Carnot Engine - Cycle, Formula, FAQs

What is a Carnot Engine?

A Carnot engine is an idealized heat engine that operates reversibly between:

• A hot reservoir at temperature $T_h$
• A cold reservoir at temperature $T_c$

It converts heat into work with maximum theoretical efficiency and consists of four reversible thermodynamic processes-two isothermal and two adiabatic.

Carnot Heat Engine: Working Principle

In a Carnot engine:

• Heat $Q_h$ is absorbed from the hot reservoir
• Part of this heat is converted into work (W)
• Remaining heat $Q_c$ is rejected to the cold reservoir
• After completing one cycle, the system returns to its initial state

Since all processes are reversible, there is no energy loss due to friction or heat leakage.

What is Carnot Cycle ?

The Carnot cycle is an ideal thermodynamic cycle that describes the maximum possible efficiency of a heat engine operating between two fixed temperatures. It consists of four reversible processes:

1. Isothermal expansion
2. Adiabatic expansion
3. Isothermal compression
4. Adiabatic compression

In this cycle, the engine absorbs heat from a hot reservoir, converts part of it into work, and rejects the remaining heat to a cold reservoir. The efficiency of the Carnot cycle depends only on the temperatures of the hot and cold reservoirs, not on the working substance. It provides the upper limit of efficiency for all real heat engines and is a fundamental concept in thermodynamics.

Carnot Cycle Processes

1. Isothermal Expansion (at $T_h$ )

• Gas absorbs heat $Q_h$ from the hot reservoir
• Temperature remains constant
• Gas expands and does work on surroundings

2. Adiabatic Expansion

• No heat exchange with surroundings
• Gas continues to expand
• Temperature falls from $T_h$ to $T_c$

3. Isothermal Compression (at $T_c$ )

• Gas releases heat $Q_c$ to the cold reservoir
• Temperature remains constant
• Surroundings do work on the gas

4. Adiabatic Compression

• No heat exchange
• Gas is compressed
• Temperature rises from $T_c$ back to $T_h$

Efficiency of Carnot Engine Formula (Carnot Engine Efficiency Formula)

It is the most efficient engine possible based on the premise of no accidental wasteful Carnot cycle processes such as friction and no heat conduction between various sections of the engine at different temperatures. The Carnot efficiency is defined as the ratio of energy output to energy intake.

Efficiency of Carnot engine

$\eta=1-T c / T h$
$\eta$: is the thermal efficiency of the engine
Tc: is the sink temperature
Th: is the source temperature

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Applications of Carnot Engine

• The Carnot engine gives the maximum possible efficiency of a heat engine working between two given temperatures.
• It is used as a standard of comparison to check the efficiency of real heat engines.
• The concept of the Carnot engine helps in the design and improvement of heat engines such as steam engines and internal combustion engines.
• The reverse Carnot cycle is used to explain the working of refrigerators, air conditioners, and heat pumps.
• It helps in the clear understanding of the second law of thermodynamics and the concept of entropy.

Solved Examples Based on Carnot Engine

Example 1: A Carnot engine takes $3 \times 10^6$ cal of heat from a reservoir at $627^{\circ} \mathrm{C}$, and gives it to a sink at $27^{\circ} \mathrm{C}$. The work done by the engine is

1) $4.2 \times 10^6 \mathrm{~J}$
2) $8.4 \times 10^6 \mathrm{~J}$
3) $16.8 \times 10^6 \mathrm{~J}$
4) zero

Solution:

Efficiency of a cyclic process

$\eta=\frac{\text { work done per cyclic }}{\text { gross heat supplied per cyclic }}$

wherein

Gross heat implied only part of heat absorbed.

