Carnot Cycle, Efficiency, PV, TS diagram, Theorem, Derivation

In this article, we were discussing the Carnot cycle, Carnot cycle efficiency, and other related terms for the Carnot heat engine. 

This article is quite long, but you should have a complete understanding of the terms which were used later in the Carnot cycle.

Let us begin…..

Terms used in Carnot cycle – A look-up

1. Laws of thermodynamics that are involved in the Carnot cycle.

2. What are the Spontaneous and Non-spontaneous processes?

3. What are the Reversible and irreversible processes?

4. what is the heat engine?

5. Heat engine efficiency.

6. What is a Carnot heat engine and a Carnot cycle?

7. Complete Carnot cycle processes with diagrams and derivations.

8. Carnot cycle PV diagram.

9. Carnot heat engine efficiency with examples.

10. Carnot theorem.

11. The refrigerator as a reversed Carnot cycle.

12. Co-efficient of performance.

13. Entropy and Carnot cycle.

14. Carnot cycle TS diagram.

As you know, thermodynamics is the branch of science which deals with the study of the conversion of heat energy into other forms of energy and gives information about this conversion quantitatively.

Isothermal process – Carnot Cycle 

In thermodynamics, if the temperature change in a thermodynamics process is constant, then the process is called an isothermal process.

As in Carnot heat engine, ideal gas as a working substance is used, then for an ideal gas isothermal process,

PV = Constant

Here P is the pressure of the gas,

V is the volume of the gas and,

the temperature T is constant.

Adiabatic process

In a thermodynamical system, no heat transfer from the system to the surrounding or from surrounding to the system means all heat contained in the system remains constant.

If any work is done in the adiabatic process, then this can only happen when the system’s internal energy (U) changed.

For an ideal gas,

P(V)^γ = Constant 

Here Gamma (γ)is the ratio of two specific heats of the gas at constant volume and constant pressure.

Laws of thermodynamics

Carnot heat engine obeys the first law and also the second law of thermodynamics.

While the Carnot cycle laid the foundation of the second law of thermodynamics.

let us discuss these laws.

The first law of thermodynamics

According to the first law of thermodynamics, if a certain amount of work is done, an equivalent amount of heat is produced and if a certain amount of heat is absorbed, an equivalent amount of work is done.

It means W = Q or Q = W

Q is the heat and W is the work done.

The First law of thermodynamics is also the law of conservations of energy.

Law of conservation of energy

Energy can neither be created nor be destroyed.

On behalf of this law, you can say that whole energy in the universe is conserved.

These conversions of energies are interconvertible.

The second law of thermodynamics

Many statements are given by scientists for the 2nd law of thermodynamics. Such as the statement by Lord Kelvin, Clausius, Planck-Kelvin, etc.

In thermodynamics, it was seen later that, all these statements are equivalent to each other.

Kelvin Planck’s statement for the second law of thermodynamics

It is impossible to construct a heat engine operating in cycles, which extract heat from a single thermal reservoir and converts the whole of it into work.

The second law is the most fundamental law of nature.

The limitations of the first law encouraged scientists to find the second law of thermodynamics.

As you know the first law only tells us the conversion of energies into other forms of energy.

But this law did not give information about the quantity and direction-flow of conversions and many other reverse parameters.

You can say the second law of thermodynamics is the law of nature.

Spontaneous and Non-spontaneous

Before you moved to the Carnot cycle, you should know about the spontaneous and non-spontaneous processes?

Carnot heat engine and all other heat engines follow the second law of thermodynamics.

The second law of thermodynamics is correlated with the spontaneous and non-spontaneous processes.

In a simple way, you can say that by the study of 2nd law you may tell more about the spontaneous and non-spontaneous process.

Spontaneous process

A spontaneous process, which has a tendency to do occur by itself in nature means no need for input energy from outside sources.

Also, a process is spontaneous if some initial effort is done and sometimes it is conditional.

For example

waterfalls (water itself falls from upper to lower surface) or if we move a ball on an inclined plane from top to bottom,

Iron rust, mixing sugar in water,

Melting of ice at different conditions,

The process of changing diamond into graphite at certain conditions, the process of decay of radioactive isotopes, etc.

In these examples some of the processes happen only by themselves, Such as waterfalls, mixing of sugar in water, a ball rolling top to down, etc.

These processes are naturally occurring and also condition-based.

If we move water with a spoon, the mixing rate of sugar is more than by itself.

Similarly in salt.

