Second Law of Thermodynamics and Entropy
Complete Class 11 Physics webpage for Kelvin-Planck statement, Clausius statement, reversibility, irreversibility, entropy, entropy change, applications, numericals and PYQs.
The first law says energy is conserved. The second law tells the direction in which natural processes actually occur. It explains why heat flows naturally from hot to cold, why 100% conversion of heat into work is impossible in a cyclic engine, why refrigerators need work input, and why entropy is a powerful measure of the spreading of energy.
1. Second Law Introduction
The second law of thermodynamics gives the direction, limitation and quality aspect of energy conversion.
The first law, ΔU = Q - W, tells that energy cannot be created or destroyed. But it does not tell whether a process is possible. For example, the first law does not forbid heat from flowing by itself from a cold body to a hot body if energy is conserved. In reality this never happens naturally. The second law supplies this missing direction of nature.
In simple coaching language: energy quantity is conserved, but energy quality degrades in natural processes. Mechanical work can be completely converted into heat, but heat taken from a single reservoir cannot be completely converted into work in a cyclic process. This restriction is the heart of the second law.
Direction
Natural heat flow is from high temperature to low temperature.
Limitation
A cyclic heat engine cannot convert all absorbed heat into work.
Entropy
For an isolated system, entropy never decreases: ΔS ≥ 0.
2. Kelvin-Planck Statement
It is impossible to construct a heat engine which, operating in a cycle, converts all heat taken from a single reservoir completely into work.
This statement is about heat engines. It says that a heat engine must reject some heat to a sink. A machine that absorbs heat QH from only one reservoir and gives W = QH in a cycle would violate the second law. Such a machine is called a perpetual motion machine of the second kind, and it cannot exist.
Real engines require a source and a sink. The source supplies heat, the engine converts part into work, and the sink receives the unavoidable rejected heat. Therefore efficiency is always less than one for real cyclic engines.
3. Clausius Statement
It is impossible for heat to flow from a colder body to a hotter body without external work or some other effect.
This statement is about refrigerators and heat pumps. Natural heat transfer is from hot to cold. A refrigerator transfers heat from cold to hot, but it needs work input from a compressor. Without work input, heat cannot be pumped from a lower temperature body to a higher temperature body.
In daily life, a refrigerator cools the food chamber by absorbing QC, and it rejects QH to the warmer room. The compressor supplies work W, so QH = QC + W.
4. Equivalence of Kelvin-Planck and Clausius Statements
The two statements look different but they are logically equivalent forms of the same second law.
If Kelvin-Planck statement were false, a perfect heat engine could convert all heat from a single hot reservoir into work. That work could run a refrigerator, causing heat to move from cold to hot without any net external effect. This would violate Clausius statement.
If Clausius statement were false, heat could move from cold to hot without work. Then an ordinary engine rejecting heat to a cold sink could have that rejected heat returned to the hot source automatically, giving a net effect of complete heat-to-work conversion from a single reservoir. This would violate Kelvin-Planck statement.
Violation of Kelvin-Planck
Leads to a device that can force heat from cold to hot without external work.
Violation of Clausius
Leads to an engine that converts heat completely into work in a cycle.
5. Reversible Process
A reversible process can be reversed so that both system and surroundings return exactly to their initial states.
A reversible process is an ideal limiting process. It happens infinitely slowly, passes through equilibrium states, has no friction, no turbulence, no unrestrained expansion, and heat transfer occurs through an infinitesimally small temperature difference. It gives the maximum possible work output in expansion and minimum work input in compression.
- It is quasi-static and has no dissipative effects.
- It can be represented by a well-defined path on thermodynamic diagrams.
- For a reversible cyclic process, total entropy change of universe is zero.
- Carnot cycle is the standard ideal reversible heat engine cycle.
6. Irreversible Process
An irreversible process cannot be exactly reversed without leaving changes in the system or surroundings.
Most natural processes are irreversible. Friction converts mechanical work into heat. Heat transfer through a finite temperature difference spreads energy from a hotter body to a colder body. Free expansion of gas into vacuum has no opposing pressure and cannot be retraced without external work. Mixing, diffusion, viscosity and electrical resistance are also irreversible.
- Irreversibility is caused by friction, viscosity, turbulence and finite temperature difference.
- Entropy of the universe increases in an irreversible process.
