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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.

CBSENEETJEE MainJEE AdvancedIBICSEIGCSEA-Level

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.

Meaning No cyclic engine can have 100% efficiency by using only one heat reservoir.

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.

Meaning A refrigerator cannot work with W = 0 while transferring heat from TC to TH.

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.

Definition ΔS = Qrev/T for reversible heat transfer at constant temperature.

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.

ΔS = Qrev/T

For reversible isothermal heat transfer at temperature T.

ΔS ≥ 0

For an isolated system. Equality for reversible, greater than zero for irreversible.

η = 1 - TC/TH

Maximum heat-engine efficiency for a Carnot engine.

COPR = TC/(TH - TC)

Ideal refrigerator COP between two reservoirs.

COPHP = TH/(TH - TC)

Ideal heat pump COP when heating is desired.

ΔSuniverse = ΔSsystem + ΔSsurroundings

Use this for spontaneity checks.

Reversible: ΔSuniverse = 0

No net entropy production.

Irreversible: ΔSuniverse > 0

Entropy is generated by irreversibility.

Common Entropy Changes

ProcessEntropy ChangePhysical Meaning
Melting at constant temperatureΔS = L/TSolid becomes liquid; molecular freedom increases.
Freezing at constant temperatureΔS = -L/TLiquid becomes solid; molecular freedom decreases.
Heat absorbed reversiblyPositiveSystem energy spreads more.
Heat rejected reversiblyNegativeSystem loses thermal energy.
Free expansion of gasPositiveGas 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.

Exam Line Entropy of a system can decrease, but entropy of an isolated system or universe cannot decrease.

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

Hot reservoir TH Engine Cold sink TC QH taken W delivered QC rejected Impossible: QC = 0 Kelvin-Planck: complete conversion of heat from a single reservoir into cyclic work is impossible.

Refrigerator and Clausius Statement

Hot reservoir TH Refrigerator Cold reservoir TC QC absorbed QH rejected W supplied Clausius: heat cannot move from cold to hot without work input.

Reversible Process Diagram

VP Same path can be retracedQuasi-static pathΔSuniverse = 0

Irreversible Process Diagram

Gas initially confined partition removed Free expansionCannot retrace itself without external workΔSuniverse > 0

Entropy Increase Diagram

Less dispersed natural change More dispersed Entropy increases when energy or matter spreads.

Carnot Engine Link with Entropy

Source TH CarnotEngine Sink TC QH QC W Reversible cycle: QH/TH = QC/TC, so ΔSuniverse = 0

Heat Flow from Hot to Cold Body

Hot bodyTH Cold bodyTC Q Entropy lost by hot body: -Q/TH Entropy gained by cold body: +Q/TC Since TH > TC, ΔStotal = Q/TC - Q/TH > 0

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.

Numerical 1. A system absorbs 500 J of heat reversibly at 250 K. Find entropy change.
Show Solution

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.

Numerical 2. A system rejects 900 J of heat reversibly at 300 K. Find entropy change of the system.
Show Solution

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.

Numerical 3. A Carnot engine works between 600 K and 300 K. Find maximum efficiency.
Show Solution

Given: TH = 600 K, TC = 300 K.

Formula: η = 1 - TC/TH.

Solution: η = 1 - 300/600 = 0.50.

Final Answer: 50%.

Numerical 4. A Carnot engine has source temperature 800 K and efficiency 40%. Find sink temperature.
Show Solution

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.

Numerical 5. Find ideal refrigerator COP between 300 K and 250 K.
Show Solution

Given: TH = 300 K, TC = 250 K.

Formula: COPR = TC/(TH - TC).

Solution: COPR = 250/(300 - 250) = 5.

Final Answer: 5.

Numerical 6. Find ideal heat pump COP between 300 K and 250 K.
Show Solution

Given: TH = 300 K, TC = 250 K.

Formula: COPHP = TH/(TH - TC).

Solution: COPHP = 300/50 = 6.

Final Answer: 6.

Numerical 7. Ice absorbs 334 kJ at 273 K while melting. Find entropy change.
Show Solution

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.

Numerical 8. Water freezes by releasing 334 kJ at 273 K. Find entropy change of water.
Show Solution

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.

