Second law, Carnot cycle and entropy

Kelvin–Planck and Clausius statements, reversibility, the Carnot cycle and principles, COP limits, the Clausius inequality, entropy and entropy generation, with feasibility and mixing examples.

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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 work you can get. It sets the ceiling on engine efficiency and on refrigerator COP, and it lets you reject impossible claims within a minute. Entropy, the property it introduces, is the quantity behind isentropic efficiencies of turbines and compressors, T–s diagrams of every cycle, and the irreversibility analysis in the next topic.

Key ideas

Two classical statements (they are equivalent — violating one violates the other).

  • Kelvin–Planck: no device operating in a cycle can receive heat from a single reservoir and convert all of it into work. A heat engine must reject some heat to a lower-temperature sink, so η < 100 %.
  • Clausius: no device operating in a cycle can transfer heat from a colder body to a hotter body with no other effect. A refrigerator or heat pump needs a work input.

Reversible and irreversible processes. A reversible process can be undone leaving no trace on system or surroundings. Causes of irreversibility: friction, heat transfer across a finite temperature difference, unrestrained expansion, mixing of different fluids, electrical resistance, inelastic deformation, combustion. Real processes are irreversible; reversible ones are the limits we compare against.

Carnot cycle. Two reversible isothermal processes (heat in at T_H, heat out at T_L) and two reversible adiabatic (isentropic) processes. On a T–s diagram it is a rectangle; on p–v for an ideal gas, two isotherms joined by two steeper adiabats.

Carnot principles.

  1. No engine working between two reservoirs is more efficient than a reversible engine between the same reservoirs.
  2. All reversible engines between the same two reservoirs have the same efficiency, whatever the working fluid. These let us define the thermodynamic (Kelvin) temperature scale: Q_H/Q_L = T_H/T_L for a reversible cycle. Always use kelvin.

Clausius inequality and entropy. For any cycle, ∮δQ/T ≤ 0, with equality only for a reversible cycle. Because ∮(δQ/T)_rev = 0, the quantity δQ_rev/T is the differential of a property — entropy S. Entropy is a property, so ΔS between two states is the same for any path; to compute it, imagine any reversible path between those states.

Entropy balance. For a closed system: ΔS = ∫δQ/T_b + S_gen, where T_b is the boundary temperature and S_gen ≥ 0 is entropy generated by irreversibilities. Consequences:

  • Increase-of-entropy principle: for an isolated system (or system + surroundings), ΔS_total = S_gen ≥ 0. Equals zero only for reversible processes; a negative value means the process is impossible.
  • A reversible adiabatic process is isentropic. An irreversible adiabatic process always increases entropy, which is why an actual turbine's exit state lies to the right of the isentropic one on an h–s chart.
  • Entropy of a system can decrease (e.g., a cooled gas), but only if more entropy is pushed into the surroundings.

Physical meaning. Statistically, entropy measures the number of microscopic arrangements consistent with the macroscopic state ("disorder"). For engineers the useful reading is: entropy generated measures lost work potential (W_lost = T₀ S_gen), which links to availability.

Formulas

η = W_net / Q_H = 1 − Q_L / Q_H — any heat engine. η_Carnot = 1 − T_L / T_H — reversible engine. T in K.

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

  • Q_H, Q_L: heat exchanged with hot and cold reservoirs (kJ, magnitudes), W: work (kJ), T_H, T_L: reservoir temperatures (K).

∮ δQ / T ≤ 0 — Clausius inequality.

dS = δQ_rev / T — definition of entropy. S in kJ/K; s in kJ/kg·K.

ΔS = Q / T — heat Q exchanged reversibly at constant T (a reservoir, or a phase change).

ΔS = m c ln(T₂ / T₁) — solid or liquid with constant specific heat c (kJ/kg·K).

Δs = c_p ln(T₂/T₁) − R ln(p₂/p₁) = c_v ln(T₂/T₁) + R ln(v₂/v₁) — ideal gas, constant specific heats. R: gas constant (kJ/kg·K).

T₂/T₁ = (p₂/p₁)^((γ−1)/γ) — isentropic process of an ideal gas; γ = c_p/c_v.

S_gen = ΔS_system + ΔS_surroundings ≥ 0 — second law for an isolated combination.

Worked examples

Example 1 (feasibility check). An inventor claims an engine receiving 1000 kJ from a source at 1000 K, rejecting heat at 300 K and producing 750 kJ of work. Is it possible?

  1. η_Carnot = 1 − T_L/T_H = 1 − 300/1000 = 0.70.
  2. Claimed η = 750/1000 = 0.75.
  3. 0.75 > 0.70, which violates the Carnot principle. Clausius check: Q_L = 250 kJ, so ∮δQ/T = 1000/1000 − 250/300 = 1 − 0.833 = +0.167 kJ/K > 0, also impossible.

The claim is impossible (maximum possible work is 700 kJ).

Example 2 (GATE level, entropy generation in mixing). 2 kg of water at 80 °C is mixed adiabatically with 3 kg of water at 20 °C at constant pressure. Take c = 4.18 kJ/kg·K. Find the final temperature and the entropy generated.

  1. Energy balance: T_f = (2 × 80 + 3 × 20)/(2 + 3) = 220/5 = 44 °C = 317.15 K.
  2. Hot water: ΔS₁ = m c ln(T_f/T₁) = 2 × 4.18 × ln(317.15/353.15) = 8.36 × (−0.10752) = −0.8989 kJ/K.
  3. Cold water: ΔS₂ = 3 × 4.18 × ln(317.15/293.15) = 12.54 × 0.07869 = +0.9868 kJ/K.
  4. The vessel is adiabatic, so no entropy crosses the boundary: S_gen = ΔS₁ + ΔS₂ = 0.0879 kJ/K.

