Here is the paradoxical situation.

Suppose you have two objects at temperatures $T_1$ and $T_2$, where $T_2>T_1$. Also, suppose you have some work at your disposal to power a heat pump whose efficiency is maximum. Your goal is to heat up the hotter object (initially at temperature $T_2$) using the energy extracted from the colder object (initially at temperature $T_1$), sort of like a refrigerator. Of course, you will extract heat from the colder object (seems weird, I know) by using the heat pump.

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So let's calculate the differential change in energy of the hotter object $dE_2$ upon using a small amount of work $dW$ to power the heat pump. Since the heat pump is maximally efficient, it will transfer heat to the hotter object adiabatically. This can be written as


$$\implies dQ_1=-\frac{T_1}{T_2}dQ_2$$

On the other hand, we have for the small amount of heat $dQ_2$ added to the hotter object

$$\begin{align*} dW&=dQ_1+dQ_2\\ &=\left(1-\frac{T_1}{T_2}\right)dQ_2\\ &\\ \implies dQ_2&=\left(1-\frac{T_1}{T_2}\right)^{-1}dW\\ &=\eta_{\textrm{(efficiency)}}dW \end{align*}$$

There is the paradox! The "efficiency", $\eta_{\textrm{(efficiency)}}$ of this heat pump is greater than one. You can add more energy to an object than work you put in by using a colder object!

Why this seems paradoxical to me is because this would imply the following: if you have some energy $W$, you could heat up a pot more if you use some ice cubes and a Carnot engine than if you give the energy directly to the pot. How could ice cubes help you heat a pot better?!


2 Answers 2


It may sound paradoxical, but it is true. A good way to think of the efficiency of heat engines is via the concept of entropy equal to $\displaystyle{\frac{dQ}{T}}$, as you are already doing. If you just turn $dW$ into $dQ$ and dump it into $T_2$, the entropy increases by $\displaystyle{\frac{dW}{T_2}}$. Since entropy increases, this is something that will happen spontaneously-- it's allowed.

But it's also kind of a waste, because the universe doesn't require the entropy to go up that much, it can just go up infinitesimally and the process will still happen spontaneously. So dumping work in to heat the pot is in some sense strong-arming the situation, it's more impetus than you need. Using the ice cubes allow you to take heat $dQ_1$ from somewhere else, such that $\displaystyle{\frac{dQ_1}{T_1}}$ equals $\displaystyle{\frac{(dW+dQ_1)}{T_2}}$.

You are "ahead of the game" by $\displaystyle{\frac{dW}{T_1}}$, and you start "losing ground" by $\displaystyle{dQ_1\left(\frac{1}{T_1} - \frac{1}{T_2}\right)}$ as you add in heat $dQ_1$. You could say the universe "owes you" that much entropy if you are going to drop $dW$ into $T_2$, and you can "cash that in" by extracting as much heat from something as warm as possible. So you don't do the best job by using ice cubes-- you would do even better using something very hot! But ice cubes are better than nothing if that's all you have access to.

  • $\begingroup$ please, use mathjax $\endgroup$
    – hyportnex
    Sep 19, 2016 at 5:27

In the absence of body 1 or any other energy source, increase in thermal energy of body 2 is just equal to what work is wasted on it ($dQ_2=dW$). If some other energy source is available, such as body 1 at a finite temperature, then you can also extract some energy from body 1 (by using heat pump) which is ultimately dumped into body 2, so obviously body 2 will gain more thermal energy than was the case when only work was being done. This is true whatever the temperature of body 1. So where is the paradox?

Let me give you an analogy. Say a container is being filled with water that is falling from some height. Now you could insert a turbine in between to extract work out of this falling stream of water (which wouldn't change the amount of water contributed to the container by the falling stream), and use that work to run a pump that pumps some additional water into the container from some water-source at a lower altitude. No paradox.


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