# The Second Law of Thermodynamics and Carnot Engine

Considering you have read the article of the first law of thermodynamics, you saw that the heat can be converted into work and vice versa. There are different statements of the second law of thermodynamics. We start with also known as engine statement (or Kelvin-Plank form) of the second law of thermodynamics. In terms of the conversion of heat into mechanical energy (Kelvin-Plank form) the second law of thermodynamics limits the conversion of heat into mechanical energy. It's a simple but powerful statement of the law of nature about the limitation that no heat engine is capable of converting all heat provided completely into mechanical energy.

The heat supplied to a thermodynamic system is also used to increase the internal energy of the system. The internal energy of the system does not depend on the path of the thermodynamic system but depends on the current state of the system. Take some pieces of paper in your hand and throw them in any direction. They are not ordered in a particular place on the floor instead scattered everywhere which represents the randomness or disorder of the event. All thermodynamic processes always have a tendency towards randomness or disorder of the system. We'll define the term entropy based on the randomness or disorder of a system and define the second law in terms of entropy later. Different statements of the second law ultimately mean the same thing.

## Engine Statement of the Second Law of Thermodynamics

To begin we define the engine statement of the second law of thermodynamics based on its limitation to convert heat into mechanical work. It states that no heat engine can convert the absorbed heat from a hot reservoir completely into mechanical work and come back to the original state in which it began. This statement is also called Kelvin-Plank statement of the second law of thermodynamics.

The Figure 1 shows an energy flow diagram where the engine takes heat $Q_\text{h}$ from the hot reservoir at temperature $T_\text{h}$ and converts some amount of the absorbed heat into work $W$ and some amount of the absorbed heat $Q_\text{c}$ is thrown to the cold reservoir. It means, the total heat $Q_\text{h}$ absorbed is the sum of the work done by the system and the heat thrown to the cold reservoir that is

${Q_h} = W - {Q_c} = W + \left| {{Q_{\rm{c}}}} \right| \tag{1} \label{1}$

Note that the heat thrown to the cold reservoir flows out of the system of the engine and it is negative. Therefore, $Q_\text{c}$ is already negative and negative sign with $Q_\text{c}$ is used to correct the above expression or you can add the absolute value of $Q_\text{c}$ to work done. The ratio of work done by the engine $W$ to the heat absorbed $Q_\text{h}$ is called the efficiency of the engine denoted by $e$. The efficiency $e$ is always less than unity. From Eq. \eqref{1}, $W = {Q_h} + {Q_c}$ and the efficiency is

$e = \frac{W}{{{Q_{\rm{h}}}}} = \frac{{{Q_h} + {Q_c}}}{{{Q_h}}} = 1 + \frac{{{Q_c}}}{{{Q_h}}} \tag{2} \label{2}$

## Carnot Engine

The engine statement of the second law of thermodynamics tells us that no heat engine can give $100\%$ efficiency. But we have the Carnot engine which gives the maximum efficiency possible for any two given temperatures $T_h$ and $T_c$ in agreement with the second law of thermodynamics. Here we discuss a very useful ideal engine called Carnot engine and talk about the processes involved in its cycle called Carnot cycle. Carnot engine has the maximum efficiency possible for any two given temperatures consistent with the second law of thermodynamics. To understand the operation involved in the Carnot engine we talk about the processes involved in the Carnot cycle. Carnot cycle is based on reversibility or it is a reversible cycle. It means it does not have any irreversible processes in the cycle. The reversibility of Carnot cycle makes it the maximum efficiency cycle. And reversibility is an idealization and therefore it is also an ideal cycle. A Carnot cycle involves two processes which are isothermal reversible process and adiabatic reversible process. Both isothermal and adiabatic processes in Carnot cycle are reversible, so the Carnot cycle is reversible.

