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So I know that the efficiency of a Carnot engine is given by $1-T_c/T_h$. This tells us that even with an idealised cycle we can never have a 100% efficient heat engine unless $T_c$ is $0$ K or $T_h$ approaches infinity, but I'm having a bit of trouble trying to understand this intuitively.

Perhaps my understanding of efficiency in this context is wrong, but it seems to me that if all of the heat supplied by the hot source is converted into work in an idealised reversible process, such that the final temperature is that of the cold sink ($T_c$), then this would be a 100% efficient process since there is no "waste heat" expelled.

I know that it is not possible in practice to convert all of the heat into work, but I guess my question is why, even if this were possible, would the engine still not be 100% efficient according to the Carnot efficiency?

Thanks

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  • $\begingroup$ If you look at the derivation of $1 - T_c/T_h$, you'll see why it is this way: en.wikipedia.org/wiki/Carnot_cycle#The_Carnot_cycle $\endgroup$
    – Señor O
    Aug 17 at 4:39
  • $\begingroup$ Carnot efficiency as $1-T_c/T_h$ only makes sense if you take into consideration the Second Law of Thermodynamics. In general, it is defined as the ratio between the work done and the heat extracted. $\endgroup$
    – Felipe
    Aug 17 at 4:40
  • $\begingroup$ When you say "such that the final temperature is that of the cold sink ($T_c$)". What final temperature are you referring to? The system? $\endgroup$
    – Bob D
    Aug 19 at 15:20
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From an intuitive perspective, when you introduce a hot fluid into a heat engine, that hot fluid contains a lot of thermal energy. To obtain 100% efficiency from that hot fluid, you would have to extract ALL of the energy out of it and turn that energy into work, based on the definition of 100% efficiency, which necessarily means that you would have to exhaust that fluid to a 0 K cold sink. Obviously, this isn't possible, as no working fluid can be driven to 0 K in a finite number of steps.

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  • $\begingroup$ Okay thanks, I think I see what I was missing now. Correct me if I'm wrong but I believe my confusion came from not considering the initial internal energy of the working fluid as contributing to $Q_{in}$ when calculating efficiency. $\endgroup$
    – dbp1023
    Aug 17 at 4:52
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I know that it is not possible in practice to convert all of the heat into work

It is theoretically possible to convert heat completely into work (100% efficiency) in a process. That is what happens in the reversible isothermal (constant temperature) ideal gas expansion process in a Carnot Cycle. That does not violate the second law. But it is not possible to completely convert heat into work in a cycle. That would violate the Kelvin-Planck statement second law which states:

No heat engine can operate in a cycle while transferring heat with a single heat reservoir.

but it seems to me that if all of the heat supplied by the hot source is converted into work in an idealised reversible process, such that the final temperature is that of the cold sink ($T_c$), then this would be a 100% efficient process since there is no "waste heat" expelled.

It's not clear (at least to me) what "final temperature" you are referring to, but from the first law

$$\Delta U=Q-W$$

Then, in order for all the heat to be converted to work in a process, $Q=W$ and thus $\Delta U=0$. For an ideal gas $\Delta U=0$ only if the temperature of the system is constant (a reversible isothermal expansion as discussed above).

but I guess my question is why, even if this were possible, would the engine still not be 100% efficient according to the Carnot efficiency?

Because the Carnot efficiency only applies to a heat engine operating in a cycle. It does not apply to an engine for a single process which, as indicated above, can theoretically be 100% efficient.

Hope this helps.

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I am not too sure what you mean in your second paragraph. However, why we can't have heat engines with efficiencies of 100% can be understood as follows

First, we need to consider the second law of thermodynamics

`The entropy of the universe increases in the course of any spontaneous change'.

With a heat engine, we need to think about the system and its surroundings. If we remove the cold sink from our heat engine and assume that all of the heat extracted from the hot source is converted into work, then the entire change in entropy is $dS=-dq_{\text{hot}}/T_{\text{hot}}<0$, which is a violation of the second law, where $dq_{\text{hot}}$ is the heat extracted and $T_{\text{hot}}$ is the temperature of the hot source. If we now include the cold sink, the change in entropy becomes $dS=-dq_{\text{hot}}/T_{\text{hot}}+dq_{\text{cold}}/T_{\text{cold}}$ and considering the second law must be $dS>0$, and the energy avaiable for work is $dW=dq_{\text{hot}}-dq_{\text{cold}}$. For a change in entropy to be positive, the minimum amount of heat that must be discarded into the cold sink must be such that $dq_{\text{hot}}/T_{\text{hot}}=dq_{\text{cold}}/T_{\text{cold}}$, which gives for the work $dW=dq_{\text{hot}}\left(1-T_{\text{cold}}/T_{\text{hot}}\right)$, which gives for the efficiency of the conversion $\epsilon=\left(1-T_{\text{cold}}/T_{\text{hot}}\right)$, which is Carnot's formula.

To conclude, for there to be an increase in entropy, our heat engine must have a cold sink where some heat can be discarded to provide a positive contribution to the total change in entropy. We can think of this as a sort of energy tax that the universe takes when converting one form of energy to another. And since we can not find a cold sink at absolute zero from the third law, we can not have an efficiency of 100%.

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