Second law: Kelvin-Planck and Clausius statements

Heat engines, refrigerators and heat pumps with efficiency and COP, the Kelvin–Planck and Clausius statements of the second law, their equivalence, and perpetual-motion machines of the second kind.

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Why it matters

The first law says energy is conserved; it does not say which way processes run or how much heat can become work. The second law answers both: it explains why every aircraft engine must reject heat through its exhaust, why a refrigerator or air-cycle cooling pack needs a power input, and it sets the upper limits that real designs are measured against.

Key ideas

Thermal energy reservoir. A body so large that it can absorb or supply any amount of heat without changing temperature (the atmosphere, the ocean, a large furnace). A source supplies heat; a sink absorbs it.

Heat engine. A device operating in a cycle that receives heat Q_H from a high-temperature source, converts part of it to net work W and rejects the rest Q_L to a low-temperature sink. By the first law over a cycle, W = Q_H − Q_L. Thermal efficiency η = W/Q_H = 1 − Q_L/Q_H.

Refrigerator and heat pump. Cycles that take heat Q_L from a cold region and deliver Q_H to a warm region with a work input W = Q_H − Q_L. Performance is expressed by the coefficient of performance (COP), which can exceed 1:

  • Refrigerator (purpose: remove Q_L): COP_R = Q_L/W
  • Heat pump (purpose: deliver Q_H): COP_HP = Q_H/W
  • Between the same reservoirs with the same Q and W: COP_HP = COP_R + 1

Kelvin–Planck statement. It is impossible for any device operating in a cycle to receive heat from a single reservoir and produce an equal amount of net work. Every heat engine must reject some heat, so η < 100%. A machine that would violate it is a perpetual-motion machine of the second kind (PMM2) — note that it would not violate the first law.

Clausius statement. It is impossible to construct a device operating in a cycle whose sole effect is to transfer heat from a cooler body to a hotter body. Heat flows from cold to hot only with an external effect such as work input — which is exactly what a refrigerator does, so refrigerators do not violate it.

Equivalence. The two statements are equivalent: a violation of one implies a violation of the other.

  • If a Clausius-violating device moved Q_L from sink to source unaided, pairing it with an ordinary engine that rejects Q_L would give a combined device taking net heat (Q_H − Q_L) from the source alone and producing equal work — a Kelvin–Planck violation.
  • If a Kelvin–Planck-violating engine turned Q from the source fully into work and that work drove a refrigerator, the combination would move heat from cold to hot with no net work — a Clausius violation.

What the second law adds. Processes have a natural direction (heat flows hot → cold, gases expand into vacuum, friction turns work into heat), and those spontaneous processes are irreversible. The quantitative tools — Carnot limits, the absolute temperature scale and entropy — follow in the next topics.

Formulas

W = Q_H − Q_L (cyclic device, first law) η = W / Q_H = 1 − Q_L / Q_H COP_R = Q_L / W = Q_L / (Q_H − Q_L) COP_HP = Q_H / W = Q_H / (Q_H − Q_L) COP_HP = COP_R + 1

  • Q_H: heat exchanged with the high-temperature reservoir per cycle (or rate), kJ or kW, taken positive
  • Q_L: heat exchanged with the low-temperature reservoir, kJ or kW, taken positive
  • W: net work per cycle or net power, kJ or kW
  • η: thermal efficiency (fraction); COP: dimensionless

Here all quantities are magnitudes and the direction is shown by the device diagram.

Worked examples

Example 1 (standard). (a) An engine receives 500 kJ of heat per cycle and produces 200 kJ of work. Find η and the heat rejected. (b) A refrigerator removes 5 kW from a cold store using 1.25 kW of power. Find its COP, the heat rejected, and its COP if used as a heat pump.

  1. (a) η = W/Q_H = 200/500 = 0.40; Q_L = Q_H − W = 300 kJ
  2. (b) COP_R = Q_L/W = 5/1.25 = 4.0
  3. Q_H = Q_L + W = 5 + 1.25 = 6.25 kW
  4. COP_HP = Q_H/W = 6.25/1.25 = 5.0 (= COP_R + 1 ✓)

Answer: η = 40%, Q_L = 300 kJ; COP_R = 4, Q_H = 6.25 kW, COP_HP = 5

Example 2 (GATE level). A heat engine receives 100 kW from a source at 1000 K and rejects heat to the atmosphere at 300 K. Its efficiency is 50% of the reversible (Carnot) value. All its work drives a refrigerator that keeps a space at 250 K, rejecting heat to the same atmosphere, with COP_R = 4. Find the refrigeration effect and the total heat rejected to the atmosphere, and check that the refrigerator is possible.

  1. Reversible limit: η_max = 1 − T_L/T_H = 1 − 300/1000 = 0.70 (developed in the next topic)
  2. Actual η = 0.5 × 0.70 = 0.35, so W = 0.35 × 100 = 35 kW
  3. Engine rejects Q_L,e = 100 − 35 = 65 kW
  4. Refrigeration effect Q_L,r = COP_R·W = 4 × 35 = 140 kW
  5. Refrigerator rejects Q_H,r = Q_L,r + W = 140 + 35 = 175 kW
  6. Total to atmosphere = 65 + 175 = 240 kW (check: 100 kW in + 140 kW in = 240 kW out ✓)
  7. Feasibility: the best possible COP_R = T_L/(T_H − T_L) = 250/50 = 5 > 4, so the refrigerator is possible.