The Figure 2 shows a Carnot cycle having reversible isothermal and adiabatic processes. Consider that the system of Carnot cycle contains ideal gas. We will find the efficiency of the Carnot cycle in agreement with the second law of thermodynamics. In Figure 2 the process that takes place form state $A$ to state $B$ is isothermal expansion and that takes place from $C$ to $D$ is isothermal compression. Similarly the process that takes place from state $B$ to state $C$ is adiabatic expansion and that takes place from state $D$ back to state $A$ is adiabatic compression. In isothermal expansion the system takes heat $Q_h$ from hot reservoir initially at temperature $T_h$. In adiabatic expansion the system does work spending its internal energy and the temperature of the system decreases. In isothermal compression, work is done on the system and temperature $Q_c$ is rejected or thrown in the cold reservoir initially at temperature $T_c$. Finally in adiabatic compression the work is done on the system and temperature of the system increases back to $T_h$.

In isothermal expansion from state $A$ to $B$, the volume of the system increases from $V_A$ to $V_B$ and work done $W_{AB}$ in isothermal expansion is

$W_{AB} = \int_{{V_A}}^{{V_B}} {pdV} = \int_{{V_A}}^{{V_B}} {\frac{{nRT_h}}{V}dV} = nRT_h \ln \frac{{{V_B}}}{{{V_A}}} \tag{3} \label{3}$

In above integration we replaced $p$ by the expression of $p$ from ideal gas equation. In isothermal expansion of an ideal gas the change in internal energy is zero, that is, $\Delta U = 0$, therefore according to the first law of thermodynamics we have $Q_h = W_{AB}$, so

${Q_h} = {W_{AB}} = nRT_h\ln \frac{{{V_B}}}{{{V_A}}} \tag{4} \label{4}$

Similarly if $V_C$ and $V_D$ are the volumes at state $C$ and $D$ respectively, the work done on the isothermal compression is

${Q_c} = {W_{CD}} = nRT_c\ln \frac{{{V_D}}}{{{V_C}}} = -nRT_c\ln \frac{{{V_C}}}{{{V_D}}} \tag{5} \label{5}$

As $V_D$ is less than $V_C$ in isothermal compression $Q_c$ is negative. Dividing Eq. \eqref{5} by Eq. \eqref{4}:

$\frac{{{Q_c}}}{{{Q_h}}} = -\frac{{{T_c}}}{{{T_h}}}\frac{{\ln \frac{{{V_C}}}{{{V_D}}}}}{{\ln \frac{{{V_B}}}{{{V_A}}}}} \tag{6} \label{6}$

In adiabatic process from state $B$ to $C$ and from $D$ to $A$, we know from the previous article that, ${T_B}V_B^{\gamma - 1} = {T_C}V_C^{\gamma - 1}$ and ${T_D}V_D^{\gamma - 1} = {T_A}V_A^{\gamma - 1}$. Note that $T_A$, $T_B$, $T_C$ and $T_D$ are the temperatures of the corresponding states represented by the subscripts. But $T_A = T_h$, $T_B = T_h$, $T_C = T_c$ and $T_D = T_c$ and so,

${T_h}V_B^{\gamma - 1} = {T_c}V_C^{\gamma - 1} \tag{7} \label{7}$

${T_h}V_A^{\gamma - 1} = {T_c}V_D^{\gamma - 1} \tag{8} \label{8}$

Now dividing Eq. \eqref{7} by Eq. \eqref{8}, you'll get

\begin{align*} {\left( {\frac{{{V_B}}}{{{V_A}}}} \right)^{\gamma - 1}} &= {\left( {\frac{{{V_{C}}}}{{{V_D}}}} \right)^{\gamma - 1}}\\ {\rm{or,}}\quad \frac{{{V_B}}}{{{V_A}}} &= \frac{{{V_C}}}{{{V_D}}} \tag{9} \label{9} \end{align*}

Using Eq. \eqref{9}, Eq. \eqref{6} becomes

$\frac{{{Q_c}}}{{{Q_h}}} = - \frac{{{T_c}}}{{{T_h}}} \tag{10} \label{10}$

You can find the efficiency of the Carnot engine using Eqs. \eqref{2} and \eqref{10}, that is the efficiency $e$ is

$e = 1 - \frac{{{T_c}}}{{{T_h}}} \tag{11} \label{11}$

The above equation of the efficiency of the Carnot engine depends only on the temperatures of hot and cold reservoirs. The efficiency will be $1$ if $T_c = 0\rm{K}$ but practically this is impossible. The efficiency of the Carnot engine can be increased by making the ratio $T_c/T_h$ smaller which can be achieved by increasing $T_h$ and decreasing $T_c$. As already noted a Carnot engine is reversible, therefore it is the most efficient engine than any other real engines operating between the same two temperatures.

