Second law, entropy and Carnot cycle

Why it matters

The first law says how much energy is exchanged; the second law says which direction a process can go and how much work is the most you can ever get. It sets the maximum efficiency of power cycles, the minimum work for refrigeration, separation and compression, and it is the root of the Gibbs-energy criterion used for every phase- and reaction-equilibrium calculation later in this subject.

Key ideas

Statements of the second law.

• Kelvin-Planck: no device operating in a cycle can take heat from a single reservoir and convert it completely into work. A machine that did so would be a perpetual motion machine of the second kind.
• Clausius: no cyclic device can transfer heat from a colder body to a hotter one without work input. The two statements are equivalent: violating one lets you build a device that violates the other.

Entropy. For a reversible process, dS = δQ_rev/T defines the change in entropy S, a state function. For any process in a closed system, dS ≥ δQ/T_boundary; the difference is the entropy generated by irreversibilities (friction, heat transfer across a finite temperature difference, unrestrained expansion, mixing, chemical reaction). Entropy generation is never negative: zero for a reversible process, positive for a real one. For an isolated system (or system plus surroundings), ΔS_total ≥ 0.

Because S is a state function, ΔS between two states is computed along any convenient reversible path, even if the actual process was irreversible. The entropy generated shows up in the surroundings' balance, not in the system's ΔS.

Carnot cycle. Two reversible isothermal steps (heat in at T_H, heat out at T_C) and two reversible adiabatic (isentropic) steps. Carnot's theorem: no engine working between the same two reservoirs can be more efficient than a reversible one, and all reversible engines between those reservoirs have the same efficiency, which depends only on the two temperatures. That result defines the thermodynamic (Kelvin) temperature scale. For a reversible cycle Q_H/T_H = Q_C/T_C.

Reversed Carnot cycle. Run backwards it is a refrigerator or heat pump, with coefficients of performance that rise as the temperature lift (T_H − T_C) shrinks.

Entropy balance for open systems. At steady state, the entropy leaving with streams minus that entering, minus heat-transfer entropy terms, equals the entropy generation rate, which must be ≥ 0. This is the test used to check whether a proposed device is possible.

Ideal work and lost work. The maximum work obtainable from a change of state while exchanging heat only with surroundings at T₀ is the ideal work; the difference between ideal and actual work is the lost work, equal to T₀·S_gen.

Formulas

dS = δQ_rev / T ΔS_total = ΔS_system + ΔS_surroundings ≥ 0 η_Carnot = W/Q_H = 1 − T_C/T_H Q_H/T_H = Q_C/T_C (reversible cycle) COP_R = T_C / (T_H − T_C), COP_HP = T_H / (T_H − T_C) (Carnot) ΔS = Cp·ln(T₂/T₁) − R·ln(P₂/P₁) (ideal gas, molar, constant Cp) ΔS = Cv·ln(T₂/T₁) + R·ln(V₂/V₁) (ideal gas, molar, constant Cv) ΔS = m·c·ln(T₂/T₁) (incompressible solid or liquid) ΔS_reservoir = Q/T_reservoir Ṡ_gen = Σṁ_out·s_out − Σṁ_in·s_in − Σ Q̇_j/T_j ≥ 0 (steady flow) W_lost = T₀·S_gen

Symbols: S entropy (J/K), s specific entropy (J/kg·K); Q heat (J), Q_H heat from hot reservoir, Q_C heat rejected to cold reservoir; T absolute temperature (K) — always kelvin, never °C; W net work (J); η efficiency (dimensionless); COP coefficient of performance; Cp, Cv molar heat capacities (J/mol·K); c specific heat (J/kg·K); R = 8.314 J/(mol·K); T₀ surroundings temperature (K).

Worked examples

Example 1 (standard): entropy change of an ideal gas. Given: 1 mol of nitrogen (ideal gas, Cp = 29.1 J/mol·K) goes from 300 K, 1 bar to 500 K, 5 bar. Find ΔS.

1. ΔS = Cp·ln(T₂/T₁) − R·ln(P₂/P₁).
2. Temperature term: 29.1 × ln(500/300) = 29.1 × 0.5108 = 14.865 J/mol·K.
3. Pressure term: 8.314 × ln 5 = 8.314 × 1.6094 = 13.381 J/mol·K.
4. ΔS = 14.865 − 13.381 = 1.484 J/mol·K. Answer: ΔS = +1.48 J/(mol·K) (the compression nearly cancels the heating).

Example 2 (GATE level): entropy generation and lost work. Given: a 10 kg copper block (c = 0.385 kJ/kg·K) at 500 K is dropped into a large lake at 300 K. Find the entropy change of the block, of the lake, the entropy generated, and the maximum work that could have been obtained instead with surroundings at 300 K.

1. Heat given to the lake: Q = m·c·(T₁ − T₂) = 10 × 0.385 × 200 = 770 kJ.
2. Block: ΔS_block = m·c·ln(T₂/T₁) = 3.85 × ln(300/500) = 3.85 × (−0.5108) = −1.967 kJ/K.
3. Lake (reservoir): ΔS_lake = Q/T = 770/300 = +2.567 kJ/K.
4. S_gen = ΔS_total = −1.967 + 2.567 = +0.600 kJ/K (positive: irreversible, as it must be).
5. Maximum work: run reversible engines between the cooling block and the lake. W_max = Q − T₀·|ΔS_block| = 770 − 300 × 1.967 = 180 kJ.
6. Check: the lost work in the actual process is T₀·S_gen = 300 × 0.600 = 180 kJ, the same. Answer: ΔS_block = −1.97 kJ/K, ΔS_lake = +2.57 kJ/K, S_gen = 0.600 kJ/K, W_max = 180 kJ

Common mistakes

• Using °C in η = 1 − T_C/T_H or in Q/T; temperatures must be absolute.
• Saying the entropy of a system can never decrease. It can (the copper block's does); only the total for system plus surroundings cannot.
• Computing ΔS of an irreversible process as Q_actual/T. Use a reversible path between the same states.
• Treating a finite body as a reservoir: its temperature changes, so use m·c·ln(T₂/T₁), not Q/T.
• Assuming Carnot efficiency for a real engine; it is an upper bound.
• Forgetting the minus sign on the pressure term in the ideal-gas ΔS.

For GATE CH

• NAT problems on Carnot efficiency and COP, often chained (an engine driving a refrigerator), and on ΔS of ideal gases, mixing of two liquids or a hot body cooled by a reservoir.
• Feasibility checks: given claimed heat and work flows, test whether S_gen ≥ 0 or whether η exceeds the Carnot limit.
• Conceptual MCQs on the Kelvin-Planck and Clausius statements and state versus path functions.
• Practise the entropy balance for a steady-flow device and the ideal-work/lost-work relation.

Quick check

1. A Carnot engine works between 600 K and 300 K. What is its efficiency?
2. Can the entropy of a closed system decrease? Under what condition?
3. What is the COP of a Carnot refrigerator between 250 K and 300 K?
4. An inventor claims an engine takes 1000 kJ at 500 K, rejects 500 kJ at 300 K and produces 500 kJ of work. Is it possible?
5. Which two kinds of process make up the Carnot cycle?

Answers: 1. 0.5 (50%). 2. Yes, if it rejects heat; the surroundings' entropy then rises by at least as much. 3. 5. 4. No: η = 0.5 exceeds the Carnot limit of 0.4. 5. Two reversible isothermal and two reversible adiabatic (isentropic) processes.