Second law, Carnot cycle and entropy

Kelvin–Planck and Clausius statements, reversibility, the Carnot cycle and its efficiency and COP, the Clausius inequality, entropy and entropy generation.

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Why it matters

The first law says energy is conserved; the second law says which way processes go and how much of the heat you burn can ever become work. It sets the ceiling on the efficiency of every engine and the minimum power of every refrigerator, and through entropy it measures the waste caused by friction, throttling, mixing and heat transfer across large temperature differences — the losses an engineer is paid to reduce.

Key ideas

Heat engines, refrigerators and heat pumps. A heat engine receives Q_H from a high-temperature source, delivers net work W and rejects Q_L to a low-temperature sink: W = Q_H − Q_L and thermal efficiency η = W/Q_H. A refrigerator or heat pump does the reverse, using work W to move Q_L from cold to hot; their performance is measured by the coefficient of performance (COP), which can exceed 1.

Kelvin–Planck statement. It is impossible for a device operating in a cycle to receive heat from a single reservoir and produce an equal amount of work. So no engine can have η = 100%; some heat must always be rejected. A machine that violates this is a perpetual-motion machine of the second kind (PMM-2).

Clausius statement. It is impossible for a cyclic device to transfer heat from a colder body to a hotter body with no other effect. A refrigerator needs work input. The two statements are equivalent: violating one lets you build a device that violates the other.

Reversible and irreversible processes. A reversible process can be reversed leaving no trace on the system or surroundings. Real processes are irreversible because of friction, unrestrained (free) expansion, mixing of different substances, heat transfer through a finite temperature difference, throttling, inelastic deformation and electrical resistance. Reversible processes are idealisations that give the best possible performance.

The Carnot cycle. Two reversible isothermal processes (heat added at T_H, rejected at T_L) and two reversible adiabatic (isentropic) processes. Carnot's principles: (1) no engine working between two reservoirs can be more efficient than a reversible engine working between the same reservoirs; (2) all reversible engines between the same two reservoirs have the same efficiency, which depends only on the reservoir temperatures. This defines the thermodynamic (Kelvin) temperature scale: Q_H/Q_L = T_H/T_L for a reversible cycle.

Clausius inequality. For any cycle, ∮δQ/T ≤ 0; equality holds for a reversible cycle. A claimed cycle with ∮δQ/T > 0 is impossible.

Entropy. For a reversible process, dS = δQ_rev/T defines a property S (kJ/K). Because S is a property, ΔS between two states is the same for any path; to calculate it for an irreversible process, imagine a reversible path between the same end states. For a closed system, ΔS = ∫δQ/T + S_gen, where S_gen ≥ 0 is the entropy generated by irreversibilities.

Increase-of-entropy principle. For an isolated system (or system plus surroundings), ΔS ≥ 0: zero for reversible processes, positive for real ones. A process with ΔS_universe < 0 cannot happen. Entropy generation is a direct measure of lost work potential: lost work = T₀·S_gen, where T₀ is the surroundings temperature.

T–s diagram. Area under a reversible process curve on a T–s diagram is the heat transferred; a Carnot cycle is a rectangle. An isentropic process is a vertical line.

Formulas

η = W/Q_H = 1 − Q_L/Q_H

  • Any heat engine; Q_H, Q_L heat magnitudes (kJ).

η_Carnot = 1 − T_L/T_H

  • Reversible engine; T in K.

COP_R = Q_L/W = T_L/(T_H − T_L) (Carnot) COP_HP = Q_H/W = T_H/(T_H − T_L) (Carnot), and COP_HP = COP_R + 1

  • Refrigerator and heat pump; dimensionless.

∮ δQ/T ≤ 0

  • Clausius inequality.

ΔS = Q/T

  • Heat Q reversibly transferred at constant T (kJ/K).

ΔS = m·c·ln(T₂/T₁)

  • Incompressible solid or liquid with constant c.