T_f = 44 °C, S_gen ≈ 0.088 kJ/K (positive, so the process is irreversible and possible)

Example 3 (refrigerator). A Carnot refrigerator extracts 500 J from a space at 250 K and rejects heat to surroundings at 350 K. Find the work input.

  1. COP_R = T_L/(T_H − T_L) = 250/100 = 2.5.
  2. W = Q_L / COP = 500/2.5 = 200 J. Heat rejected Q_H = 500 + 200 = 700 J.

W = 200 J

Common mistakes

  • Using °C in Carnot efficiency or COP. 1 − 27/127 is meaningless; use 300 K and 400 K.
  • Thinking a reversible engine using a "better" fluid can beat Carnot efficiency. It cannot; the fluid does not matter.
  • Concluding that entropy of a system can never decrease. Only the total for an isolated system cannot decrease.
  • Computing ΔS for an irreversible process as ∫δQ/T of the actual heat. Use a reversible path between the same end states; the actual δQ/T falls short by S_gen.
  • Mixing up COP of a heat pump and of a refrigerator, or forgetting COP_HP = COP_R + 1.
  • Assuming adiabatic means isentropic. Only a reversible adiabatic process is isentropic.

For GATE ME

Typical questions: Carnot efficiency and COP with reservoir temperatures; checking whether a claimed engine or refrigerator is possible; an engine driving a refrigerator or heat pump in combination; entropy change of solids, liquids and ideal gases; entropy generated by heat transfer across a finite ΔT or by mixing; and T–s areas as heat. Practise using the Clausius inequality as a quick feasibility test and computing ΔS along a convenient reversible path.

Quick check

  1. A reversible engine works between 900 K and 300 K. What is its efficiency?
  2. What is COP_HP for a Carnot heat pump between 270 K and 300 K?
  3. 1000 kJ of heat flows from a reservoir at 800 K to one at 400 K. Find the entropy generated.
  4. Can the entropy of a system decrease? Under what condition?
  5. Is every adiabatic process isentropic?

Answers: 1. 66.7 %. 2. 300/30 = 10. 3. −1000/800 + 1000/400 = 1.25 kJ/K. 4. Yes, if heat leaves the system; the surroundings then gain at least as much entropy. 5. No; only a reversible adiabatic process is isentropic.

Interview questions

All Thermal Engineering interview questions

Try answering each one aloud before you open it.

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

    It states the direction of natural processes and the limit on converting heat into work. Kelvin–Planck: no cyclic device can turn all the heat from a single reservoir into work, so an engine must reject heat to a sink. Clausius: heat cannot pass from a colder to a hotter body without some other effect, such as work input. In entropy form, the total entropy of an isolated system never decreases; it stays constant only for reversible processes.

  2. 2.Explain the Carnot cycle and its significance in thermal engineering.Concept

    The Carnot cycle is a fully reversible cycle of two isothermal processes (heat added at T_H, rejected at T_L) and two reversible adiabatic (isentropic) processes; on a T–s diagram it is a rectangle. Its efficiency 1 − T_L/T_H depends only on the reservoir temperatures, and no engine between the same reservoirs can beat it. It is the benchmark for real cycles and shows that efficiency improves by adding heat at a higher temperature or rejecting it at a lower one.

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

    Entropy is a property defined by dS = δQ_rev/T; it follows from the Clausius inequality, and statistically it measures the number of microscopic arrangements behind a macroscopic state. Its practical importance is that the entropy generated by irreversibilities is never negative, which tells you whether a process is possible and in which direction it runs. Entropy generated times the ambient temperature equals the work potential lost, and isentropic processes give the ideal reference for turbines, compressors and nozzles.

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

    Every process in it is reversible: heat is exchanged with each reservoir across an infinitesimal temperature difference, and the expansion and compression are frictionless and adiabatic. It therefore gives the highest efficiency any engine can have between the same two reservoir temperatures. It is impractical because isothermal heat transfer across a vanishing ΔT needs infinite time or area, and the net work per cycle is small compared with the gross work.

  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 is dependent on the temperature difference between the hot and cold reservoirs, and a smaller difference results in lower efficiency.

  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 use work input to remove heat from the interior (cold reservoir) and expel it to the surroundings (hot reservoir), thus maintaining a lower temperature inside.

  7. 7.Why can't a real engine be as efficient as a Carnot engine?Application

    A real engine cannot be as efficient as a Carnot engine because real processes involve irreversibilities such as friction, heat losses, and non-instantaneous heat transfer. These factors increase entropy and reduce the efficiency compared to the idealized, reversible processes of a Carnot cycle.

  8. 8.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 η of a Carnot engine is given by η = 1 - (T_cold / T_hot). Substituting the given temperatures, η = 1 - (300 / 500) = 1 - 0.6 = 0.4 or 40%.

  9. 9.A heat engine absorbs 1500 J of heat from a hot reservoir and expels 900 J to a cold reservoir. Calculate the engine's efficiency.Numerical

    The efficiency η of a heat engine is given by η = (W_out / Q_in), where W_out is the work done by the engine and Q_in is the heat absorbed. Here, W_out = Q_in - Q_out = 1500 J - 900 J = 600 J. Thus, η = 600 J / 1500 J = 0.4 or 40%.

  10. 10.Explain how entropy change can be used to determine the feasibility of a thermodynamic process.Application

    Add the entropy change of the system and of everything it exchanges heat with (system plus surroundings, treated as isolated). If the total is positive, the process is possible and irreversible; if zero, it is reversible; if negative, it cannot happen. The same test in cycle form is the Clausius inequality ∮δQ/T ≤ 0, which quickly rules out claimed engines or refrigerators that beat Carnot limits.

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