To prove that a Carnot engine is the most efficient engine than any other real engine operating between the same two temperatures, first you should know what a refrigerator is.

## Refrigerators

What a refrigerator does is the opposite of what a heat engine does. Therefore, a refrigerator takes in heat $Q_c$ from cold reservoir and throws heat $Q_h$ to hot reservoir. The work $W$ should be done to take the heat $Q_h$ to the hot reservoir. All natural processes are irreversible processes so the heat flow from a hotter region to a colder region across a finite temperature difference is an irreversible process (as already noted that a reversible process is an idealization that is takes place in infinitesimal steps). And some work must be done to take the heat from the cold reservoir to the hot reservoir. The Figure 3 shows the energy flow diagram of a refrigerator. The energy relationship for a refrigerator is

${Q_h} = W - {Q_c} \tag{12} \label{12}$

Note that the heat $Q_c$ flows into the system so it is positive, and heat $Q_h$ flows out of the system and it is negative. The work $W$ is done on the system and therefore it is negative. Considering the sign conventions the above equation for the energy relationship of a refrigerator is valid. You can also understand the above equation in terms of absolute values, that is $\left| {{Q_h}} \right| = \left| W \right| + \left| {{Q_c}} \right|$.

The ratio $Q_c/W$ or the ratio of heat taken from the cold reservoir to the mechanical work done by external source is called the coefficient of performance or COP of a refrigerator. The same idea for the efficiency of a heat engine and the coefficient of performance of a refrigerator is to find the ratio of energy gained (what you gain) to expenditure of energy (what you lose).

${\rm{CO}}{{\rm{P}}_r} = \frac{{{Q_c}}}{{\left| W \right|}} \tag{13} \label{13}$

The subscript $r$ in the above equation represents refrigerator and the Eq. \eqref{13} gives the coefficient of performance of the refrigerator.

Now we continue to prove that a Carnot engine is the most efficient engine than any other real heat engine operating between the same two temperatures. To prove this consider that a real engine has efficiency $e$ greater than the efficiency of a Carnot engine $e_c$, that is $e > e_c$. We temporarily call the heat engine having greater efficiency than the efficiency of the Carnot engine as super engine. The Carnot engine takes heat $Q_h$ from the hot reservoir, does work $W$ and rejects heat $Q_c$ to the cold reservoir. The Eq. \eqref{2} suggests that the super engine takes greater amount of heat than the Carnot engine for the same heat $Q_c$ rejected by both engines. Therefore, for example, the super engine takes heat $Q_h + Q'$ from the hot reservoir and rejects heat $Q_c$ to the cold reservoir and does greater amount of work $W + W'$. If $W$ is the work done by the super engine taking heat $Q_h$, then $W + W'$ is the work done by the engine taking heat $Q_h + Q'$. You know that a Carnot refrigerator is reversible and therefore it can be reversed which does the opposite of what a Carnot engine does, that is, takes heat $Q_c$ from the cold reservoir and throws heat $Q_h$ to the hot reservoir but you need do work $W$ from external source to run a Carnot refrigerator. The interesting thing is that if you couple the super engine and the Carnot refrigerator together to form a single machine, the net output of the coupled engine has $100\%$ efficiency ($Q_c = 0$). It means that there is no loss of energy and you don't need to add energy to run this machine; it always runs after it's started. In other words the work $W$ done by the super engine is used to run Carnot refrigerator and so the extra heat $Q'$ the super engine takes is converted completely into mechanical work $W'$. The output of the coupled engine violets the second law of thermodynamics and therefore no real engine can have efficiency greater than a Carnot engine operating between the same two temperatures.

Thermodynamics
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