Intermediate

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Heat Engines, Refrigerators, Carnot Cycle & Second Law

Master Carnot Cycle and Second Law for JEE & NEET. Covers heat engine efficiency, Carnot theorem, COP of refrigerator and heat pump, Kelvin-Planck Clausius statements, entropy.

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If the First Law tells us that energy is conserved, the Second Law of Thermodynamics tells us something even more profound: not all energy conversions are created equal. Heat flows spontaneously from hot to cold — never the reverse — and no machine can convert heat entirely into work. These fundamental asymmetries of nature are captured by the Second Law, which governs the operation of heat engines, refrigerators, and heat pumps. The Carnot cycle establishes the theoretical upper limit of efficiency for any engine operating between two temperatures — a benchmark no real engine can exceed. For JEE and NEET, this topic provides direct numerical questions on Carnot efficiency, COP of refrigerators, and the entropy concept, as well as conceptual MCQs on the Second Law statements.

1. Heat Engines

A heat engine is a device that converts heat energy into mechanical work by operating in a cycle between a hot source and a cold sink.

$Q_{1} \to Engine \to W + Q_{2}$

$Q_{1}$ = heat absorbed from hot source; $Q_{2}$ = heat rejected to cold sink; $W$ = net work output.

Energy conservation: $W = Q_{1} - Q_{2}$

Thermal Efficiency of a Heat Engine

$η = \frac{W}{Q_{1}} = \frac{Q_{1} - Q_{2}}{Q_{1}} = 1 - \frac{Q_{2}}{Q_{1}}$

$η < 1$ always (Second Law — some heat must always be rejected).
$η = 1$ only if $Q_{2} = 0$ — impossible (Kelvin-Planck statement).

2. Refrigerators and Heat Pumps

A refrigerator is the reverse of a heat engine — it uses work input to transfer heat from a cold body to a hot body.

$Q_{2} + W = Q_{1}$

$Q_{2}$ = heat absorbed from cold reservoir; $Q_{1}$ = heat rejected to hot reservoir; $W$ = work input.

Coefficient of Performance (COP)

Refrigerator COP: Ratio of heat extracted from cold body to work done:

$β_{ref} = \frac{Q_{2}}{W} = \frac{Q_{2}}{Q_{1} - Q_{2}}$

Heat Pump COP: Ratio of heat delivered to hot body to work done:

$β_{hp} = \frac{Q_{1}}{W} = \frac{Q_{1}}{Q_{1} - Q_{2}} = β_{ref} + 1$

Device	Purpose	COP	Desired effect
Heat engine	Convert heat to work	$η = W / Q_{1}$	Maximum work output
Refrigerator	Cool cold body	$β = Q_{2} / W$	Maximum heat removed per unit work
Heat pump	Heat hot body	$β = Q_{1} / W$	Maximum heat delivered per unit work

3. Second Law of Thermodynamics

The Second Law has two equivalent classical statements:

Kelvin-Planck Statement

It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the absorption of heat from a single reservoir and the performance of an equivalent amount of work.

Simply: No engine can have 100% efficiency. Some heat must always be rejected to a cold sink.

Clausius Statement

It is impossible to construct a device that, operating in a cycle, produces no effect other than the transfer of heat from a colder body to a hotter body.

Simply: Heat cannot flow spontaneously from cold to hot. A refrigerator requires external work input.

The two statements are equivalent — violation of one implies violation of the other.

4. Reversible and Irreversible Processes

Property	Reversible process	Irreversible process
Restorability	System and surroundings can be exactly restored	Cannot be exactly undone — permanent change in universe
Process type	Quasi-static, infinitely slow	Fast, non-quasi-static; involves friction, turbulence, mixing
Entropy change	$Δ S_{universe} = 0$	$Δ S_{universe} > 0$
Reality	Ideal (never truly achieved)	All real processes

5. Carnot Cycle and Carnot Engine

The Carnot cycle is a theoretical reversible cycle operating between two temperatures $T_{1}$ (hot) and $T_{2}$ (cold). It consists of four quasi-static steps:

Step	Process	Heat exchange	$Δ U$
A → B	Isothermal expansion at $T_{1}$	$Q_{1}$ absorbed from hot source	0
B → C	Adiabatic expansion ( $T_{1} \to T_{2}$ )	$Q = 0$	$- W_{B C}$
C → D	Isothermal compression at $T_{2}$	$Q_{2}$ rejected to cold sink	0
D → A	Adiabatic compression ( $T_{2} \to T_{1}$ )	$Q = 0$	$+ W_{D A}$

Carnot Efficiency

$η_{Carnot} = 1 - \frac{T_{2}}{T_{1}} = \frac{T_{1} - T_{2}}{T_{1}}$

Key point: Carnot efficiency depends only on the temperatures of the hot and cold reservoirs — not on the working substance. All temperatures must be in Kelvin.

