# Need of the cold body in heat engine

Purpose of heat engine is to transfer heat into the useful work, basically. Working object either gas or liquid takes heat from a thermal energy reservoir as a result of which their internal energy increases and this energy is somehow used to do work. Why would we need a cold body then?

• – user36790 Nov 13 '16 at 17:28
• It is important to note, that the 2nd Law of thermodynamics plays no fundamental role in answering this question; we need a heat sink because the entropy is a state function, and at the end of the reversible process (which is visualized through the Carnot cycle diagram relevant for this problem), the entropy value of the system must return to the value it had originally. – Aritro Pathak Nov 13 '16 at 18:18

The heat engine is assumed to be brought back to it's original thermodynamic state, after each cycle of operation. That is, in particular, the operation is reversible, with the net change in entropy of the engine being 0.

The net change in internal energy $U$ and entropy $S$ have to be $0$, after a cycle of operation. The net energy change is indeed zero, considering that the work done is equivalent to the net heat taken in. For the entropy change to be zero, it has to get rid of some of the heat it took in, in the form of heat itself.

In order to maximize the amount of work done, it may expel heat at a very low temperature, so it has to give out as little heat as possible; but give out some heat it must, to the cold body.

• +1 very nice succinct summary of the reason. Maybe if you have time you could rewrite the Wikipedia page, which in my opinion, is very unclear on entropy en.wikipedia.org/wiki/Entropy_(energy_dispersal) – user108787 Nov 13 '16 at 19:36
• Thank you. There should be many qualified people who can contribute and make sure this Wikipedia page is improved. I had made specific contributions to the articles on Carnot Engine and the Clausius inequality. – Aritro Pathak Nov 13 '16 at 19:55
• @AritroPathak you have written in second para, " The net energy change is indeed zero, considering that the work done is equivalent to the net heat taken in. For the entropy change to be zero, it has to get rid of some of the heat it took in, in the form of heat itself." If the net heat energy is equivalent to the work done, What heat is that which is rejected? – user104909 Nov 15 '16 at 12:18
• The net heat taken in, in one cycle is $\delta Q= Q_{in}-Q_{out}$, where $Q_{in}$ is the heat the engine takes in from the hot reservoir, and $Q_{out}$ is the heat given out to the lower temperature. The work done $W$ by the engine, on the working system is $W=\delta Q$. The net entropy change being 0, we require $\frac{Q_{in}}{T_{high}}=\frac{Q_{out}}{T_{low}}$. The net energy change is $W-\delta Q=0$, and $Q_{out}$ is the heat that is rejected. – Aritro Pathak Nov 15 '16 at 12:54

The cold reservoir provides a heat sink for the excess heat that is rejected after some of the heat is converted into work. We know from the second law of thermodynamics that 100% conversion of heat to work is impossible hence a colder sink is necessary to allow a heat engine to work in the first place.

Entropy and the need to minimise its creation is the reason for the cold reservoir, $R_c$.

We need a heat sink because the entropy is a state function, and at the end of the reversible process (which is visualized through the Carnot cycle diagram relevant for this problem), the entropy value of the system must return to the value it had originally. Thanks to  Aritro Pathak for this

Since the state of the engine must be unchanged at the end of a cycle, the entropy it expels must be at least as much as the entropy it absorbs. It expels entropy, through its working fluid, usually a gas, to $R_c$.

The efficiency of a Carnot engine, for example is $(1 -\frac {R_c}{R_h})$, which is as high as you can get, but at the cost of a very "slow" engine. If you fitted it to a car, people could easily walk past you, even at your top speed.

Every engine utilises a working fluid, that absorbs heat from the hot reservoir,$R_h$, expels waste heat and does useful work. In say, a Carnot engine, the working fluid that operates the piston in the "middle" of the engine, needs a source of heat. As it absorbs heat from $R_h$, the entropy of $R_h$ decreases, while the entropy of the working fluid increases. We need to keep the working fluid at a slightly lower temperature than $R_h$, as otherwise there will be no energy transfer between two objects at the same temperature. We keep the gas at this temperature by letting it expand isothermally as it absorbs the energy. Image Source: NASA Carnot Engine

During the part of the cycle where the working fluid is in contact with the cold reservoir, $R_c$, we would like it's temperature to be just slightly higher than that of $R_c$, so to avoid creating any new entropy. As the heat leaves the gas, we want to compress it isothermally, to maintain its temperature.

• "Since the state of the engine must be unchanged at the end of a cycle, the entropy it expels must be at least as much as the entropy it absorbs." : rather, because it is a cyclic process, we must have that the amount of "entropy expelled" is exactly equal to the amount of "entropy gained". The 2nd Law of thermodynamics plays no fundamental role in this question; we need a heat sink because the process is cyclic,and the engine has to ideally return to it's original state at the end of the cycle. The 2nd law of thermodynamics only says that we may not exactly reach precisely the original state. – Aritro Pathak Nov 13 '16 at 18:16