Second Law, Entropy and Availability

The first law says energy is conserved; the second law says which way processes go and how much work can be obtained. It introduces entropy and sets the maximum efficiency of any engine.

Statements of the second law

• Kelvin–Planck: no cyclic device can take heat from one reservoir and convert it all into work.
• Clausius: heat cannot flow from a colder body to a hotter one without external work.
• The two statements are equivalent.

Heat engines, refrigerators and heat pumps

• Heat engine efficiency η = W/Q_H = 1 − Q_L/Q_H.
• Refrigerator COP_R = Q_L/W. Heat pump COP_HP = Q_H/W = COP_R + 1.

Carnot (reversible) limits between reservoirs at T_H and T_L (in kelvin):

• η_max = 1 − T_L/T_H
• COP_R,max = T_L/(T_H − T_L); COP_HP,max = T_H/(T_H − T_L)

No engine between the same reservoirs can beat Carnot, and all reversible engines between them have the same efficiency.

Entropy

• Clausius inequality: ∮δQ/T ≤ 0 (= 0 for reversible cycles).
• dS = δQ_rev/T. For any process, ΔS ≥ ∫δQ/T.
• Principle of increase of entropy: for an isolated system (or system + surroundings), ΔS_total ≥ 0.

Entropy change of an ideal gas:

Δs = c_p ln(T₂/T₁) − R ln(p₂/p₁) = c_v ln(T₂/T₁) + R ln(v₂/v₁)

For a solid or liquid with specific heat c: Δs = c ln(T₂/T₁). For heat Q added to a reservoir at T: ΔS = Q/T.

A reversible adiabatic process is isentropic: for an ideal gas, pv^γ = constant and T₂/T₁ = (p₂/p₁)^((γ−1)/γ).

Availability (exergy)

The available energy of heat Q at temperature T, with surroundings at T₀: AE = Q(1 − T₀/T). The rest, Q T₀/T, is unavailable.

Irreversibility (lost work) I = T₀ ΔS_gen (Gouy–Stodola theorem). Second-law efficiency = actual useful work / maximum possible work.

Worked example

1 kg of water at 90 °C is mixed with 1 kg at 10 °C (c = 4.18 kJ/kg·K). Find the entropy generated.

• Final temperature 50 °C = 323.15 K.
• ΔS = 4.18 [ln(323.15/363.15) + ln(323.15/283.15)] = 4.18 × (−0.1167 + 0.1321) = 0.0645 kJ/K (positive, as it must be).

Practice