- Real machines are irreversible, so they are less efficient than ideal reversible machines.
- Irreversible processes proceed naturally in one preferred direction.
7. Entropy Concept
Entropy is a thermodynamic state function that measures energy dispersal and tells whether a process is naturally possible.
For Class 11 and competitive exams, entropy should be understood in two connected ways. First, it is defined quantitatively through reversible heat transfer. Second, it increases when energy spreads, matter disperses, or a system moves toward more probable states.
Entropy is a state function, so ΔS depends only on initial and final states, not on the actual path. If the actual process is irreversible, we still calculate entropy change by imagining a reversible path between the same initial and final states.
Entropy unit is J K-1. If a system absorbs heat reversibly at temperature T, its entropy increases. If it rejects heat reversibly, its entropy decreases. For an isolated system, the total entropy change is never negative.
8. Entropy Change
Entropy change is calculated from reversible heat transfer even when the actual process is irreversible.
For reversible isothermal heat transfer at temperature T.
For an isolated system. Equality for reversible, greater than zero for irreversible.
Maximum heat-engine efficiency for a Carnot engine.
Ideal refrigerator COP between two reservoirs.
Ideal heat pump COP when heating is desired.
Use this for spontaneity checks.
No net entropy production.
Entropy is generated by irreversibility.
Common Entropy Changes
| Process | Entropy Change | Physical Meaning |
|---|---|---|
| Melting at constant temperature | ΔS = L/T | Solid becomes liquid; molecular freedom increases. |
| Freezing at constant temperature | ΔS = -L/T | Liquid becomes solid; molecular freedom decreases. |
| Heat absorbed reversibly | Positive | System energy spreads more. |
| Heat rejected reversibly | Negative | System loses thermal energy. |
| Free expansion of gas | Positive | Gas occupies larger volume; number of accessible states increases. |
9. Entropy in Reversible and Irreversible Processes
Entropy separates ideal processes from natural real processes.
For a reversible process, the system may gain entropy and surroundings may lose exactly the same entropy, so total entropy of universe remains unchanged. For an irreversible process, the entropy gained by one part is larger than the entropy lost by another part, so total entropy of universe increases.
Reversible Heat Transfer
Heat flows through an infinitesimal temperature difference. Entropy is transferred, not produced. ΔSuniverse = 0.
Irreversible Heat Transfer
Heat flows through a finite temperature difference. Entropy is produced. ΔSuniverse > 0.
NCERT-Style Diagrams
All diagrams use black axes or bodies, red arrows and red labels to show physically correct heat and work directions.
Heat Engine and Kelvin-Planck Statement
Refrigerator and Clausius Statement
Reversible Process Diagram
Irreversible Process Diagram
Entropy Increase Diagram
Carnot Engine Link with Entropy
Heat Flow from Hot to Cold Body
10. Applications
Heat Engine
Second law explains why a heat engine must reject heat and cannot be 100% efficient.
Refrigerator
Clausius statement explains why work input is needed to move heat from cold chamber to warm room.
Melting of Ice
When ice melts at 0°C, entropy increases because liquid water has greater molecular freedom.
Free Expansion
Gas spreading into vacuum is irreversible and entropy increases even if Q = 0 and W = 0.
Hot to Cold Heat Flow
Total entropy increases because the cold body gains more entropy than the hot body loses.
Daily Life
Cooling of tea, mixing of gases, frictional heating and aging of machines are entropy-increasing processes.
Case Study: Heat Engine
A thermal power plant takes heat from burning fuel or nuclear reactions and rejects part of it to cooling water or atmosphere. Even with perfect engineering, the second law requires rejected heat, so maximum efficiency is limited by reservoir temperatures.
Case Study: Refrigerator
A refrigerator keeps food cold by evaporating refrigerant at low temperature. The compressor supplies work, and condenser coils reject more heat to the room than was absorbed from the food chamber because QH = QC + W.
Case Study: Melting of Ice
When ice at 0°C melts into water at 0°C, heat is absorbed without temperature rise. The entropy change is positive because the liquid state has more accessible molecular arrangements.
Case Study: Free Expansion
In free expansion, gas expands into vacuum. No work is done and no heat is supplied for an ideal gas, yet entropy increases because the final volume offers more possible positions for molecules.