The efficiency of a Carnot cycle
$
\eta=\frac{W}{Q_1}=1-\frac{T_2}{T_1}
$

$T_1$ and $T_2$ are in kelvin
wherein
$T_1=$ Source temperature
$T_2=$ Sink Temperature

$
\left(T_1>T_2\right)
$
Given,

$
\begin{aligned}
& \eta=1-\frac{T_2}{T_1}=\frac{W}{Q_1} \\
& =1-\frac{300}{900}=\frac{W}{3 \times 10^6 \mathrm{cal}} \\
& \frac{\mathrm{W}}{\text { or } 3 \times 10^6 \mathrm{cal}}=\frac{2}{3} \\
& \text { or } \mathrm{W}=2 \times 10^6 \mathrm{kcal} \\
& =8.4 \times 10^6 \mathrm{Joule}
\end{aligned}
$

Hence, the answer is the option (2).

Example 2: Even the Carnot engine cannot give 100% efficiency because we cannot

1) prevent radiation

2) find ideal sources

3) reach absolute zero temperature

4) eliminate friction.

Solution:

The efficiency of a Carnot cycle
$
\eta=\frac{W}{Q_1}=1-\frac{T_2}{T_1}
$

$T_1$ and $T_2$ are in kelvin
wherein
$T_1=$ Source temperature
$T_2=$ Sink Temperature

$
\left(T_1>T_2\right)
$
For Carnot Cycle

Efficiency

$
\eta=1-\frac{T 2}{T 1}
$
Efficiency will be $100 \%$ only if $\mathrm{T}_2=\mathrm{OK}$

But this is not practically possible.

Hence the answer is the option (3).

Example 3: Two Carnot engines $A$ and $B$ are operated in series. Engine $A$ receives heat from a reservoir at 600 K and rejects heat to a reservoir at temperature $T$. Engine $B$ receives heat rejected by engine $A$ and $\underline{\eta_B}$
in turn rejects it to a reservoir at 100 K . If the efficiencies of the two engines $A$ and $B$ are represented by $\eta A$ and $\eta B$, respectively, then what is the value of $\eta_A$ ? (Work output is same for both)

1) $\frac{12}{7}$
2) $\frac{7}{12}$
3) $\frac{12}{5}$
4) $\frac{5}{12}$

Solution:

Efficiency of a Carnot cycle
$
\eta=\frac{W}{Q_1}=1-\frac{T_2}{T_1}
$

$T_1$ and $T_2$ are in kelvin wherein
$T_1=$ Source temperature
$T_2=$ Sink Temperature

$
\left(T_1>T_2\right)
$

$
\begin{aligned}
& \eta_A=1-\frac{T}{600}=\frac{w}{Q_1} \rightarrow(1) \\
& \eta_B=1-\frac{100}{T}=\frac{w}{Q_2}=\frac{w}{Q_1-w} \\
& \eta_B=1-\frac{100}{T}=\frac{1}{\frac{Q_1}{w}-1}
\end{aligned}
$

from eqn1:

$
\begin{aligned}
& \eta_B=1-\frac{100}{T}=\frac{1}{\frac{1}{1-\frac{T}{600}}-1}=\frac{1-\frac{T}{600}}{\frac{T}{600}} \\
& \eta_B=\frac{600}{T}-1=1-\frac{100}{T} \text { or } \frac{700}{T}=2 \text { or } T=350 \mathrm{~K} \\
& \frac{\eta_B}{\eta_A}=\frac{1-\frac{100}{T}}{1-\frac{T}{600}}=\frac{12}{7}
\end{aligned}
$

Hence the answer is the option (1).

Example 4: Three Carnot engines operate in series between a heat source at a temperature $T_1$ and a heat sink at a temperature $T_4$ (see figure). There are two other reservoirs at temperature $T_2$ and $T_3$, as shown, with $T_1>T_2>T_3>T_4$. The three engines are equally efficient if:

1) $
\mathrm{T}_2=\left(\mathrm{T}_1 \mathrm{~T}_4\right)^{1 / 2} ; \mathrm{T}_3=\left(\mathrm{T}_1^2 \mathrm{~T}_4\right)^{1 / 3}
$

2) $
\mathrm{T}_2=\left(\mathrm{T}_1^2 \mathrm{~T}_4\right)^{1 / 3} ; \mathrm{T}_3=\left(\mathrm{T}_1 \mathrm{~T}_4^2\right)^{1 / 3}
$