ΔS = m·[c_p·ln(T₂/T₁) − R·ln(p₂/p₁)] = m·[c_v·ln(T₂/T₁) + R·ln(V₂/V₁)]

  • Ideal gas, any process.

S_gen = ΔS_system + ΔS_surroundings ≥ 0

  • Entropy generation (kJ/K).

Worked examples

Example 1 (standard): Carnot engine and the reversed cycle. Given: a Carnot engine receives 1000 kJ from a source at 800 K and rejects heat to a sink at 300 K. Find η, W and Q_L. Then find the COP if the same cycle is run in reverse as a heat pump.

  1. η = 1 − 300/800 = 0.625
  2. W = η·Q_H = 0.625 × 1000 = 625 kJ
  3. Q_L = Q_H − W = 375 kJ; check Q_L/Q_H = 300/800 = 0.375 ✓
  4. COP_HP = T_H/(T_H − T_L) = 800/500 = 1.6 (= 1/η) Answer: η = 62.5%, W = 625 kJ, Q_L = 375 kJ; COP_HP = 1.6

Example 2 (GATE level): entropy generated by mixing water. Given: 2 kg of water at 80 °C is mixed in an insulated vessel with 3 kg of water at 20 °C. c = 4.18 kJ/kg·K. Find the final temperature and the entropy generated.

  1. Energy balance (no heat, no work): T_f = (2 × 353.15 + 3 × 293.15)/5 = 317.15 K (44 °C).
  2. Hot water: ΔS₁ = 2 × 4.18 × ln(317.15/353.15) = −0.8989 kJ/K
  3. Cold water: ΔS₂ = 3 × 4.18 × ln(317.15/293.15) = +0.9868 kJ/K
  4. Insulated, so no entropy transfer with surroundings: S_gen = ΔS₁ + ΔS₂ = 0.0879 kJ/K Answer: T_f = 44 °C; S_gen ≈ 0.088 kJ/K (positive, so the process is irreversible, as expected)

Example 3 (quick): checking an inventor's claim. An engine is claimed to take 1000 kJ at 500 K and reject 550 kJ at 300 K. Claimed η = 45% but η_Carnot = 1 − 300/500 = 40%. Equivalently ∮δQ/T = 1000/500 − 550/300 = +0.167 kJ/K > 0. The claim is impossible.

Common mistakes

  • Using °C instead of K in Carnot efficiency, COP or entropy formulas.
  • Thinking an adiabatic process is always isentropic; it is isentropic only if it is also reversible.
  • Assuming entropy of a system can never decrease; it can, if enough entropy leaves with heat. Only the total for an isolated system cannot decrease.
  • Confusing COP_R and COP_HP (they differ by exactly 1 for the same cycle).
  • Calculating ΔS for an irreversible process as Q/T with the actual heat; use a reversible path between the same states.

For GATE PI

Expect Carnot efficiency and COP numericals, combined engine–refrigerator arrangements, feasibility checks using Carnot's limit or the Clausius inequality, entropy change for heating, mixing and ideal-gas processes, and conceptual one-markers on the Kelvin–Planck and Clausius statements and causes of irreversibility. Practise testing a claim three ways (efficiency, Clausius integral, entropy generation) — they always agree.

Quick check

  1. A Carnot refrigerator works between 270 K and 300 K. What is its COP?
  2. Can an adiabatic process have an entropy increase?
  3. 500 kJ of heat is added reversibly at a constant 500 K. What is ΔS?
  4. Which statement of the second law forbids a 100%-efficient engine?

Answers: 1. 270/30 = 9 2. Yes, if it is irreversible 3. 1 kJ/K 4. Kelvin–Planck

Interview questions

All Thermal and Fluids Engineering interview questions

Try answering each one aloud before you open it.