Case Study: Heat Transfer from Hot to Cold Body
For heat Q transferred from TH to TC, entropy change is -Q/TH + Q/TC. Since TC is smaller, Q/TC is larger; hence total entropy increases.
Case Study: Entropy in Daily Life
A room becomes messy naturally because there are many more disordered arrangements than ordered ones. This is only an analogy; thermodynamic entropy is calculated by heat and temperature, but the idea of more probable spread-out states is useful.
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11. Numericals: 30 Solved Problems
Each numerical includes given data, formula, solution and final answer. Try before opening the solution.
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Given: Qrev = 500 J, T = 250 K.
Formula: ΔS = Qrev/T.
Solution: ΔS = 500/250 = 2 J K-1.
Final Answer: 2 J K-1.
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Given: Qrev = -900 J, T = 300 K.
Formula: ΔS = Qrev/T.
Solution: ΔS = -900/300 = -3 J K-1.
Final Answer: -3 J K-1.
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Given: TH = 600 K, TC = 300 K.
Formula: η = 1 - TC/TH.
Solution: η = 1 - 300/600 = 0.50.
Final Answer: 50%.
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Given: TH = 800 K, η = 0.40.
Formula: η = 1 - TC/TH.
Solution: 0.40 = 1 - TC/800, so TC = 480 K.
Final Answer: 480 K.
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Given: TH = 300 K, TC = 250 K.
Formula: COPR = TC/(TH - TC).
Solution: COPR = 250/(300 - 250) = 5.
Final Answer: 5.
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Given: TH = 300 K, TC = 250 K.
Formula: COPHP = TH/(TH - TC).
Solution: COPHP = 300/50 = 6.
Final Answer: 6.
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Given: Q = 334000 J, T = 273 K.
Formula: ΔS = Q/T.
Solution: ΔS = 334000/273 = 1223.4 J K-1 approximately.
Final Answer: 1.22 × 103 J K-1.
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Given: Q = -334000 J, T = 273 K.
Formula: ΔS = Q/T.
Solution: ΔS = -334000/273 = -1223.4 J K-1 approximately.
Final Answer: -1.22 × 103 J K-1.
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Given: Q = 100 J, TH = 500 K, TC = 250 K.
Formula: ΔStotal = -Q/TH + Q/TC.
Solution: ΔStotal = -100/500 + 100/250 = -0.2 + 0.4 = 0.2 J K-1.
Final Answer: 0.2 J K-1.
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Given: Q = 200 J, TH = 400 K, TC = 300 K.
Formula: ΔSuniverse = -Q/TH + Q/TC.
Solution: ΔS = -200/400 + 200/300 = -0.5 + 0.667 = 0.167 J K-1.
Final Answer: 0.167 J K-1 approximately.
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Given: QH = 1200 J, QC = 800 J.
Formula: W = QH - QC, η = W/QH.
Solution: W = 400 J. η = 400/1200 = 0.333.
Final Answer: 33.3%; it obeys second law because QC is not zero.
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Given: QH = 1000 J, W = 1000 J, QC = 0.
Formula: Kelvin-Planck statement.
Solution: A cyclic engine cannot convert all heat from a single reservoir into work.
Final Answer: Impossible; violates Kelvin-Planck statement.
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Given: QC = 1500 J, W = 300 J.
Formula: QH = QC + W, COP = QC/W.
Solution: QH = 1800 J. COP = 1500/300 = 5.
Final Answer: QH = 1800 J; COP = 5.
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Given: COPR = 4, TC = 240 K.
Formula: COPR = TC/(TH - TC).
Solution: 4 = 240/(TH - 240), so TH - 240 = 60 and TH = 300 K.
Final Answer: 300 K.
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Given: TC = 280 K, TH = 350 K.
Formula: COPHP = TH/(TH - TC).
Solution: COPHP = 350/(350 - 280) = 350/70 = 5.
Final Answer: 5.
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Given: Q = 2000 J, T = 400 K.
Formula: ΔS = Q/T.
Solution: ΔS = 2000/400 = 5 J K-1.
Final Answer: 5 J K-1.
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Given: Q = -2000 J, T = 500 K.
Formula: ΔS = Q/T.
Solution: ΔS = -2000/500 = -4 J K-1.
Final Answer: -4 J K-1.
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Given: QH = 600 J, TH = 600 K, TC = 300 K.