In the case of a ball, if we give a little push to the ball, after then the ball start continues to rolling itself on the inclined plane from top to down.

Non-spontaneous process

If we reverse the spontaneous process, which is not possible by itself, then it called a non-spontaneous process.

These processes need external effort in the form of energy or work.

Example

The water does not move by itself from down to top.

You need a water pump to continue this process.

Reversible and Irreversible process

A Carnot heat engine can work in a reverse direction, so here, we take a short view of reversible and irreversible processes.

Reversible process

A process is reversible if the system and surroundings came back to their original state in the opposite direction, after passing through the same intermediate conditions as during the direct process.

There are infinitesimal changes during the reversible process.

A reversible process can take an infinite amount of time to complete.

In actual practice, a perfect reversible process is impossible.

Examples

Some processes can be considered as a reversible process,

such as an isothermal expansion and compression in the case of an ideal gas at a constant temperature in a cyclic process.

If we compress a spring at a slow rate, then the work done by spring during expansion is equal to the applied work during compression.

Irreversible process

Any process, which is not reversible is called an irreversible process.

Examples

There are lots of examples of irreversible processes.

Some of these are,

1. Transfer of heat by radiation.
2. The heat produced during the current flow through a resistance.
3. The heat produced during relative motion with friction.
4. Throttling
5. The plastic deformation
6. Diffusion, etc.

Now you understood a few concepts and terms used in heat engines or Carnot cycle.

Let us talk about the heat engine and the Carnot heat engine process, by keeping these terms in mind.

What is a heat engine?

A heat engine is a device or machine, which converts heat energy into useful work.

Heat engine examples

The internal combustion engine,

The thermal power plant,

A steam locomotive engine, etc.

Heat engine construction and working

In a heat engine, a source (Heat reservoir) is used to transfer heat to Engine (E)at high-temperature T_1.

Let Q₁ be the heat absorbed by the engine from the source.

Work is performed by the heat engine = W

According to the second law of thermodynamics,

“All work can be converted into heat, but all heat can not be converted into work”

It means some amount of heat is always lost.

Let the heat loss is equal to Q₂.

Therefore work done is, W = Q₁-Q₂ 

To full fill, this requirement a sink (Heat reservoir) at low-temperature T₂ is used in a heat engine.

As you know heat engines obey both the first law of thermodynamics and the second law of thermodynamics.

From the above figure, it is clear that a heat engine is a device, which converts heat into work.

The term “law of conversation of energy” is not violated by heat engines.

On another hand, we can say that the heat engine obeys the first law of thermodynamics.

According to the first law of thermodynamics, conversions of energies are interconvertible.

Hence, W = Q  ⇒ Work is done = Heat produced.

According to the Clausius, the derivation of the first law,

As in heat engines, the working is a cyclic process, therefore,

internal energy change (ΔU) = 0

This is because for a cyclic process change in internal energy (ΔU) is always Zero (0).

Hence proved, Q = W  

Heat engine efficiency

The efficiency of a heat engine is the ratio of output mechanical energy to the input heat energy.

The efficiency of heat engines is denoted by a symbol named as, eta  = η

The formula of heat engine efficiency (η)  =  W/Q₁

Here W = output mechanical work during the complete cycle.

Q₁ = Heat absorbing from the source at a high temp during the complete cycle.

As we know W = Q  ⇒  Q₁-Q₂ 

after using this, η = Q₁−Q₂/Q₁

What is a Carnot heat engine?

The Carnot heat engine is a hypothetical heat engine, which is given by A French mechanical engineer Nicolas Léonard Sadi Carnot.

He was a physicist, military scientist in the French army.

The Carnot Sadi was born on 1 June 1796 and died at the age of 36 years on 24 August 1832 in France.

He published a book at the age of 27 years in 1824 named “Reflections on the Motive Power of Fire”.

The Carnot heat engine is a reversible heat engine working between cycles of operations.

So, in the Carnot heat engine, the ideal gas, which is used as a working substance comes back to its initial state at the end of each cycle.

What is the Carnot cycle?

The Carnot heat engine works on the four reversible processes.

Isothermal gas expansion

Adiabatic gas expansion

Isothermal gas compression

Adiabatic gas compression

In the Carnot heat engine, the working series of these four processes is called the Carnot cycle.

The Carnot cycle construction

As you know that the Carnot heat engine works on the four operations, called cycle.

