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The net work of a Carnot cycle, $W$, is $$\tag{1}W=Q_{h}-Q_c.$$

This work also satisfies $$\tag{2}W=Q_{h}\bigg ( 1-\frac{T_c}{T_h} \bigg )$$

Let us fix $Q_{h}$, the amount of heat initially extracted from the heat bath with temperature $T_h$.

$Q_c$ is exactly equal to the amount of work $W_3$ we put in during the isothermal compression, i.e., $Q_c = W_3$. Therefore, $W$ in eq. $(1)$ can be made bigger by reducing $W_3$, in other words $W$ depends on us.

Now, eq. $(2)$ says a completely different story. Given $Q_h$, which we already fixed, we no longer have any control over $W$. It is a constant that is determined by the temperatures of the heat baths and the initial amount of heat exctracted $Q_h$.

Where does this apparent contradiction come from?

P.S. We assume that we are working with a perfect gas.

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2 Answers 2

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The quantities $Q_h$, $Q_c$, and $W$ all depend on how "big" the Carnot cycle is, i.e. its size on a PV diagram. Once you fix any one of them, you fix all of them.

A Carnot cycle has four steps: (1) isothermal expansion, (2) adiabatic expansion, (3) isothermal contraction, and (4) adiabatic contraction. If you have fixed $Q_h$, then you have already fixed step (1). Then step (2) is fixed by the fact that the final temperature must be $T_c$. Naively you have freedom in step (3) to contract the gas as much as you want, but if you don't contract it just the right amount, you won't return to the initial state in step (4), so you won't have a cycle at all.

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𝑄𝑐 is exactly equal to the amount of work π‘Š3 we put in during the isothermal compression, i.e., 𝑄𝑐=π‘Š3. Therefore, π‘Š in eq. (1) can be made bigger by reducing π‘Š3, in other words π‘Š depends on us.

Keep in mind that in order to reduce the isothermal compression work (lower $Q_c$) you need to lower the temperature of the low temperature reservoir, $T_c$. That increases the Carnot efficiency ΞΆ in equation (2)

$$ΞΆ=1-\frac{T_c}{T_h}$$

Now, eq. (2) says a completely different story. Given π‘„β„Ž, which we already fixed, we no longer have any control over π‘Š. It is a constant that is determined by the temperatures of the heat baths and the initial amount of heat exctracted π‘„β„Ž. Where does this apparent contradiction come from?

There is no contradiction. Equations (1) and (2) are not independent of one another. As i indicated above, in order to lower $Q_c$ in equation (1) you need to lower $T_c$ in equation (2). Lowering $T_c$ while keeping $T_h$ fixed in equation (2) increases the efficiency and therefore increases the net work done.

"in order to reduce the isothermal compression work (lower ) you need to lower the temperature of the low temperature reservoir, " Would you mind explaining why this is the case? I don't really see why the amount of work that we put in depends on the temperature of the thermal reservoir.

To help visualize why this is the case, see the Pv and Ts diagrams for the Carnot Cycle.

The area enclosed in the cycle represents the net work done in the cycle. Note that in order to increase this area, you need to raise the isothermal expansion path and/or lower the isothermal compression path. To do this you need to increase the work done in the isothermal expansion (path 1-2), which means increasing $T_h$ and/or decrease the work done in the isothermal compression (path 4-1), which means lowering $T_c$. Either approach increases the Carnot efficiency and thus the net work done.

Hope this helps.

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  • $\begingroup$ Bob D "in order to reduce the isothermal compression work (lower ) you need to lower the temperature of the low temperature reservoir, " Would you mind explaining why this is the case? I don't really see why the amount of work that we put in depends on the temperature of the thermal reservoir. Why can't we just compress the gas by an amount of work less than the first $Q_c$? $\endgroup$
    – Hilbert
    Jan 12, 2020 at 10:34
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    $\begingroup$ @Hilbert look at the Pv diagram of the Carnot Cycle. The net work done in the cycle is the area enclosed by the cycle. In order to increase the area you have to either raise the isothermal expansion curve (increase isothermal expansion work) which means increasing $T_H$, or lowering the isothermal compression curve (decreasing isothermal compression work) which means decreasing $T_C$ (or a combination of both). $\endgroup$
    – Bob D
    Jan 12, 2020 at 10:55
  • $\begingroup$ I think it's more clearer now with the area/work method. If you don't mind, I have a small side question, . The ingenuity of the Carnot engine lies in the fact that we need an amount of work $Q_c$ lower than the produced heat /work $Q_h$ in order to restore the engine to its initial state, isn'it? $\endgroup$
    – Hilbert
    Jan 12, 2020 at 11:03
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    $\begingroup$ @Hilbert Actually you need to reject heat at a lower temperature than the temperature at which you received heat for any cycle, not just Carnot cycle. The ingenuity of the Carnot is its efficiency depends only on the temperature spread between the two reservoirs, as shown in the equation for the Carnot efficiency. Hope this helps. $\endgroup$
    – Bob D
    Jan 12, 2020 at 11:12
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    $\begingroup$ @Hilbert I've added some diagrams of the cycle to further help explain. Hope it helps. $\endgroup$
    – Bob D
    Jan 12, 2020 at 15:50

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