Formula: QC/QH = TC/TH.
Solution: QC = 600 × 300/600 = 300 J.
Final Answer: 300 J.
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Given: QC = 400 J, TC = 320 K, TH = 640 K.
Formula: QC/QH = TC/TH, W = QH - QC.
Solution: 400/QH = 320/640 = 1/2, so QH = 800 J. W = 400 J.
Final Answer: QH = 800 J; W = 400 J.
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Given: QH = 1000 J, TH = 500 K, TC = 300 K.
Formula: QC/QH = TC/TH; ΔS = Q/T.
Solution: QC = 1000 × 300/500 = 600 J. Source loses entropy -1000/500 = -2 J K-1. Sink gains 600/300 = 2 J K-1. Total = 0.
Final Answer: ΔSuniverse = 0.
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Given: QH = 1000 J, QC = 700 J, TH = 500 K, TC = 300 K.
Formula: ΔSuniverse = -QH/TH + QC/TC.
Solution: ΔS = -1000/500 + 700/300 = -2 + 2.333 = 0.333 J K-1.
Final Answer: 0.333 J K-1; irreversible.
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Given: Q = 60 J, T = 300 K.
Formula: ΔS = Q/T.
Solution: ΔS = 60/300 = 0.2 J K-1.
Final Answer: 0.2 J K-1.
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Given: TH = 900 K, TC = 300 K.
Formula: η = 1 - TC/TH.
Solution: η = 1 - 300/900 = 2/3 = 0.667.
Final Answer: 66.7%.
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Given: TC = 27°C = 300 K, TH = 127°C = 400 K.
Formula: η = 1 - TC/TH.
Solution: η = 1 - 300/400 = 0.25.
Final Answer: 25%.
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Given: TC = 250 K, TH = 300 K.
Formula: COPR = TC/(TH - TC).
Solution: COP = 250/(300 - 250) = 5.
Final Answer: 5.
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Given: TC = 270 K, TH = 300 K.
Formula: COPHP = TH/(TH - TC).
Solution: COPHP = 300/30 = 10.
Final Answer: 10.
Show Solution
Given: m = 0.01 kg, L = 3.34 × 105 J kg-1, T = 273 K.
Formula: Q = mL, ΔS = Q/T.
Solution: Q = 0.01 × 3.34 × 105 = 3340 J. ΔS = 3340/273 = 12.23 J K-1.
Final Answer: 12.23 J K-1.
Show Solution
Given: Q = 500 J, TH = 1000 K, TC = 500 K.
Formula: ΔS = -Q/TH + Q/TC.
Solution: ΔS = -500/1000 + 500/500 = -0.5 + 1 = 0.5 J K-1.
Final Answer: 0.5 J K-1.
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Given: System is isolated and process is reversible.
Formula: For reversible isolated process, ΔS = 0.
Solution: No entropy is exchanged and no entropy is produced.
Final Answer: ΔS = 0.
Show Solution
Given: System is isolated and process is irreversible.
Formula: Second law: ΔS ≥ 0 for isolated system.
Solution: For irreversible isolated process, entropy increases strictly.
Final Answer: ΔS > 0.
12. PYQs and Question Bank: 55 Hidden Answers
Mixed CBSE, NEET, JEE, IB, ICSE, IGCSE, A-Level, assertion-reason, true/false, case-study and difficult conceptual questions.
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Answer: It tells the natural direction of processes and states that total entropy of an isolated system cannot decrease.
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Answer: No cyclic heat engine can convert all heat taken from a single reservoir completely into work.
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Answer: Heat cannot flow from colder body to hotter body without external work or another external effect.
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Answer: It must reject some heat to a cold sink during cyclic operation according to Kelvin-Planck statement.
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Answer: A process that can be reversed leaving no change in system and surroundings.
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Answer: A process that cannot be exactly reversed without leaving changes in system or surroundings.
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Answer: Entropy is a state function whose change for reversible heat transfer is ΔS = Qrev/T.
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Answer: J K-1.
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Answer: (B) Kelvin-Planck statement.
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Answer: (B) Hot to cold.
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Answer: (B) zero or positive.
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Answer: (C) Q/T.
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Answer: (B) TC/(TH-TC).
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Answer: η = 1 - TC/TH.
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Answer: Carnot efficiency is based on thermodynamic absolute temperature, so kelvin must be used.