In the making of this type of cycle, the Sadi Carnot uses two different heat reservoir (Source and sink) at different temperature range, T₁ and T_2. 

A cylinder, a piston, and also an insulating stand.

Let us talk about these requirements in detail.

♦ Cylinder 

A cylinder that has perfectly non-conducting walls and the base of the cylinder is perfectly conducting.

A frictionless, weightless, and Perfectly Non-conducting piston is used to fit inside the cylinder. 

An ideal gas is used in the cylinder as a working substance.

♦ Source

It is an infinite thermal capacity heat reservoir at a fixed high-temperature T₁.

Infinite thermal capacity means during adding and subtracting of heat from the reservoir does not change its temperature T₁.

♦ Sink

It is a second heat reservoir, which has also infinite heat capacity at a fixed low-temperature T₂.

♦ Perfectly Non-Conducting stand

A perfectly Non-Conducting stand is used during operations in the Carnot cycle process. 

Carnot cycle PV diagram

The processes during the Carnot cycle and total work done can be shown by the PV diagram.

In this pressure and volume graph, change in pressure shown vertically, and change in volume shown horizontally directions.

This Carnot cycle PV diagram shows all four reversible processes and their changing values.

The total area ABCDA inclosed by the PV diagram is the work done during a complete Carnot cycle.

Carnot cycle working

For changing the pressure during four processes, some weights were added on the piston and some weight removed from the piston.

1 Reversible Isothermal expansion

 

In this process, the cylinder is placed on the source.

As the base of the cylinder is perfectly conducting, then let Q₁ be the heat absorbed from the source at a temperature of T₁ by the gas. 

Now slowly remove the weight.

Due to this gas expand at a constant temperature T₁, as a result, the volume increases and the pressure decreases.

As you know, if the gas is expanding slowly at a constant temperature and the pressure-volume changes then the process is an isothermal expansion.

In this process, work W is done by the gas on the surroundings is equal to Q heat absorbed from the source.

Let in the initial state the gas has P₁, V₁, T₁ are the respectively the pressure, volume, and temperature.

See in the PV diagram

After expansion, the pressure decreases from P₁ to P₂, the volume increases from V₁ to V₂, and the temperature remains constant at T₁.

The work done 

W = Q − ΔU

But in this process, internal energy (U) is not changing, as the temperature is constant, the work is done only due to heat absorbed.

The internal energy is temperature-dependent.

Therefore, the change in internal energy (ΔU) is equal to zero (0).

W = Q − 0  ⇒ W = Q

As we are in the first process of the cycle then,

ΔU = 0 can be denoted by ΔU₁ = 0

W₁ = Q₁  ⇒ Work done for curve AB on the PV diagram.

Now applying the work done formula for the Isothermal process in the ideal gas case.

Which is W = nRT log_e V₂/V₁ 

(Where n = no. of moles of gas, R is gas constant, and V₂, V₁ are volume changes and T is the temperature).

Hence, W₁ = Q₁ =nRT₁ log_e V₂/V₁      (For first process temperature T is T₁)

This is the work done in the first process.

2 Reversible Adiabatic Expansion

 

In this process, the cylinder removed from the source and placed on the perfectly Non-conducting stand.

As the stand is perfectly non-conducting, no heat transfer takes place during this process.

Now remove the weight fast and gas start expanding.

Now the work is done due to a decrease in internal energy (U) of gas and because of it, the temperature starts falling.

As you know, if the gas expanding and no heat transfer takes place from the system to the surrounding or from the surrounding to the system, then the process is an adiabatic expansion.

Removing of weight is until we reach to the sink’s temperature, which is T₂.

Let, due to a decrease in internal energy (U) the temperature falls from T₁ to T₂ equal to sink temperature.

The volume and pressure also change, volume increases from V₂ to V₃ and pressure decreases from P₂ to P₃.

The work done 

W = Q − ΔU,

W = 0 − ΔU ⇒ W = −ΔU,       (Q = 0, as no heat transfer)

As we are in the second process of the cycle then,

W₂ = −ΔU₂ ⇒  Work done for curve BC on the PV diagram.

The Work done formula in terms of internal energy change for an adiabatic process in the ideal gas case.

W = −nR (T₂−T₁)/1−γ

Here ΔU = nC_vΔT ⇒ nR (T_low−T_high)

For ideal gas case ⇒ nR (T₂−T₁)/1−γ

In term of the second process, the work done is,

W₂ = −ΔU₂ ⇒ −nR (T₂−T₁)/1−γ

This is the work done in the second process.