Numerical 9. 100 J of heat flows from a 500 K body to a 250 K body. Find total entropy change.
Show Solution

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.

Numerical 10. 200 J of heat flows from 400 K to 300 K. Find entropy change of universe.
Show Solution

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.

Numerical 11. A heat engine absorbs 1200 J and rejects 800 J. Find efficiency and comment on second law.
Show Solution

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.

Numerical 12. A claimed engine absorbs 1000 J from a single reservoir and gives 1000 J work in a cycle. Is it possible?
Show Solution

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.

Numerical 13. A refrigerator removes 1500 J from cold chamber and uses 300 J work. Find heat rejected and COP.
Show Solution

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.

Numerical 14. An ideal refrigerator has COP 4 and TC = 240 K. Find TH.
Show Solution

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.

Numerical 15. An ideal heat pump works between 280 K and 350 K. Find COP.
Show Solution

Given: TC = 280 K, TH = 350 K.

Formula: COPHP = TH/(TH - TC).

Solution: COPHP = 350/(350 - 280) = 350/70 = 5.

Final Answer: 5.

Numerical 16. A body absorbs 2 kJ reversibly at 400 K. Find entropy change.
Show Solution

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.

Numerical 17. A body loses 2 kJ reversibly at 500 K. Find entropy change.
Show Solution

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.

Numerical 18. A reversible engine absorbs 600 J at 600 K. Find heat rejected at 300 K.
Show Solution

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.

Numerical 19. A reversible engine rejects 400 J at 320 K and works with source at 640 K. Find heat absorbed and work.
Show Solution

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.

Numerical 20. A Carnot engine absorbs 1000 J at 500 K and rejects heat at 300 K. Find entropy changes of source, sink and universe.
Show Solution

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.

Numerical 21. A real heat engine absorbs 1000 J at 500 K and rejects 700 J at 300 K. Find entropy change of universe.
Show Solution

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.

Numerical 22. A heat transfer of 60 J occurs reversibly at 300 K. Find entropy transferred.
Show Solution

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.

Numerical 23. A heat engine works between 900 K and 300 K. What is maximum possible efficiency?
Show Solution

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%.

Numerical 24. A student uses 27°C and 127°C directly in Carnot formula. Correct the temperatures and find efficiency.
Show Solution

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%.

Numerical 25. A refrigerator has TC = -23°C and TH = 27°C. Find ideal COP.
Show Solution

Given: TC = 250 K, TH = 300 K.

Formula: COPR = TC/(TH - TC).

Solution: COP = 250/(300 - 250) = 5.

Final Answer: 5.

Numerical 26. A heat pump has TC = 270 K and TH = 300 K. Find ideal COP.
Show Solution

Given: TC = 270 K, TH = 300 K.

Formula: COPHP = TH/(TH - TC).

Solution: COPHP = 300/30 = 10.

Final Answer: 10.

Numerical 27. 10 g of ice melts at 273 K. Given latent heat of fusion = 3.34 × 105 J kg-1. Find entropy change.
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.

Numerical 28. 500 J of heat flows from 1000 K to 500 K. Find total entropy change.
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.

Numerical 29. An isolated system undergoes a reversible process. What is ΔS?
Show Solution

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.

Numerical 30. An isolated system undergoes an irreversible process. What can be said about ΔS?
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.

Q1 CBSE. State the second law of thermodynamics in simple words.
Show Answer

Answer: It tells the natural direction of processes and states that total entropy of an isolated system cannot decrease.

Q2 CBSE. Write Kelvin-Planck statement.
Show Answer

Answer: No cyclic heat engine can convert all heat taken from a single reservoir completely into work.

Q3 CBSE. Write Clausius statement.
Show Answer

Answer: Heat cannot flow from colder body to hotter body without external work or another external effect.

Q4 CBSE. Why can a heat engine not be 100% efficient?
Show Answer

Answer: It must reject some heat to a cold sink during cyclic operation according to Kelvin-Planck statement.

Q5 CBSE. What is a reversible process?
Show Answer

Answer: A process that can be reversed leaving no change in system and surroundings.

Q6 CBSE. What is an irreversible process?
Show Answer

Answer: A process that cannot be exactly reversed without leaving changes in system or surroundings.

