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Looking at the equation for Carnot efficiency, I notice that as the temperature of the heat sink approaches zero, the efficiency approaches unity:

$$ \eta_{rev} = 1 - \frac{0}{T_H} = 1 $$

Seeing as the efficiency of a heat engine is the ratio between the heat it is absorbing and the work it outputs, an efficiency of 1 indicates that all heat absorbed is output as work. By first law, this implies that the engine is rejecting no heat to the low temperature sink.

This result doesn't make any sense to me. Why would a decreasing heat sink temperature result in less heat rejected?

To explain my confusion somewhat hand-wavily: if the temperature of the two reservoirs is equal, we end up with no heat transfer, and therefore $Q_L$ is zero. As we deviate from this case of reservoirs with equal temperature (which is what happens if you decrease $T_L$ while holding $T_H$ constant), why is it that we once again approach the case of $Q_L$ equals zero?

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  • $\begingroup$ I think it's important to remember here that a heat sink at absolute zero is physically impossible. One expects impossible objects to exhibit paradoxical behaviors and contradictory properties. $\endgroup$ – David H Nov 26 '13 at 7:15
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    $\begingroup$ @DavidH It wasn't actually paradoxical, I was just being dumb. When we speak of high and low temperature in a Carnot engine, we have to remember that the heat exchange is occurring reversibly, i.e. the temperature of the working fluid must not differ from the temperature of the reservoirs it is exchanging heat with by more than an infinitesimal amount. That means, if we say $T_H$ is zero, the temperature of the working fluid must also be zero as it interacts with the heat sink. A zero temperature fluid contains no heat, and therefore cannot dump any heat. $\endgroup$ – Asad Saeeduddin Nov 26 '13 at 7:35
  • $\begingroup$ @DavidH This notion of the fluid having been robbed of all thermal energy is of course consistent with $\eta = 1$. $\endgroup$ – Asad Saeeduddin Nov 26 '13 at 7:45
  • $\begingroup$ Correction to the comment above: "That means, if we say $T_L$ is zero..." $\endgroup$ – Asad Saeeduddin Oct 24 '14 at 0:11
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The answer can be found by looking at the details of how Carnot engines, with their ideal gas ($PV=NkT$) working fluid, work. Suppose we start at the end of the isothermal expansion. The next step is to adiabatically expand the piston until $T_c$ is reached. When $T_c=0$, though, that requires $V\rightarrow \infty$. We are then supposed to isothermally compress the gas until it is on the desired adiabatic curve. The ideal gas adiabats obey $PV^\gamma=\mathrm{constant}$, with $\gamma > 0$, so all of the adiabats intersect at $V=\infty$. Because of this property the isothermal compression is, effectively, cut out of the process. Doing this required that the piston have infinite volume with zero outside pressure into which it could expand, and the ability to do it in a finite amount of time without letting the internal gas reach an out of equilibrium state.

From a more realistic perspective, where $T_h \gg T_c>0$, it's just a question of the adiabatic expansion removing most of the internal energy of the working fluid, so there isn't much left to reject into the cold bath to get onto the compression adiabat.

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Think about this is physical rather than algebraic terms for a moment. Notice that the term $\frac{T_L}{T_H}$ is the ratio between the highest and lowest temperature. This ratio tells you how well heat can flow from $T_H$ to $T_L$. The absolute temperatures don't actually matter, only their ratio.

So what is the physical meaning of $T_L = 0$? It means it has infinite heat absorption capacity. You can just keep dumping heat from $T_H$ to $T_L$ because when $T_L$ is $0$ the ratio between the two is infinite (mathematically the ratio is undefined but that doesn't matter here).

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    $\begingroup$ One reaches the same conclusion with $T_H \to \infty$, which is perhaps easier to understand. $\endgroup$ – lionelbrits Nov 22 '13 at 0:36
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    $\begingroup$ Okay, so a 0 K reservoir is extremely good at absorbing heat. This is precisely why my brain refuses to accept that it is passed absolutely no heat whatsoever. In other words, as $\Delta T$ increases, the rate of heat transfer between the two reservoirs ($Q_L$) decreases. $\endgroup$ – Asad Saeeduddin Nov 22 '13 at 0:37
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    $\begingroup$ This does not look correct to me. Why should $T_L=0$ imply infinite heat capacity? Temperature and heat capacity are not related. The former is an intensive property, the latter is extensive. And why is the ratio $T_L/T_H$ undefined when $T_L=0$? $0/0$ is undefined, $0/1=0$. $\endgroup$ – sammy gerbil Jul 28 '18 at 16:15
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If there is more difference in temperature, there is more difference in heat, which means more work, since $W= Q_1- Q_2$. Since a reversible process is by definition a very slow process that converts almost all heat to work, we can't consider the heat transfer rate idea of temp. Gradient in this

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This is my perspective: A carnot engine always operates between TWO temperatures only. AND, heat exchange ONLY happens when the working fluid is at the same temperature as the heat sink or source. In your argument, the heat sink is at zero kelvin. This implies that the working fluid is somehow brought to zero kelvin before it can exchange heat with the sink. Since the fluid is already at zero kelvin, there is nothing to exchange, as the entropy of a system at absolute zero is zero.

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This is related to the third law which states that entropy is zero at $T = 0$.

Since the carnot engine implies a reversible process, the second law says that $dQ = T\ dS$. So both $T$ and $dS$ (by the third law) go to zero and so a zero temperature reservoir cannot accept or give heat.

Notice that the efficiency of the refrigerator (engine running backwards) is $Q_C/W$, which goes to zero. So more and more work is required to extract heat from cold things. Which sort of proves the unattainability of zero temperature.

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