3 Reversible Isothermal Compression

In this process, the cylinder removed from the stand and placed on the sink.

Some weights were added on the piston.

In this process, the temperature (T₂) of the gas inside the cylinder is the same as the temperature (T₂) of the sink.

It means no temperature change during the process, hence temperature is constant.

The work is done in the form of heat rejected to the sink.

Let Q₂ be the heat rejected to the sink at constant temperature T₂ as the base of the cylinder is conducting.

During this process, the volume starts to decrease from V₃ to V₄ and the pressure starts to increase from P₃ to P₄.

We know that if the gas compress at a constant temperature and the value of its pressure and volume changes then the process is Isothermal compression.

As the temperature is constant then there is no change in internal energy (U).

The work done

W = Q − ΔU ⇒  W = Q − 0  ⇒ W = Q    (As the ΔU is equal to Zero)

As we are in the third process of the cycle then,

ΔU = 0 can be denoted by ΔU₃ = 0

W₃ = Q₂  ⇒ nRT₂ log_e V₄/V₃

W₃ = nRT₂ log_e V₄/V₃  ⇒ Work done for curve CD on the PV diagram.

This is the work done for the third process.

4 Reversible Adiabatic Compression

In this process, the cylinder removed from the sink and placed again on the stand.

Weights are added on the piston.

The work is done during this process is due to an increase in the internal energy (U) of gas and because of it, the temperature starts increasing.

Let the temperature change from T₂ to T₁, which is the same as the initial state of the gas.

As the stand is perfectly Non-conducting, no heat transfer takes place during this process.

As you know, if the gas compress and no heat transfer takes place from the system to the surrounding or from surrounding to the system, then the process is an adiabatic compression.

In this process, the volume decreases from V₄ to V_1, and respectively the pressure increases from P₄ to P_1, and the gas comes back to its initial state.

The work done

W = Q − ΔU,

W = 0 − ΔU ⇒ W = −ΔU       (Q = 0, as no heat transfer)

As we in the fourth process of the cycle then,

W₄ = −ΔU₄ ⇒  Work done for curve DA on the PV diagram.

Here temperature changes from T₂ to T₁ then,

W₄ = −nR (T₁−T₂)/1−γ

Here ΔU = nC_vΔT ⇒ nR(T_high−T_low),   

W₄ = −ΔU₄ ⇒  −nR(T₁−T₂)/1−γ

This is the work done by the fourth process of the cycle.

Total work done by the Carnot cycle. 

When you study the PV diagram of the Carnot heat engine, you see, that the total

area enclosed by the PV diagram is equal to the total work done by the Carnot cycle.

Now the total area enclosed by the PV diagram = Area ABCDA.

The total work done by the Carnot cycle = W

W = W₁ + W₂ + W₃ + W₄

Now adding the terms,

W₁ = nRT₁ log_e V₂/V₁

W₂ = −nR (T₂−T₁)/1−γ

W₃ = nRT₂ log_e V₄/V₃

W₄ = −nR (T₁−T₂)/1−γ

Here W₂ + W₄  = 0   (After solving)

Hence W = w₁ + W₃

W = nRT₁log_eV₂/V₁ − nRT₂log_eV₃/V₄   (After solving)

After solving the curves AB, BC, CD, DA in terms of isothermal and adiabatic processes, we get,

V₂/V₁ = V₃/V₄

Substitute this, W = nRT₁log_eV₂/V₁ − nRT₂log_eV₂/V₁ 

Hence total work done by the carnot cycle W = nRlog_eV₂/V₁[T₁−T₂]

In this complete cycle net amount of heat absorbed Q = Q₁-Q₂

Hence from the first law of thermodynamics, W = Q

Then, W = nR log_eV₂/V₁ [T₁−T₂] = Q = Q₁-Q₂

Carnot cycle efficiency

The efficiency of any heat engine (η)  =  W/Q₁

Here, W = Q = Q₁-Q₂

= nR log_e V₂/V₁ [T₁−T₂] 

Putting these valves in the heat engine efficiency formula,

η = nR log_e V₂/V₁ [T₁−T₂]/nRT₁ log_e V₂/V₁

η = [T₁−T₂]/T₁ = Carnot cycle efficiency.

From the above relations,

η = [T₁−T₂]/T₁ = [Q₁-Q₂]/Q₁ = Carnot cycle efficiency.