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Answer: Zero.
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Answer: Positive.
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Answer: Friction and heat transfer through finite temperature difference. Other examples: viscosity, turbulence, free expansion, mixing.
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Answer: Zero, because entropy is a state function.
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Answer: Yes, if surroundings gain more entropy. But entropy of an isolated system/universe cannot decrease.
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Answer: There is no opposing pressure and the gas spreads spontaneously; returning it to initial volume requires external work and changes surroundings.
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Answer: ΔS = -Q/TH + Q/TC. Since TH > TC, Q/TC > Q/TH, so ΔS > 0.
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Answer: Entropy increases when energy becomes more spread out among accessible microscopic states.
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Answer: Entropy is a state function.
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Answer: Because it uses external work input to transfer heat from cold to hot.
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Answer: Infinitely slow frictionless isothermal expansion is an ideal reversible process.
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Answer: Free expansion of a gas, frictional motion, or heat flow from hot body to cold body.
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Answer: Entropy increases because liquid has more molecular freedom than solid.
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Answer: Heat flows from hot tea to cooler surroundings, increasing total entropy.
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Answer: No, that would violate the Clausius statement.
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Answer: The net work output obtained from part of the heat absorbed from the hot source.
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Answer: QH/TH = QC/TC for magnitudes in a reversible Carnot cycle.
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Answer: It has no entropy production due to irreversibility, so it wastes the least availability of energy.
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Answer: Refrigerator COP uses desired cooling QC/W, while heat pump COP uses desired heating QH/W.
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Answer: Both are true and the reason correctly explains the assertion.
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Answer: Assertion is false; in a reversible process ΔSuniverse = 0. Reason is true.
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Answer: Both true and reason correctly explains assertion.
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Answer: False.
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Answer: False.
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Answer: True.
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Answer: True.
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Answer: True.
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Answer: It must reject some heat to the surroundings, so efficiency cannot be 100%.
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Answer: It absorbs QC from inside, receives work W from compressor, and rejects QH = QC + W to the room.
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Answer: Positive.
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Answer: Entropy increases; the process is irreversible.
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Answer: Total entropy increases until thermal equilibrium is reached.
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Answer: The definition uses reversible heat transfer. For irreversible process, calculate ΔS using a reversible path between the same states.
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Answer: Friction converts organized mechanical energy into disordered thermal energy, spreading energy microscopically.
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Answer: Entropy increases because friction generates entropy even though Q = 0.
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Answer: ΔS = 300/300 = 1 J K-1.
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Answer: η = 1 - 350/700 = 0.5 = 50%.
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Answer: COPR = 240/(300 - 240) = 4.
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Answer: ΔS = -500/500 + 500/250 = -1 + 2 = 1 J K-1.
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Answer: 0.25 = 1 - 300/TH, so TH = 400 K.
13. Revision Notes
Must-Remember Statements
- Kelvin-Planck: no 100% efficient cyclic heat engine using a single reservoir.
- Clausius: heat cannot flow from cold to hot without work input.
- Both statements are equivalent forms of the second law.
- For an isolated system, ΔS ≥ 0.
- Entropy is a state function, not a path function.
Must-Remember Formulas
- ΔS = Qrev/T
- ΔS ≥ 0 for isolated system
- η = 1 - TC/TH
- COPR = TC/(TH - TC)
- COPHP = TH/(TH - TC)
Common Mistakes
- Using Celsius instead of kelvin in Carnot efficiency or ideal COP.
- Thinking entropy of a system can never decrease. It can decrease, but total entropy of isolated system cannot.
- Writing ΔS = Q/T for irreversible actual path without using reversible heat.
- Confusing refrigerator COP with heat engine efficiency.
- Assuming a reversible process is easy to achieve in real life. It is an ideal limit.
One-Line Exam Answers
- Heat flows naturally from hot to cold because total entropy increases.
- A refrigerator needs work because it transfers heat against natural direction.
- Carnot efficiency depends only on reservoir temperatures.
- Free expansion is irreversible and increases entropy.
- For a reversible cycle, entropy change of working substance and universe are zero.
If Second Law of Thermodynamics is not clear and you are looking for a Physics Tutor, contact Kumar Sir.
Phone: +91-9958461445 Email: kumarsirphysics@gmail.com