3) $\mathrm{T}_2=\left(\mathrm{T}_1 \mathrm{~T}_4^2\right)^{1 / 3} ; \mathrm{T}_3=\left(\mathrm{T}_1^2 \mathrm{~T}_4\right)^{1 / 3}$

4) $\mathrm{T}_2=\left(\mathrm{T}_1^3 \mathrm{~T}_4\right)^{1 / 4} ; \mathrm{T}_3=\left(\mathrm{T}_1 \mathrm{~T}_4^3\right)^{1 / 4}$

Solution:

Efficiency of a Carnot cycle
$
\eta=\frac{W}{Q_1}=1-\frac{T_2}{T_1}
$

$T_1$ and $T_2$ are in kelvin wherein
$T_1=$ Source temperature
$T_2=$ Sink Temperature
$\left(T_1>T_2\right)$

$
\begin{gathered}
\varepsilon_1=1-\frac{T_2}{T_1} \\
\varepsilon_2=1-\frac{T_3}{T_2} \\
\varepsilon_3=1-\frac{T_4}{T_3} \\
\varepsilon_1=\varepsilon_2=\varepsilon_3 \\
\Rightarrow 1-\frac{T_2}{T_1}=1-\frac{T_3}{T_2}=1-\frac{T_4}{T_3}
\end{gathered}
$
$
\begin{aligned}
& \frac{T_2}{T_1}=\frac{T_3}{T_2}=\frac{T_4}{T_3} \\
& \Rightarrow \frac{T_2}{T_1}=\frac{T_3}{T_2} \\
& \Rightarrow T_2=\sqrt{T_1 T_3} \cdots \cdots(1) \\
& \text { SIMILARLY } \\
& T_3=\sqrt{T_2 T_4} \cdots \cdots(2) \text { PUT IN } \\
& T_2=\sqrt{T_1 T_3} \\
& T_{2=}=\sqrt{T_1 \sqrt{T_2 T_4}} \\
& T_2=T_1^{\frac{2}{3}} T_4^{\frac{1}{3}}
\end{aligned}
$

Hence the answer is the option (2).

Example 5: A Carnot engine has an efficiency of $1 / 6$. When the temperature of the sink is reduced $62^{\circ} \mathrm{C}$, its efficiency is doubled. The temperatures of the source and the sink are, respectively,

1) $62^{\circ} \mathrm{C}, 124^{\circ} \mathrm{C}$
2) $99^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}$
3) $124^{\circ} \mathrm{C}, 62^{\circ} \mathrm{C}$
4) $37^{\circ} \mathrm{C}, 99^{\circ} \mathrm{C}$

Solution:

The efficiency of a Carnot cycle

$\begin{aligned} & \eta=\frac{W}{Q_1}=1-\frac{T_2}{T_1} \\ & T_1 \text { and } T_2 \text { are in kelvin } \\ & \text { wherein } \\ & T_1=\text { Source temperature } \\ & T_2=\text { sink Temperature } \\ & \left(T_1>T_2\right) \\ & \text { given } \\ & \frac{1}{6}=1-\frac{T \sin k}{T \operatorname{sourse}} \\ & \frac{T \sin k}{T \operatorname{sourse}}=\frac{5}{6} \cdots \text { (1) }\end{aligned}$

also
$
\begin{aligned}
& \frac{2}{6}=1-\frac{T \sin k-62}{T \text { sourse }} \\
& \frac{1}{3}=1-\frac{5}{6}+\frac{62}{\text { Tsourse }} \\
& \frac{1}{6}=\frac{62}{\text { Tsourse }} \\
& \text { Tsourse }=372 \mathrm{~K}=99^{\circ} \mathrm{C} \\
& \text { Tsin } k=\frac{5}{6} \times 372=310 \mathrm{~K}=37^{\circ} \mathrm{C}
\end{aligned}
$

Hence, the answer is the option (2).