Q₂/Q₁ = T₂/T₁

From this equation, we can say that the efficiency of the Carnot heat engine or the Carnot cycle efficiency depends upon the temperature of the source and sink,

it is independent of the nature of the working substance.

Carnot Cycle efficiency calculator

Why no heat engine can have an efficiency of 100% or 1?

The efficiency of a heat engine η = [T₁−T₂]/T₁ = [Q₁-Q₂]/Q₁

If you want to gain the efficiency equal to 1 or 100%,

then you must have T₂ = 0 Kelvin and also, Q₂ = 0

The T₂ = 0 Kelvin, only when you have the value of heat Q₂= 0

This means no loss of heat and all heat can be completely converted into work.

Which not possible, as it violates the second law of thermodynamics.

Hence, no Q₂ = 0 and thus no T₂ = 0, and no heat engine can have an efficiency of 1 or 100%.

Carnot theorem

The above relation for Carnot cycle efficiency is, Q₂/Q₁ = T₂/T₁ 

It can be seen from the relation, that heat engines operating in cycles between the same

two temperatures T₁ and T₂ can not have efficiency more than the Carnot reversible heat engine.

This is the Carnot theorem, that is, the efficiency of a Carnot reversible heat engine is maximum.

Another form of Carnot theorem in the same sense,

According to the Carnot theorem, the efficiency of all reversible heat engines operating between the same two temperatures is the same (equally efficient) and no irreversible heat engines operating between the same two temperatures can have efficiency more than the Carnot heat engine. 

The refrigerator as a reversed Carnot cycle

As you know, the processes in the Carnot cycle are reversible.

Therefore, the Carnot cycle can be used as a reverse Carnot cycle.

After this, the reversed Carnot cycle worked as a refrigerator.

For a Carnot cycle, the Q₁ heat is absorbed from the source at temperature T₁ and Q₂ heat is rejected to the sink at temperature T₂.

The second law of thermodynamics not violated here.

But in the case of a refrigerator or reverse Carnot cycle,

the Q₂ heat absorbed from the sink at temperature T₂ and Q₁ heat is rejected to the source at temperature T₁.

This is a clear violation of the second law of thermodynamics.

As the heat can not transfer by itself from a lower temperature to a higher temperature, without any external agency.

Therefore in the case of a refrigerator, some external work is done to transfer heat from a low temp body to a high temp body.

For refrigerators, this work is performed by the compressor of the refrigerator by the use of electricity.

An interesting fact in the refrigerator

In the refrigerator, the amount of rejected heat Q₁ is more than the amount of heat absorbed Q₂.

This is because the amount of rejected heat Q₁ is equal to the sum of work done W and heat absorbed Q₂.

that is,   Q₁ = W + Q₂

Coefficient of Performance COP

The efficiency of the refrigerator is measured in terms of the Coefficient of performance.

The coefficient of performance is the ratio of the amount of heat absorbed from the sink to the work done by an external agency.

That is,   COP (k) = Q₂/W

From the first law of thermodynamics, W = Q

Also, Q = Q₁−Q₂

k = Q₂/Q₁−Q₂

Interesting fact!

The efficiency of all heat engines cannot be more than 100% but, the coefficient of performance can be 100% or more than 100%.

Example of Coefficient performance, in terms of Numerical question.

Question:

A Carnot’s refrigerator takes heat from water at 0°C and leaves it in the room. The temperature of the room is 27°C. 100 grams of water at 0°C is to be changed into ice at 0°C.

Find, how many calories of heat are discarded to the room? Also, find the Coefficient of performance? Given, one gram of heat = 80 calories.

Answer:

Her given that the temperature of water = 0°C = T₂ = Sink’s temp.

T₂ = 0°C + 273 K = 273 Kelvin

Here, Source temp = Room’s temp = T1 = 27°C, Given

T₁ = 27°C + 273 K = 300 Kelvin,

Let Q₂ and Q₁ be the heat absorbed from the sink and heat rejected to the source.

Given, that Q₂ = 100 gram of water = 100 × 80 calories = 8000 calories,

Applying carnot cycle efficiency equation,

Q₂/Q₁ = T₂/T₁ 

8000/Q₁ = 273 K/300 K

Q₁ = approximately 8791 calories.

The work is done by the compressor of the refrigerator

W = Q₁−Q₂ = 8791−8000 = 791

Then the coefficient of performance COP,

K = Q₂/W = 8000/791 = 10.1

Also, Q₁ = 8791 calories and Q₂ = 8000 calories

From this example, we can say that the quantity of rejected heat by the refrigerator is much higher than the total quantity of heat absorbed from the sink.

