The automobiles such as a motorcycle, car, lorry, bus, train etc. use heat engines. Heat engines are the devices that convert heat energy into kinetic energy. Some more examples of heat engines are are gas turbine, steam engine, jet engine and rocket engine. Heat engine is a device that convert heat continuously into mechanical work. There are three essential part of any heat engine Working substance: The material, which on being given heat, performs mechanical work, is known as the working substance. Source: A hot body at a high fixed temperature $T_{1}$ from which the heat engin can draw heat is known as the source. Sink: A cold body at a fixed lower temperature $T_{2}$, to which any amount of heat can be rejected, is known as the sink. In a heat engine, the working substances in takes the heat($Q_{1}$) from the source and the part of it will converted into external work ($W$), the rest ($Q_{2}$) gives to the sink and returns to its initial states. This serious of operations constitutes a cycle. The work can be continuously obtained by performing the same cycle over and over again.

Heat Engine Efficiency is defined as the the ratio of the net work obtained in the circle (output) to the heat absorbed by the working substance from the source (input) expressed in the same units.
Thus efficiency,
$\eta$ = $\frac{Work\ output}{Heat\ input}$ = $\frac{W}{Q_{1}}$ = $\frac{Q_{1}Q_{2}}{Q_{1}}$
$\eta$ = 1  $\frac{Q_{2}}{Q_{1}}$
It is clear from this expression that $\eta$ will be unity for 100% of efficiency, that is $Q_{2}$ is zero or, in other words, if an engine could be built to operate in such a way that no heat is at all rejected by the working substances in a cycle, there will be 100% conversion of heat into work. This, however, is not possible in practice.
Reversible process : A reversible process is the one which can be retraced in opposite direction in such a way that all changes taking place in the direct process are exactly repeated in the opposite order and inverse sense, and no changes are left in any of the bodies taking part in the process or in the surroundings. There is no wastage of energy at all in a reversible process. These conditions are never realized in practice. Hence a reversible process is only an ideal conception.
Irreversible Process : Any process which is not reversible exactly is an irreversible process. It can not be retraced in the opposite order by reversing the controlling factors.
Heat engines are broadly classified into two,
 Internal Combustion Engines
 External Combustion Engines
Internal Combustion Engines : In this engine, the combustion of the fuel in presents of air takes place inside the the engine cylinder or the product of combination enter into the cylinder as a working fluid. The examples are petrol and diesel engines.
External Combustion Engines : In this engine, the combustion of the fuel in presents of air takes place outside the engine cylinder. Steam engines is one of the example for this. Steam is a working fluid , which enter in the cylinder of a steam engine to perform mechanical work. Here the combination of fuel do not enter into the engine cylinder and hence they do not form the working fluid. We have more examples of External Combustion Engines, they are steam turbine, hot air engine and closed cycle gas turbine.
Normally a heat engine operates in a cyclic manner. This engine uses the energy in the form of heat. Using the first and second law of thermodynamics, heat engine works. First law tell about the conservation of energy whereas, second law helps to determine the efficiency and energy flow direction. The illustration of the heat engine cycle is given below:
As the condition of reversibility cannot be realised in practice, a reversible heat engine is only an ideal conception. Sadi Carnot conceived an ideal theoretical engine free from all imperfections of actual engines. Though it could not be realised in practice, Carnot engine is taken as standard against which the performance of actual engines is judged.
A reversible cycle is an ideal cycle in which all the processes in the cycle are reversible. The carnot engine is an good example for this cycle. The cycle consists of two isothermal processes, ab and cd, together with two adiabatic processes, bc and da. No practical engine was ever built to run on this cycle.
The sequence of processes in the carnot cycle is given by,
 Reversible isothermal heat absorption
 Reversible adiabatic work done
 Reversible isothermal heat rejection
 Reversible adiabatic work supply
An engine which executes carnot cycle is called as the carnot heat engine. It is important because it is composed of reversible processes, which are the most thermodynamically efficient processes, so its thermal efficiency establishes the maximum thermal efficiency possible within the temperature limits of the cycle. By calculating the thermal efficiency of this cycle it is possible to establish the maximum possible efficiency between the temperature limits taken.
The following are the examples of heat engine. The efficiency of the Carnot's engine is given as,
$\eta$ = 1  $\frac{Q_{2}}{Q_{1}}$ = 1  $\frac{T_{2}}{T_{1}}$
 Refrigerator
 Heat pump
 Diesel engine
 Gasoline engine
 Carnot engine
Solved Examples
Question 1: Find the efficiency of the Carnot’s engine working between the steam point and the ice point.
Solution:
Given,
$T_{1}$ = 273+100 = 373 K
$T_{2}$ = 273+0 = 273 K
We have
$\eta$ =1 $\frac{T_{2}}{T_{1}}$
= 1 $\frac{273}{373}$ = 10.7319=0.2681
% efficiency = 0.2681 $\times$ 100
$\eta$ = 26.81%
Solution:
Given,
$T_{1}$ = 273+100 = 373 K
$T_{2}$ = 273+0 = 273 K
We have
$\eta$ =1 $\frac{T_{2}}{T_{1}}$
= 1 $\frac{273}{373}$ = 10.7319=0.2681
% efficiency = 0.2681 $\times$ 100
$\eta$ = 26.81%
Question 2: Find the efficiency of a Carnot engine working between 127°C and 27°C?
Solution:
Given,
$T_{1}$ = 273+127 =400 K
$T_{2}$= 273+27=300 K
We have
$\eta$ =1 $\frac{T_{2}}{T_{1}}$
= 1 $\frac{300}{400}$ = 10.75=0.25
% efficiency = 0.25 $\times$ 100
$\eta$ = 25%
Solution:
Given,
$T_{1}$ = 273+127 =400 K
$T_{2}$= 273+27=300 K
We have
$\eta$ =1 $\frac{T_{2}}{T_{1}}$
= 1 $\frac{300}{400}$ = 10.75=0.25
% efficiency = 0.25 $\times$ 100
$\eta$ = 25%