Q7 CBSE. Define entropy.
Show Answer

Answer: Entropy is a state function whose change for reversible heat transfer is ΔS = Qrev/T.

Q8 CBSE. Write SI unit of entropy.
Show Answer

Answer: J K-1.

Q9 NEET MCQ. A heat engine working in a cycle absorbs heat from one reservoir and converts it completely into work. It violates: (A) First law (B) Kelvin-Planck statement (C) Zeroth law (D) Gas law
Show Answer

Answer: (B) Kelvin-Planck statement.

Q10 NEET MCQ. Natural heat flow is from: (A) Cold to hot (B) Hot to cold (C) Low entropy to high entropy always (D) Low pressure to high pressure only
Show Answer

Answer: (B) Hot to cold.

Q11 NEET MCQ. For an isolated system, entropy change is: (A) always negative (B) zero or positive (C) always zero (D) independent of process
Show Answer

Answer: (B) zero or positive.

Q12 NEET MCQ. For reversible heat transfer Q at temperature T, entropy change is: (A) QT (B) T/Q (C) Q/T (D) Q + T
Show Answer

Answer: (C) Q/T.

Q13 NEET MCQ. COP of ideal refrigerator is: (A) TH/(TH-TC) (B) TC/(TH-TC) (C) 1 - TC/TH (D) TH/TC
Show Answer

Answer: (B) TC/(TH-TC).

Q14 JEE Main. A Carnot engine operates between TH and TC. Write its efficiency.
Show Answer

Answer: η = 1 - TC/TH.

Q15 JEE Main. Why must temperatures be in kelvin in Carnot formula?
Show Answer

Answer: Carnot efficiency is based on thermodynamic absolute temperature, so kelvin must be used.

Q16 JEE Main. What is entropy change of universe for a reversible cycle?
Show Answer

Answer: Zero.

Q17 JEE Main. What is entropy change of universe for an irreversible process?
Show Answer

Answer: Positive.

Q18 JEE Main. Name two causes of irreversibility.
Show Answer

Answer: Friction and heat transfer through finite temperature difference. Other examples: viscosity, turbulence, free expansion, mixing.

Q19 JEE Advanced. A system returns to initial state after a cycle. What is ΔS of the system?
Show Answer

Answer: Zero, because entropy is a state function.

Q20 JEE Advanced. Can entropy of a system decrease?
Show Answer

Answer: Yes, if surroundings gain more entropy. But entropy of an isolated system/universe cannot decrease.

Q21 JEE Advanced. Why is free expansion irreversible?
Show Answer

Answer: There is no opposing pressure and the gas spreads spontaneously; returning it to initial volume requires external work and changes surroundings.

Q22 JEE Advanced. For heat Q flowing from TH to TC, show why total entropy increases.
Show Answer

Answer: ΔS = -Q/TH + Q/TC. Since TH > TC, Q/TC > Q/TH, so ΔS > 0.

Q23 IB. Explain entropy using energy dispersal.
Show Answer

Answer: Entropy increases when energy becomes more spread out among accessible microscopic states.

Q24 IB. Is entropy a path function or state function?
Show Answer

Answer: Entropy is a state function.

Q25 IB. Why is a refrigerator not a violation of Clausius statement?
Show Answer

Answer: Because it uses external work input to transfer heat from cold to hot.

Q26 ICSE. Give one example of a reversible process.
Show Answer

Answer: Infinitely slow frictionless isothermal expansion is an ideal reversible process.

Q27 ICSE. Give one example of an irreversible process.
Show Answer

Answer: Free expansion of a gas, frictional motion, or heat flow from hot body to cold body.

Q28 ICSE. What happens to entropy during melting?
Show Answer

Answer: Entropy increases because liquid has more molecular freedom than solid.

Q29 IGCSE. Why does a hot cup of tea cool down naturally?
Show Answer

Answer: Heat flows from hot tea to cooler surroundings, increasing total entropy.

Q30 IGCSE. Can heat spontaneously flow from cold air to hot tea?
Show Answer

Answer: No, that would violate the Clausius statement.

