Second law, heat engines, Carnot cycle and entropy

Why it matters

The first law says energy is conserved; the second law says how much of it can actually be turned into work, and in which direction processes run. It sets the ceiling on the efficiency of every engine and the COP of every vehicle air-conditioner, and entropy is the tool used to locate turbine and compressor end states and to measure losses.

Key ideas

Thermal reservoir. A body so large that it can supply or absorb heat without its temperature changing (the atmosphere, a furnace, the combustion gases treated as a source).

Heat engine. A device working in a cycle that receives heat Q_H from a high-temperature reservoir, produces net work W and rejects heat Q_L to a low-temperature reservoir. Thermal efficiency η = W/Q_H. A refrigerator or heat pump runs the other way: it takes in work to move heat from cold to hot.

Statements of the second law.

• Kelvin–Planck: no device operating in a cycle can receive heat from a single reservoir and produce an equivalent amount of work. Every engine must reject some heat; η < 100%. A machine that violated this would be a perpetual-motion machine of the second kind (PMM2).
• Clausius: no device operating in a cycle can transfer heat from a colder to a hotter body without any other effect (i.e. without 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 system or surroundings. Irreversibilities: friction, heat transfer through a finite temperature difference, unrestrained expansion, mixing of different fluids, electrical resistance, combustion. Real processes are all irreversible; reversible ones are the ideal limit.

Carnot cycle. Two reversible isothermal processes (heat added at T_H, rejected at T_L) and two reversible adiabatic (isentropic) processes. On the T–s diagram it is a rectangle.

Carnot theorems. (1) No engine operating between two given reservoirs can be more efficient than a reversible engine between the same reservoirs. (2) All reversible engines between the same two reservoirs have the same efficiency, independent of the working fluid. This allows the thermodynamic (Kelvin) temperature scale to be defined by Q_H/Q_L = T_H/T_L.

Why practical engines are not Carnot engines. Isothermal heat transfer at finite rate needs a temperature difference, so a reversible isothermal process is infinitely slow; with a gas, the Carnot cycle has a very small net work per cycle compared with its peak pressures (low mean effective pressure); and with steam, compressing a wet mixture is impractical. Real cycles (Otto, Diesel, Rankine, Brayton) replace isotherms with constant-pressure or constant-volume processes.

Clausius inequality and entropy. For any cycle, ∮δQ/T ≤ 0: equal to zero for a reversible cycle, less than zero for an irreversible one, and a cycle giving > 0 is impossible. Because ∮(δQ/T)_rev = 0, the quantity dS = (δQ/T)_rev is a property, entropy S (J/K).

Increase of entropy principle. For any process, ΔS = ∫δQ/T + S_gen, with S_gen ≥ 0. For an isolated system (or system + surroundings) ΔS ≥ 0: entropy can be created, never destroyed. S_gen = 0 only for reversible processes, so entropy generation measures irreversibility. Entropy of a system can decrease if heat is removed from it, but the total for the universe still rises.

Physical meaning. On a T–s diagram the area under a reversible process line is the heat transferred. A reversible adiabatic process is isentropic; an irreversible adiabatic process always has s₂ > s₁.

Formulas

η = W / Q_H = 1 − Q_L / Q_H — any heat engine. Q_H, Q_L: heat magnitudes, J; W: net work, J.

η_Carnot = 1 − T_L / T_H — reversible engine; T in K only.

COP_R = Q_L / W = T_L / (T_H − T_L) (reversible); COP_HP = Q_H / W = T_H / (T_H − T_L) (reversible); COP_HP = COP_R + 1.

∮ δQ / T ≤ 0 — Clausius inequality.

dS = (δQ / T)_rev; ΔS = Q / T for a reversible isothermal process (or a reservoir). S: J/K.

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

Δs = c_p ln(T₂/T₁) − R ln(P₂/P₁) = c_v ln(T₂/T₁) + R ln(v₂/v₁) — ideal gas, kJ/kg·K.