Carnot COP (Refrigerator)

$β_{Carnot,ref} = \frac{T_{2}}{T_{1} - T_{2}}$
$β_{Carnot,hp} = \frac{T_{1}}{T_{1} - T_{2}}$

Worked Example

A Carnot engine operates between $T_{1} = 500 K$ and $T_{2} = 300 K$ . It absorbs $Q_{1} = 1000 J$ per cycle. Find $η$ , $W$ , and $Q_{2}$ .

$η = 1 - \frac{300}{500} = 1 - 0.6 = 0.4 = 40 %$

$W = η Q_{1} = 0.4 \times 1000 = 400 J$

$Q_{2} = Q_{1} - W = 1000 - 400 = 600 J$

COP of refrigerator operating between same temperatures $= \frac{T_{2}}{T_{1} - T_{2}} = \frac{300}{200} = 1.5$

6. Carnot's Theorem

Statement: (i) No engine working between two temperatures can be more efficient than a Carnot engine operating between the same temperatures. (ii) All reversible engines operating between the same two temperatures have the same efficiency (= Carnot efficiency).

Consequence: $η_{real} < η_{Carnot}$ always, because real engines are irreversible.

7. Entropy

Entropy ( $S$ ) is a state function that measures the degree of disorder of a system. For a reversible process:

$Δ S = \frac{Q_{rev}}{T}$

Second Law in terms of entropy: For any natural (irreversible) process, the total entropy of the universe increases: $Δ S_{universe} \geq 0$ .

Reversible process: $Δ S_{universe} = 0$
Irreversible (natural) process: $Δ S_{universe} > 0$
No process: $Δ S_{universe} < 0$ (would violate Second Law)

Example: Isothermal expansion of gas at $T = 300 K$ absorbing $Q = 1000 J$ :

$Δ S = \frac{Q}{T} = \frac{1000}{300} = 3.33 {J K}^{- 1}$

Carnot Efficiency — Instant Calculation Guide

Given	Find	Formula
$T_{1}$ , $T_{2}$	Efficiency $η$	$η = 1 - T_{2} / T_{1}$
$η$ , $Q_{1}$	Work $W$	$W = η Q_{1}$
$Q_{1}$ , $W$	Heat rejected $Q_{2}$	$Q_{2} = Q_{1} - W$
$T_{1}$ , $T_{2}$	Refrigerator COP	$β = T_{2} / (T_{1} - T_{2})$
$β_{ref}$	Heat pump COP	$β_{hp} = β_{ref} + 1$

JEE / NEET Power Tips — Carnot & Second Law

Carnot efficiency uses absolute (Kelvin) temperatures — never Celsius. A very common error: using $η = 1 - T_{2} / T_{1}$ with Celsius values. Always convert: $T (K) = T (° C) + 273$ before substituting.
COP of heat pump = COP of refrigerator + 1. This is a direct consequence of energy conservation: $Q_{1} = Q_{2} + W$ . Dividing by $W$ : $Q_{1} / W = Q_{2} / W + 1$ . So $β_{hp} = β_{ref} + 1$ . Always $β_{hp} > β_{ref}$ .
Carnot efficiency depends only on $T_{1}$ and $T_{2}$ . It does NOT depend on the working substance, the number of moles, or the process details. Two engines with the same source and sink temperatures have the same maximum (Carnot) efficiency, regardless of what gas they use.
To increase Carnot efficiency: either increase $T_{1}$ or decrease $T_{2}$ . $η = 1 - T_{2} / T_{1}$ increases when $T_{2} / T_{1}$ decreases. In practice, increasing $T_{1}$ (source temperature) is more effective and economical than decreasing $T_{2}$ .
Entropy of universe never decreases. $Δ S_{universe} \geq 0$ is the mathematical statement of the Second Law. For a reversible process, $Δ S_{universe} = 0$ (entropy is conserved). For all real processes, $Δ S_{universe} > 0$ (entropy is created). This is why time flows in one direction.