Q31 IGCSE. What is meant by useful work in a heat engine?
Show Answer

Answer: The net work output obtained from part of the heat absorbed from the hot source.

Q32 A-Level. For a Carnot cycle, what is the entropy relation between QH, QC, TH, and TC?
Show Answer

Answer: QH/TH = QC/TC for magnitudes in a reversible Carnot cycle.

Q33 A-Level. Why is a reversible engine most efficient?
Show Answer

Answer: It has no entropy production due to irreversibility, so it wastes the least availability of energy.

Q34 A-Level. State the difference between COP of refrigerator and COP of heat pump.
Show Answer

Answer: Refrigerator COP uses desired cooling QC/W, while heat pump COP uses desired heating QH/W.

Q35 Assertion-Reason. Assertion: A real process is generally irreversible. Reason: Real processes involve dissipative effects like friction and finite temperature gradients.
Show Answer

Answer: Both are true and the reason correctly explains the assertion.

Q36 Assertion-Reason. Assertion: Entropy of the universe increases in a reversible process. Reason: Reversible processes are quasi-static.
Show Answer

Answer: Assertion is false; in a reversible process ΔSuniverse = 0. Reason is true.

Q37 Assertion-Reason. Assertion: A refrigerator needs work input. Reason: It transfers heat from lower temperature to higher temperature.
Show Answer

Answer: Both true and reason correctly explains assertion.

Q38 True/False. A heat engine can convert heat completely into work in a cycle.
Show Answer

Answer: False.

Q39 True/False. Entropy of an isolated system can decrease in a natural process.
Show Answer

Answer: False.

Q40 True/False. For a reversible process, entropy change of universe is zero.
Show Answer

Answer: True.

Q41 True/False. COP of refrigerator can be greater than one.
Show Answer

Answer: True.

Q42 True/False. A cyclic process has zero change in entropy of the working substance.
Show Answer

Answer: True.

Q43 Case Study. A car engine becomes hot and gives useful motion. Which second-law idea limits its performance?
Show Answer

Answer: It must reject some heat to the surroundings, so efficiency cannot be 100%.

Q44 Case Study. A refrigerator back side becomes warm while inside becomes cold. Explain.
Show Answer

Answer: It absorbs QC from inside, receives work W from compressor, and rejects QH = QC + W to the room.

Q45 Case Study. Ice at 0°C melts in a room. What is the sign of entropy change of ice?
Show Answer

Answer: Positive.

Q46 Case Study. Gas expands freely into vacuum in an insulated container. What happens to entropy?
Show Answer

Answer: Entropy increases; the process is irreversible.

Q47 Case Study. Two metal blocks at different temperatures are placed in contact in insulation. What happens to total entropy?
Show Answer

Answer: Total entropy increases until thermal equilibrium is reached.

Q48 Difficult Conceptual. Why is ΔS = Q/T not used directly for an irreversible path?
Show Answer

Answer: The definition uses reversible heat transfer. For irreversible process, calculate ΔS using a reversible path between the same states.

Q49 Difficult Conceptual. Why does friction increase entropy?
Show Answer

Answer: Friction converts organized mechanical energy into disordered thermal energy, spreading energy microscopically.

Q50 Difficult Conceptual. A system is compressed adiabatically with friction. What happens to entropy?
Show Answer

Answer: Entropy increases because friction generates entropy even though Q = 0.

Q51 Difficult Numerical. A body absorbs 300 J at 300 K reversibly. What is ΔS?
Show Answer

Answer: ΔS = 300/300 = 1 J K-1.

Q52 Difficult Numerical. A Carnot engine works between 700 K and 350 K. What is efficiency?
Show Answer

Answer: η = 1 - 350/700 = 0.5 = 50%.

Q53 Difficult Numerical. An ideal refrigerator works between 240 K and 300 K. Find COP.
Show Answer

Answer: COPR = 240/(300 - 240) = 4.

Q54 Difficult Numerical. 500 J heat flows from 500 K to 250 K. Find ΔStotal.
Show Answer

Answer: ΔS = -500/500 + 500/250 = -1 + 2 = 1 J K-1.

Q55 Difficult Numerical. A Carnot engine has efficiency 25% and TC = 300 K. Find TH.
Show Answer

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.

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