Kelvin-Planck statement: No process whose sole result is to extract heat from a reservoir and convert it entirely into work. Clausius statement: No process whose sole result is to transfer heat from cold to hot reservoir.
-- NCERT Class 11 Physics, Ch. 11, p. 236Second Law Thermodynamics
8 MCQs9-step worked example
Source: NCERT ThermodynamicsPYQ coverage: NEET 2023Official key: NTA-verifiedLast updated: 26 Sep 2026
Second Law Thermodynamics, explained for NEET
The second law of thermodynamics addresses what the first law cannot: direction. The first law says energy is conserved — it does not say which processes actually happen. The second law fills that gap.
Kelvin–Planck statement: No process is possible whose sole result is the absorption of heat from a reservoir and its complete conversion into work (NCERT Class 11 Physics Chapter 11, page 236). A heat engine must reject some heat to a cold reservoir; 100% efficiency is impossible.
Clausius statement: No process is possible whose sole result is the transfer of heat from a colder body to a hotter body. Spontaneous heat flow goes hot → cold; reversing it requires external work (a refrigerator).
These two statements are equivalent — violating one implies violating the other. NEET occasionally tests this equivalence as a conceptual recall item.
Why it matters for NEET: Questions on the second law typically test whether you can distinguish what the law forbids from what it permits. A common confusion is assuming the second law forbids heat flowing from cold to hot entirely — it does not. It forbids this as the sole result, without any other change. A refrigerator transfers heat from cold to hot, but it consumes work to do so, which is permitted.
Another frequent confusion: aspirants conflate "efficiency < 100%" (a consequence of the second law) with "efficiency depends on the working substance." The Carnot theorem — also a consequence of the second law — states that efficiency depends only on reservoir temperatures, not on the working substance, for reversible engines.
Watch out: when a question asks "which of the following violates the second law," look for the option claiming a sole result. If external work or another compensating change is mentioned, the process may be allowed.
Can you answer these Second Law Thermodynamics MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which of the following is the Kelvin–Planck statement of the second law of thermodynamics?
Show answer and why every option is right or wrong
Answer: C. C is correct. The Kelvin–Planck statement states that no process can have as its sole result the absorption of heat from a single reservoir and its complete conversion into work (NCERT Class 11 Physics Chapter 11, page 236).
Why A is wrong: A describes an incomplete version of the Clausius statement — the Clausius statement specifies 'sole result,' not an absolute prohibition. This is not the Kelvin–Planck statement.
Why B is wrong: B is the first law of thermodynamics (energy conservation), not the second law.
Why D is wrong: D is incorrect — the second law implies entropy of an isolated system never decreases (it increases or stays constant), not that it 'always decreases.'
MCQ 2Easy RecallPractice
The Clausius statement of the second law of thermodynamics states that:
Show answer and why every option is right or wrong
Answer: A. A is correct. The Clausius statement asserts that no process can have as its sole result the transfer of heat from a colder body to a hotter body (NCERT Class 11 Physics Chapter 11, page 236).
Why B is wrong: B omits the critical qualifier 'sole result.' Heat can flow from cold to hot (e.g., in a refrigerator) as long as external work is supplied — the second law does not forbid that.
Why C is wrong: C is true (e.g., friction converts work to heat) but is a consequence of the first law, not a statement of the second law.
Why D is wrong: D is false — even reversible engines operating between two reservoirs have efficiency less than 100% unless the cold reservoir is at 0 K (which is unattainable).
MCQ 3Easy RecallPractice
The Kelvin–Planck and Clausius statements of the second law of thermodynamics are:
Show answer and why every option is right or wrong
Answer: D. D is correct. The two statements are logically equivalent. It can be shown that if the Kelvin–Planck statement is violated, the Clausius statement is also violated, and vice versa (NCERT Class 11 Physics Chapter 11, page 236).
Why A is wrong: A is incorrect — the two statements are not contradictory; they express the same physical law in different terms (one about engines, one about heat transfer).
Why B is wrong: B is incorrect — the statements are logically connected; a proof of their equivalence exists and is part of the NCERT treatment.
Why C is wrong: C is incorrect — the second law applies to all thermodynamic systems, not just ideal gases.
MCQ 4Concept TrapPractice
A refrigerator transfers heat from a cold compartment to a warmer room. Does this violate the second law of thermodynamics?
Show answer and why every option is right or wrong
Answer: C. C is correct. The Clausius statement forbids heat transfer from cold to hot only as the sole result. A refrigerator consumes electrical work to drive the transfer, so there is an additional change in the surroundings, and the second law is not violated.
Why A is wrong: A misreads the Clausius statement — it forbids cold-to-hot transfer as the sole result. Since the refrigerator requires work input, this is not the sole result.
Why B is wrong: B confuses the two statements. The Kelvin–Planck statement concerns heat engines (converting heat fully to work), not refrigerators.
Why D is wrong: D is incorrect — the second law applies to all thermodynamic processes including refrigerators, heat pumps, and natural processes.
MCQ 5Concept TrapPractice
An inventor claims to have built a heat engine that absorbs 1000 J of heat from a single thermal reservoir and converts all of it into 1000 J of work, with no other effect. This claim:
Show answer and why every option is right or wrong
Answer: B. B is correct. The claim describes complete conversion of heat into work as the sole result, which directly violates the Kelvin–Planck statement. Note that it does not violate the first law (energy is conserved: 1000 J in = 1000 J out).
Why A is wrong: A is partially right that the first law is satisfied (energy is conserved), but the claim still violates the second law — so 'consistent with the first law only' is misleading but the real problem is the second law violation, which this option ignores.