S_gen = ΔS_system + ΔS_surroundings ≥ 0

S_gen = Q̇ (1/T_L − 1/T_H) — heat flow Q̇ (W) through a wall from T_H to T_L, result in W/K.

Worked examples

Example 1 (standard). A Carnot engine operates between 1000 K and 300 K and receives 600 kJ of heat per cycle. Find its efficiency, work and heat rejected. If it were reversed as a heat pump between the same temperatures, what would its COP be?

1. η = 1 − T_L/T_H = 1 − 300/1000 = 0.70 (70%).
2. W = η Q_H = 0.70 × 600 = 420 kJ.
3. Q_L = Q_H − W = 600 − 420 = 180 kJ. Check: Q_L/Q_H = 180/600 = 0.3 = T_L/T_H.
4. COP_HP = T_H/(T_H − T_L) = 1000/700 = 1.43 (= 1/η, as expected for a reversed Carnot engine).

Example 2 (GATE level). A reversible engine works between 800 K and 300 K and drives a reversible refrigerator that keeps a cold space at 250 K, also rejecting heat to the 300 K atmosphere. The engine receives 1000 kJ from the 800 K source. Find the heat removed from the cold space and the total heat rejected to the atmosphere.

1. Engine: η = 1 − 300/800 = 0.625, so W = 0.625 × 1000 = 625 kJ; engine rejects 1000 − 625 = 375 kJ.
2. Refrigerator: COP_R = T_L/(T_H − T_L) = 250/(300 − 250) = 5.
3. Heat removed from cold space: Q_L = COP_R × W = 5 × 625 = 3125 kJ.
4. Refrigerator rejects Q_L + W = 3125 + 625 = 3750 kJ.
5. Total to the atmosphere: 375 + 3750 = 4125 kJ. Energy check: in = 1000 + 3125 = 4125 kJ. Consistent.

Example 3 (entropy and the Clausius test). (a) 2 kg of water (c = 4.18 kJ/kg·K) is heated from 20 °C to 80 °C. ΔS = m c ln(T₂/T₁) = 2 × 4.18 × ln(353.15/293.15) = 1.557 kJ/K. (b) An inventor claims an engine receives 1000 kJ at 500 K and rejects 400 kJ at 300 K per cycle. ∮δQ/T = 1000/500 − 400/300 = 2.0 − 1.333 = +0.667 kJ/K > 0, so the claim is impossible (its efficiency 60% also exceeds Carnot's 40%).

Common mistakes

• Using °C in 1 − T_L/T_H or in ΔS = Q/T. Always convert to kelvin.
• Applying η_Carnot to a real engine's measured heat flows; for a real engine use η = 1 − Q_L/Q_H.
• Thinking entropy can never decrease. The entropy of a system can fall (when cooled); only the total entropy of system plus surroundings cannot.
• Writing ΔS = Q/T for a body whose temperature changes; integrate (m c ln T₂/T₁).
• Mixing up COP_R and COP_HP, or forgetting COP_HP = COP_R + 1 holds only when both are for the same machine.
• Signing Q in the Clausius inequality wrongly: heat in positive, heat out negative.

For GATE ME

Expect checks of claimed engines and refrigerators against Carnot limits and the Clausius inequality, coupled engine–refrigerator or engine–heat-pump systems, entropy change of solids, liquids and ideal gases, entropy generated in heat transfer or mixing, and conceptual MCQs on the Kelvin–Planck and Clausius statements and on reversibility. Practise drawing the cycle on a T–s diagram and reading heat as area.

Quick check

1. A Carnot engine works between 600 K and 300 K. What is its efficiency?
2. A reversible refrigerator keeps 250 K with surroundings at 300 K. What is its COP?
3. 500 J of heat is added reversibly at a constant 250 K. What is ΔS?
4. Can the entropy of a closed system decrease?

Answers: 1. 0.5 (50%). 2. 250/50 = 5. 3. 2 J/K. 4. Yes, if it rejects heat; the entropy of system plus surroundings still does not decrease.