Practice Questions

Q1 (JEE Main / NEET): A Carnot engine operates between a source at 500 K and a sink at 300 K. It absorbs 1000 J of heat per cycle. Find: (i) efficiency, (ii) work done, (iii) heat rejected.

Explanation:

(i) Efficiency ( $η$ ):
$η = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{300}{500} = 1 - 0.60$
$= 0.40 (or 40 %)$

(ii) Work done ( $W$ ):
$W = η Q_{1} = 0.40 \times 1000 = 400 J$

(iii) Heat rejected ( $Q_{2}$ ):
$Q_{2} = Q_{1} - W = 1000 - 400 = 600 J$

Q2 (JEE Main): The efficiency of a Carnot engine is 40%. If the source temperature is 500 K, find the sink temperature.

Explanation:

Using the efficiency formula:
$η = 1 - \frac{T_{2}}{T_{1}} ⟹ 0.40 = 1 - \frac{T_{2}}{500}$

Rearranging to solve for $T_{2}$ :
$\frac{T_{2}}{500} = 1 - 0.40 = 0.60$
$T_{2} = 0.60 \times 500 = 300 K$

Q3 (NEET MCQ): A refrigerator operates between -23°C and 27°C. Its coefficient of performance is:

A) 5
B) 1
C) 0.2
D) 6

Answer: A) 5.

Explanation: First, convert temperatures to Kelvin:
Sink temp $T_{2} = - 23 + 273 = 250 K$
Source temp $T_{1} = 27 + 273 = 300 K$

The Coefficient of Performance ( $β$ ) for an ideal refrigerator is:
$β = \frac{T_{2}}{T_{1} - T_{2}} = \frac{250}{300 - 250} = \frac{250}{50} = 5$

Q4 (JEE Main): Two Carnot engines A and B operate in series. Engine A absorbs heat at $T_{1} = 900 K$ and rejects it to a temperature $T$ ; Engine B absorbs heat at $T$ and rejects it at $T_{2} = 400 K$ . If both engines have equal efficiency, find $T$ .

Explanation:

Efficiency of Engine A: $η_{A} = 1 - \frac{T}{T_{1}} = 1 - \frac{T}{900}$

Efficiency of Engine B: $η_{B} = 1 - \frac{T_{2}}{T} = 1 - \frac{400}{T}$

Given that their efficiencies are equal ( $η_{A} = η_{B}$ ):
$1 - \frac{T}{900} = 1 - \frac{400}{T}$

$\frac{T}{900} = \frac{400}{T}$

Cross-multiply to solve for $T$ :
$T^{2} = 900 \times 400 = 360, 000$
$T = \sqrt{360, 000} = 600 K$

Q5 (Board): State the two statements of the Second Law of Thermodynamics and explain why they are equivalent.

Explanation:

Kelvin-Planck Statement: No heat engine operating in a cycle can absorb heat from a single thermal reservoir and convert it entirely into work. (Some heat must be rejected to a colder sink; 100% efficiency is impossible).

Clausius Statement: No device operating in a cycle can transfer heat spontaneously from a cold body to a hot body without any external work input. (Refrigerators require a compressor/power source).

Equivalence: They are logically equivalent because a violation of one leads to a violation of the other.
Suppose we violate Kelvin-Planck by building an engine that converts 100% of absorbed heat into work ( $W = Q_{1}$ , with zero rejection). We could use this free work to drive a standard refrigerator (which transfers heat from cold to hot). The net result of combining these two devices would be the spontaneous transfer of heat from a cold body to a hot body with no net external work input — directly violating the Clausius statement. The reverse is also true.

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DifficultyIntermediate

Est. Read Time26 minutes

CategoryThermodynamics

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Heat Engines, Refrigerators, Carnot Cycle & Second Law

1. Heat Engines

Thermal Efficiency of a Heat Engine

2. Refrigerators and Heat Pumps

Coefficient of Performance (COP)

3. Second Law of Thermodynamics

Kelvin-Planck Statement

Clausius Statement

4. Reversible and Irreversible Processes

5. Carnot Cycle and Carnot Engine

Carnot Efficiency

Carnot COP (Refrigerator)

Worked Example

6. Carnot's Theorem

7. Entropy

Carnot Efficiency — Instant Calculation Guide

Practice Questions

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Practice Questions

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