Entropy and Carnot cycle

As you know dS = dQ/T    ⇒   dQ = TdS

The above relation is also the second law of thermodynamics, in fact, it is the mathematical formulations of the second law of thermodynamics.

In the Carnot, cycle entropy is used to draw a TS diagram to find Carnot cycle efficiency in terms of temp and entropy.

Also, the area of the TS diagram of the Carnot cycle is equal to the area of the PV diagram for the same Carnot cycle.

dQ = dW + dU    ⇒ Frist law of thermodynamics,

dQ = TdS           ⇒ Second law of thermodynamics,

dW + dU = TdS  

dU is zero as internal energy is a state function and change in internal energy for a reversible cycle is 0.

Hence mathematically,

∫PdV = ∫TdS

Carnot cycle TS digram

The above diagram represents a graph between temperature T and entropy S for a reversible Carnot cycle.

Like the PV diagram, the processes for a TS diagram of a complete Carnot cycle is similar, such as,

Line AB = Isothermal expansion, where the temperature T₁ = Constant.

Line BC = Adiabatic expansion, where the Entropy S₂ = Constant.

CD line = Isothermal compression, where the temperature T₂ = Constant.

and line DA = Adiabatic compression, where the Entropy S₁ = Constant.

Both the Lines AB (Isothermal expansion) and CD (Isothermal compression) are parallel to the entropy axis.

Here change in temperature ΔT = 0

The lines BC (Adiabatic expansion) and DA (Adiabatic compression) are parallel to the temperature axis.

Here change in entropy ΔS = 0

Now the quantity of heat absorbed by the source during the complete Carnot cycle = Q₁

Q₁ = +ΔS

For curve AB on the TS diagram.

The first process, which is reversible Isothermal expansion Q₁ = TΔS

Q₁ = T₁(S₂−S₁)

For the curve BC on the TS diagram.    ⇒ (Second process, Adiabatic expansion)

ΔS = 0 (constant)

For the curve CD on the TS diagram.    ⇒ (Third process, Isothermal compression)

Q₂ = −ΔS = −T₂(S₁−S₂) = T₂(S₂−S₁)

For the curve DA on the TS diagram     ⇒ (Fourth process, Adiabatic compression)

ΔS = 0 (constant)

Net work done during complete process W = W₁ + W₃  = Q₁ − Q₂

W = (T₁−T₂) (S₂−S₁)    ⇒  Enclosed Area of complete ABCDA Carnot cycle on the TS diagram.

Carnot cycle TS diagram efficiency.

The Carnot cycle efficiency in terms of TS diagram

The efficiency of a heat engine (η) = W/Q₁

As, W = Q₁−Q₂

η = Q₁−Q₂/Q₁

(T₁−T₂) (S₂−S₁)/T₁(S₂−S₁) = T₁T₂/T₁ = Carnot cycle efficiency.

Carnot cycle Conclusion

In the Carnot cycle, all processes are reversible but in nature, no process is perfectly reversible. So, the concept of the Carnot cycle is hard to digest. 

The idea of the weightless or frictionless piston is also imaginary (hypothetical). 

Hence, you can say that the setup of the Carnot heat engine Purely imaginary concept, and in real life, no heat engine exists.

That’s why the Carnot cycle is not used in power plants, automobile industries, etc.

Although the Carnot heat engine or Carnot cycle is completely imaginary,

but it established a maximum efficiency concept for all heat engines operating between the same T₁(High) and T₂(Low) temperatures.

That’s the End Of The Carnot heat engine!

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FAQ related to Carnot Cycle

There are few questions that are somehow co-relate to the Carnot Heat engine or if we say can be asked to know more about the Carnot Cycle. 

1 What is the Carnot cycle in conventional and non-conventional?

According to us, after study complete the Carnot cycle we can say that (personal opinion) Carnot Cycle is non-conventional. As it broke our known-thermodynamic physics standards. 

References

1. https://www.grc.nasa.gov
2. https://en.wikipedia.org
3. https://chem.libretexts.org
4. http://hyperphysics.phy
5. https://www.sciencedirect.com

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1. Latent Heat definition, example, unit, latent heat of water, fusion
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3. What is Crystalline Solid, Definition, Properties, Characteristics, Examples
4. What is an Amorphous Solid, Definition, Properties, Characteristics, Examples
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