  1. 1.What is the second law of thermodynamics?Concept

    The second law says processes have a natural direction and that heat cannot be fully converted into work in a cycle. Kelvin–Planck: no cyclic device can take heat from a single reservoir and convert all of it into work. Clausius: heat cannot flow from a colder to a hotter body without some other effect, such as work input. In entropy form, the entropy of an isolated system never decreases; it stays constant only for reversible processes. Consequences include the Carnot limit η = 1 − T_L/T_H on engine efficiency.

  2. 2.Explain the Carnot cycle and its significance in thermodynamics.Concept

    The Carnot cycle is a theoretical thermodynamic cycle that provides the maximum possible efficiency for a heat engine operating between two temperature reservoirs. It consists of two isothermal processes and two adiabatic processes. The significance of the Carnot cycle lies in its ability to set an upper limit on the efficiency of real engines, serving as a benchmark for evaluating their performance.

  3. 3.What is entropy, and why is it important in thermodynamics?Concept

    Entropy is a measure of the disorder or randomness in a system. It is a central concept in the second law of thermodynamics, indicating the direction of spontaneous processes and the feasibility of energy conversions. Entropy is important because it helps predict the efficiency of thermodynamic cycles and the irreversibility of real-world processes.

  4. 4.Why is the Carnot cycle considered an ideal cycle?Application

    The Carnot cycle is considered an ideal cycle because it assumes no energy losses due to friction, unrestrained expansion, or heat transfer across finite temperature differences. It operates between two thermal reservoirs and achieves maximum efficiency by being completely reversible. This idealization helps in understanding the limits of efficiency for real engines.

  5. 5.What happens to the efficiency of a Carnot engine if the temperature of the cold reservoir is increased?Application

    If the temperature of the cold reservoir is increased while keeping the hot reservoir temperature constant, the efficiency of the Carnot engine decreases. This is because the efficiency of a Carnot engine is given by the formula η = 1 - (T_cold/T_hot), where T_cold and T_hot are the absolute temperatures of the cold and hot reservoirs, respectively.

  6. 6.How does the second law of thermodynamics apply to refrigerators?Application

    The second law of thermodynamics applies to refrigerators by dictating that work must be done to transfer heat from a colder body to a hotter body. Refrigerators operate by removing heat from the interior (cold reservoir) and expelling it to the surroundings (hot reservoir), requiring external work input to achieve this transfer against the natural direction of heat flow.

  7. 7.Calculate the efficiency of a Carnot engine operating between a hot reservoir at 500 K and a cold reservoir at 300 K.Numerical

    The efficiency of a Carnot engine is calculated using the formula η = 1 - (T_cold/T_hot). Here, T_hot = 500 K and T_cold = 300 K. Substituting these values, η = 1 - (300/500) = 1 - 0.6 = 0.4 or 40%.

  8. 8.A heat engine absorbs 1000 J of heat from a hot reservoir and expels 600 J to a cold reservoir. Calculate the engine's efficiency.Numerical

    The efficiency of a heat engine is given by η = (W_out/Q_in), where W_out is the work output and Q_in is the heat input. Here, W_out = Q_in - Q_out = 1000 J - 600 J = 400 J. Therefore, η = 400 J / 1000 J = 0.4 or 40%.

  9. 9.Explain why real engines cannot achieve the efficiency of a Carnot engine.Application

    Real engines cannot achieve the efficiency of a Carnot engine because they experience irreversibilities such as friction, heat losses, and non-instantaneous heat transfer. These factors lead to energy dissipation and deviations from the ideal reversible processes assumed in the Carnot cycle. As a result, real engines operate at lower efficiencies than the theoretical maximum set by the Carnot cycle.

  10. 10.What is the relationship between entropy and the reversibility of a process?Concept

    The relationship between entropy and the reversibility of a process is that reversible processes do not result in a net change in entropy, while irreversible processes increase the entropy of the system and its surroundings. In a reversible process, the system is always in equilibrium, and entropy changes are balanced by the surroundings, whereas irreversible processes lead to an overall increase in disorder.

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