling the hot reservoir transfers out entropy, but work can't carry entropy. Since entropy can't be destroyed, it must be sent somewhere; we call this location the cold reservoir. $\endgroup$ Commented Sep 12, 2018 at 16:19
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    $\begingroup$ The second law diesn’t say you can’t do what you say. It says you can’t do it in a cycle. If you think you can, just try to conceive of such s cycle. $\endgroup$ Commented Sep 12, 2018 at 16:36
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    $\begingroup$ your correction is correct now; and this is why I told you to read the quote again so that you notice that $Q_L$ is negative while all the other terms are positive, and one cannot just divide an inequality with a negative number without flipping the sign. But you should also study @Chester_Miller's comment because that inequality (Clausius) $Q_L/T_L + Q_H/T_H \le 0$ you are quoting is true only for an isothermal-adiabatic-isothermal-adiabatic thermodynamic cycle. $\endgroup$
    – hyportnex
    Commented Sep 13, 2018 at 23:24

1 Answer 1


The maximum efficiency is that of a Carnot Cycle and is given by

$$1-\frac {T_L}{T_H}$$

In order to have a 100 % efficient cycle (use all the heat from the high temperature source to produce work), $T_L$ would need to be zero. I believe current consensus is that absolute zero Kelvin is not attainable.

Hope this helps


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