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A heat pump has an efficiency greater than 100%. Does this violate the laws of thermodynamics?
My answer:
I'm assuming that "heat pump" means heat engine.
$e>1$
$\displaystyle\implies\frac{W}{Q_H}>1$
$\implies W>Q_H$
$\implies W-Q_H>0$
$\implies (Q_H+Q_C)-Q_H>0$
$\implies Q_C>0$
This violates the laws of thermodynamics.
Is this correct?
How is the efficiency of a heat engine related to the entropy produced during the process?
The maximum efficiency for any heat engine operating between two temperature $T_H$ and $T_C$ is the Carnot efficiency, given by $$e_C = 1 -\frac{T_C}{T_H}.$$
Such a heat engine produces no entropy, because we can show that the entropy lost by the hot reservoir is exactly equal to the entropy gain of the cold reservoir, and of course, the system's entropy on the net doesn't change because the system undergoes a cycle.
Any heat engine operating between the same two temperatures whose efficiency is less than $e_C$ necessarily increases the entropy of the universe; in particular, the total entropy of the reservoirs must increase. This increase in entropy of the reservoirs is called entropy generation.
Finally, the efficiency of the perfect engine is less than one, necessarily, because the entropy "flow" into the system from the hot reservoir must be at least exactly balanced by the entropy "flow" out of the system into the cold reservoir (because the net change in system entropy must be zero in the cycle), and this necessitates waste heat from the system into the cold reservoir. The fact that $e_C$ goes to one in the limit of small ratios $T_C/T_H$ is a consequence of the fact that $Q_C$ is small compared to $Q_H$. It is not a consequence of the fact that entropy generation is small in this case, because entropy generation is already zero for the Carnot cycle.
Let's concentrate first on the interaction between the system and the hot reservoir. An amount $\delta Q_H$ of energy flows into the system from the hot reservoir, which means that the system's entropy changes by $$\mathrm dS_\text{sys} = \frac{\delta Q_H}{T_\text{sys}},$$ and correspondingly, the reservoir's entropy changes by $$\mathrm dS_\text{hot} = -\frac{\delta Q_H}{T_{H}}.$$ It is straight-forward to show then, that the total change in entropy of system plus environment satisfies $$\mathrm dS = \mathrm dS_\text{hot}+\mathrm dS_\text{sys} \geq0,$$ with equality holding if and only if the system and environment exchange energy via heating when they have equal temperatures, $T_\text{sys} = T_H$.
As a consequence, in order to minimize entropy production (and, in fact zero it out completely) during this process, we want $T_\text{sys} = T_H$, and the net change in system entropy during this process can then be written as $$\Delta S_\text{sys} = \int \frac{\delta Q_H}{T_\text{sys}} = \frac{Q_H}{T_{H}},$$ since we are assuming that the temperature of the reservoir doesn't change at all during the cycle.
Now, since the system operates on a thermodynamic cycle, and since the system entropy $S_\text{sys}$ is a state variable (state function/$dS$ is an exact differential, etc.), it must be true that $$\mathrm dS_\text{sys,cycle}=0.$$ Therefore, there must be some other process during which the system expels an amount of energy $Q_C$ to some other reservoir via heating in such a way that the change in system entropy during this new process is the negative of the change in system entropy that we calculated before. By the same argument as above, it must be that this change in entropy is $$\Delta S_2 = -\frac{Q_C}{T_C},$$ where $T_C$ is the temperature of the cold reservoir.
Finally, then, since system entropy is a state variable, $$0 = \Delta S + \Delta S_2 = \frac{Q_H}{T_H}-\frac{Q_C}{T_C}.$$ Another way of looking at this equation is that the net change in entropy of the hot reservoir is negative the net change in entropy of the cold reservoir during the cycle, and hence the net change in entropy of the universe is zero during the cycle.
Now, none of this seemed related to the fact that efficiency goes to 1 as the ratio of $T_C$ to $T_H$ goes to zero. This comes in in the following way. First, the net work output during one cycle is $$W_\text{out} = Q_H-Q_C,$$ and hence the efficiency of the engine that we've just made is $$e = \frac{W_\text{out}}{Q_H} = 1 - \frac{T_C}{T_H},$$ after some algebra. Based on our calculation above, this must be the maximum efficiency of any engine operating between these two temperatures. However, if we change the temperatures, then we can change the efficiency. The reason the efficiency goes up as the temperature ratio goes down is that $W_\text{out}$, being the difference between the heat flows, must go up if, say, we lower $T_C$ (because then $Q_C$ goes down) or if we raise $T_H$ (because then $Q_H$ goes up).
In some sense, this part really doesn't have much to do with entropy at all, because from the thermodynamic perspective, entropy production (which is the increase in entropy of an isolated system) is a measure of how much work we could have done if we had done the process reversibly, but we have already designed the perfect engine operating between those two particular temperatures above, so entropy doesn't have anything else to say.
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).
EDIT:
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.
EDIT 2
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?
Best Answer
A heat pump is not a heat engine. It is a heat engine in reverse. It's Coefficient of Performance (the term used instead of efficiency) is normally greater than 1 and does not violate the laws of thermodynamics. Your first equation is for a heat engine. For the heat pump it is $\frac{Q_H}{W}$.
The figures below show a heat engine and heat pump operating between two temperatures.
A heat engine takes heat $Q_H$ from a higher temperature environment $T_H$, produces output work $W_{OUT}$, and rejects heat $Q_L$ to a lower temperature environment, where
$$W_{OUT}=Q_{H}-Q_{L}$$
The efficiency of a heat engine is the work output divided by the gross heat input, or
$$e=\frac{W_{OUT}}{Q_H}$$
It's efficiency is always less than 1. An efficiency of more than 1 violates the first law of thermodynamics (conservation of energy). The second law limits the maximum efficiency to $1-\frac{T_L}{T_H}$
A heat pump takes heat $Q_L$ from a lower temperature environment, uses work input $W_{IN}$ to move heat $Q_H$ to a higher temperature environment $T_H$. The desired output of the heat pump is the heat $Q_H$ and is
$$Q_{H}= Q_{L}+W_{IN}$$
We don't use the term efficiency for a heat pump, but rather the Coefficient of Performance $COP$. The COP is the desired heat transferred to the higher temperature environment divided by the work in required, or
$$COP=\frac{Q_H}{W_{IN}}$$
The COP is normally greater than 1 and does not violate the laws of thermodynamics.
Hope this helps.