Answer: refrigeration effect 140 kW; 240 kW rejected to the atmosphere

Common mistakes

  • Thinking a refrigerator violates the Clausius statement — it uses work, so the transfer is not its sole effect.
  • Believing a PMM2 violates the first law. It conserves energy; it violates the second law.
  • Expecting COP to be below 1 like an efficiency. COP values of 3–5 are normal.
  • Using COP_R where COP_HP is asked (they differ by exactly 1 for the same machine).
  • Forgetting that the statements apply to devices working in a cycle. A single isothermal expansion of an ideal gas converts heat entirely to work, but the gas does not return to its initial state.

For GATE AE

Expect one-mark questions on which statement a described device violates, PMM1 versus PMM2, and the relation COP_HP = COP_R + 1. Numericals combine heat engines with refrigerators or heat pumps (engine-driven refrigerators, checking a claimed device against the limits). Practise drawing each device with its reservoirs and arrows before writing energy balances.

Quick check

  1. An engine absorbs 600 J and rejects 400 J per cycle. Find its efficiency.
  2. A refrigerator has COP_R = 3. What is its COP when used as a heat pump between the same reservoirs?
  3. Which statement does a device violate if it converts all heat from one reservoir into work in a cycle?
  4. Does a PMM2 violate the first law?
  5. A refrigerator removes 150 J from its interior and rejects 200 J. Find the work input.

Answers: 1. 33.3%. 2. 4. 3. Kelvin–Planck. 4. No. 5. 50 J.

Interview questions

All Engineering Thermodynamics interview questions

Try answering each one aloud before you open it.

  1. 1.What is the Kelvin-Planck statement of the second law of thermodynamics?Concept

    The Kelvin-Planck statement of the second law of thermodynamics asserts that it is impossible to construct a heat engine that operates in a cycle and produces no effect other than the extraction of heat from a single reservoir and the performance of an equivalent amount of work. This means that no heat engine can be 100% efficient, as some energy must always be rejected to a lower temperature reservoir.

  2. 2.Explain the Clausius statement of the second law of thermodynamics.Concept

    The Clausius statement of the second law of thermodynamics states that it is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a cooler body to a hotter body. This implies that heat cannot spontaneously flow from a colder region to a hotter region without external work being performed on the system.

  3. 3.How are the Kelvin-Planck and Clausius statements of the second law of thermodynamics related?Concept

    The Kelvin-Planck and Clausius statements are two equivalent expressions of the second law of thermodynamics. They are related in that a violation of one would imply a violation of the other. Both statements emphasize the impossibility of achieving certain types of perpetual motion machines, highlighting the inherent limitations in energy conversion processes.

  4. 4.Why is the second law of thermodynamics important in the design of heat engines?Application

    The second law of thermodynamics is crucial in the design of heat engines because it sets the fundamental limits on the efficiency of energy conversion processes. It informs engineers that no engine can be 100% efficient, as some energy will always be lost as waste heat. This understanding helps in optimizing engine designs to achieve the best possible efficiency within these constraints.

  5. 5.What would happen if a heat engine violated the Kelvin-Planck statement?Application

    It would exchange heat with a single reservoir and convert all of that heat into net work over a cycle, with no heat rejected — a perpetual-motion machine of the second kind (PMM2). Such a machine would not violate the first law, since energy is conserved; it violates the second law. Coupled to a refrigerator, it would also let heat flow from cold to hot with no net external effect, so it would violate the Clausius statement too. No such device has ever been built, which is the experimental basis of the law.

  6. 6.Can a refrigerator work without violating the Clausius statement? Explain.Application

    Yes, a refrigerator can work without violating the Clausius statement. Refrigerators operate by using external work to transfer heat from a cooler region (inside the refrigerator) to a warmer region (outside the refrigerator). This process does not violate the Clausius statement because it involves external work, which is necessary for heat to flow from a cold to a hot body.

  7. 7.How does the concept of entropy relate to the second law of thermodynamics?Concept

    Entropy is a measure of the disorder or randomness in a system, and it is central to the second law of thermodynamics. The second law states that the total entropy of an isolated system can never decrease over time. This implies that natural processes tend to move towards a state of maximum entropy, or disorder, and this principle helps explain the directionality of thermodynamic processes.

  8. 8.A heat engine absorbs 500 J of heat from a hot reservoir and rejects 300 J to a cold reservoir. Calculate the efficiency of the engine.Numerical

    The efficiency (η) of a heat engine is given by the formula η = (W_out / Q_in) × 100%, where W_out is the work output and Q_in is the heat input. Here, W_out = Q_in - Q_out = 500 J - 300 J = 200 J. Therefore, η = (200 J / 500 J) × 100% = 40%. The efficiency of the engine is 40%.

  9. 9.If a refrigerator removes 150 J of heat from its interior and expels 200 J to the surroundings, how much work does it require?Numerical

    Over a cycle the first law gives Q_H = Q_L + W, so the work input is W = Q_H − Q_L = 200 − 150 = 50 J. This is work done on the refrigerator, typically by the compressor motor. Its coefficient of performance is COP_R = Q_L/W = 150/50 = 3.

  10. 10.What is the significance of the Carnot cycle in relation to the second law of thermodynamics?Concept

    The Carnot cycle is significant because it represents an idealized heat engine cycle that operates with maximum possible efficiency between two temperature reservoirs. According to the second law of thermodynamics, no real engine can be more efficient than a Carnot engine operating between the same two reservoirs. The Carnot cycle provides a benchmark for evaluating the performance of real engines and highlights the theoretical limits imposed by the second law.

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