Why C is wrong: C has it backwards — the first law is satisfied (energy is conserved at 1000 J), but the second law is violated.
Why D is wrong: D is incorrect — while the first law is satisfied, the second law is violated because the sole result is complete conversion of heat to work from a single reservoir.
MCQ 6Direct ApplicationPractice
A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. What is the maximum efficiency of this engine?
Show answer and why every option is right or wrong
Answer: A. A is correct. The Carnot efficiency is η = 1 − T_cold/T_hot = 1 − 300/600 = 0.50 = 50%. This is the maximum efficiency any engine can achieve between these reservoirs, as a consequence of the second law.
Why B is wrong: B (75%) does not follow from the Carnot expression 1 − T_cold/T_hot = 1 − 300/600 = 50%; reaching 75% would need a cold reservoir at 150 K.
Why C is wrong: C is incorrect — 100% efficiency would mean complete conversion of heat to work, violating the Kelvin–Planck statement. This is only possible if T_cold = 0 K, which is unattainable.
Why D is wrong: D (25%) is half the correct value. The Carnot efficiency is 1 − T_cold/T_hot = 1 − 300/600 = 0.50, i.e. 50%.
MCQ 7Direct ApplicationPractice
A Carnot engine has an efficiency of 40%. If the temperature of the hot reservoir is 500 K, what is the temperature of the cold reservoir?
Show answer and why every option is right or wrong
Answer: B. B is correct. η = 1 − T_cold/T_hot → 0.40 = 1 − T_cold/500 → T_cold/500 = 0.60 → T_cold = 300 K.
Why A is wrong: A (200 K) results from computing T_cold = η × T_hot = 0.40 × 500 = 200 K, which confuses the efficiency fraction with the temperature ratio.
Why C is wrong: C (250 K) results from computing T_cold = T_hot/2 = 250 K, ignoring the given efficiency value.
Why D is wrong: D (350 K) results from computing T_cold = T_hot − η × T_hot incorrectly as 500 − 0.30 × 500 = 350 K, using the wrong coefficient.
MCQ 8Direct ApplicationPractice
For a Carnot engine operating between 800 K and 400 K, the engine absorbs 2000 J from the hot reservoir per cycle. How much heat is rejected to the cold reservoir per cycle?
Show answer and why every option is right or wrong
Answer: D. D is correct. Carnot efficiency η = 1 − 400/800 = 0.50. Work done W = η × Q_H = 0.50 × 2000 = 1000 J. Heat rejected Q_C = Q_H − W = 2000 − 1000 = 1000 J.
Why A is wrong: A (500 J) results from squaring the temperature ratio: Q_C = Q_H × (T_C/T_H)² = 2000 × 0.25. The ratio enters once, Q_C/Q_H = T_C/T_H.
Why B is wrong: B (2000 J) would mean zero work is done (W = 0), which implies 0% efficiency — this contradicts the temperature ratio.
Why C is wrong: C (4000 J) results from inverting the temperature ratio, Q_C = Q_H × T_H/T_C = 2000 × 2. The cold reservoir receives less heat than the hot one supplies: Q_C = Q_H × T_C/T_H.
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How do you solve a Second Law Thermodynamics question? A worked example
- 1
Given
A Carnot engine operates between a source at T_H = 727 °C and a sink at T_C = 227 °C. The engine absorbs Q_H = 1.00 × 10³ J per cycle from the source.
- 2
Required
(a) The efficiency of the engine.
(b) The work done per cycle.
(c) The heat rejected to the sink per cycle. - 3
Concept
The second law implies that no engine operating between two reservoirs can exceed the Carnot efficiency, which depends only on the reservoir temperatures (not the working substance).
- 4
Formula
Carnot efficiency: η = 1 − T_C / T_H (temperatures must be in kelvin).
Work: W = η × Q_H.
Heat rejected: Q_C = Q_H − W. - 5
Substitution
Convert temperatures: T_H = 727 + 273 = 1000 K, T_C = 227 + 273 = 500 K.
η = 1 − 500/1000 = 1 − 0.500 = 0.500 (50.0%).
W = 0.500 × 1.00 × 10³ = 5.00 × 10² J.
Q_C = 1.00 × 10³ − 5.00 × 10² = 5.00 × 10² J. - 6
Calculation
The arithmetic is completed above. Note: 273 (the Celsius-to-Kelvin offset) and the temperature values 727, 227 are exact given values in this problem; they do not limit significant figures. The heat Q_H = 1.00 × 10³ J has 3 significant figures, so answers are reported to 3 significant figures.
- 7
Final answer
(a) η = 50.0%
(b) W = 5.00 × 10² J
(c) Q_C = 5.00 × 10² J - 8
Common trap
Forgetting to convert °C to K before applying the Carnot formula. Using T_H = 727 and T_C = 227 directly gives η = 1 − 227/727 ≈ 68.8%, which is wrong. Always convert to absolute temperature first.
- 9
Similar NEET-style question
A Carnot engine works between 527 °C and 127 °C. If 600 J of heat is supplied per cycle, find the work output and the heat rejected.
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What to remember before solving Second Law Thermodynamics questions
More in Thermodynamics: 2 exam traps and mistakes · 4 formulas · 2 question patterns from its other lessons.
Second Law Thermodynamics questions from past NEET papers
1 question from NEET 2023. Answers verified against NTA official keys. — click to collapse
NEET 2023
115°C
2100°C
3200°C
427°C
NTA Answer: Option 4(final)
Sources
NCERT refs: Class 11 Physics Chapter 11, p.236
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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