Calculating entropy changes & irreversibility

Drop a 0.500 kg copper block at 400 K into a large lake at 300 K. While it cools, its surface is colder than its core and the water beside it is warmer than the lake, so neither body has a single temperature and the real process has no ∫dQ/T to evaluate. Entropy is a state function: ΔS depends only on the two equilibrium end states, block at 400 K and block at 300 K. So replace the real process with a reversible surrogate joining the same states. Imagine the block handed down a ladder of reservoirs, each a hair cooler than the last, so that at every instant its temperature is defined and the heat it absorbs is dQᵣₑᵥ = C dT, negative while it cools. Along that ladder the integral is ∫C dT/T, and it gives the answer for the real process too. The lake needs its own surrogate: it is so large that its temperature stays at 300 K, so its change is simply Q/300 K. The rule for the rest of the topic: system and surroundings each get a reversible route to their own final state, and the two routes need not fit together into one reversible process. That mismatch is exactly where entropy gets generated.

For a body of constant heat capacity C the surrogate integral is ∫C dT/T = C ln(T₂/T₁), and the logarithm matters. Take the copper block: C = 0.500 kg × 385 J kg⁻¹ K⁻¹ = 192.5 J K⁻¹, so warming it from 300 K to 400 K gives ΔS = 192.5 ln(4/3) = 55.4 J K⁻¹. The heat involved is Q = 192.5 × 100 = 19 250 J. Divide that by the starting temperature and you get 64.2 J K⁻¹; divide by the final temperature and you get 48.1 J K⁻¹; the truth sits between them, because each parcel of heat is weighted by the temperature at which it arrived and the early parcels arrive cold. Dividing by the arithmetic mean 350 K gives 55.0 J K⁻¹, within 1%, which is why the shortcut Q/Tₘₑₐₙ survives in first-year work for small ratios: expand ln(1 + x) for small x and the two agree to second order. Double the temperature and the shortcut is 4% low; multiply it by ten and the shortcut is 29% low. The log form also fails if C moves across the range, as it does for a solid well below its Debye temperature; then C(T) stays inside the integral and the logarithm is replaced by whatever ∫C(T) dT/T evaluates to.

At a transition the temperature holds still while the heat flows, so ∫dQᵣₑᵥ/T becomes Q/Tₜᵣ with nothing left to integrate. Freezing 0.100 kg of water at 0 °C removes 0.100 × 334 000 J = 33.4 kJ at 273.15 K, so the water's entropy falls by 33 400/273.15 = 122 J K⁻¹. A negative ΔS for the system is perfectly legal; it is the total that may not fall, and freezer coils that took the heat at, say, −18.0 °C gained 33 400/255.15 = 131 J K⁻¹, more than the water lost. Because every stage is a reversible integral between equilibrium states, stages simply add. Ice at −10 °C becoming water at 20 °C is three terms: m c_ice ln(273.15/263.15) for warming the ice, mL_f/273.15 for melting it, and m c_water ln(293.15/273.15) for warming the water. What you may not do is skip a stage or let one end away from equilibrium: the surrogate must pass through the same intermediate states the sum assumes, and it must use the latent heat at the actual transition temperature, not a value averaged over the whole range.

For n moles of ideal gas along any reversible path, the first law reads dQᵣₑᵥ = dU + p dV = nCᵥ dT + (nRT/V) dV. Divide by T and both terms become exact differentials: dS = nCᵥ dT/T + nR dV/V. Integrate from state 1 to state 2 with Cᵥ constant and ΔS = nCᵥ ln(T₂/T₁) + nR ln(V₂/V₁). Nothing about the path survived: the two logarithms depend only on where the gas started and where it ended, which is the state-function property made explicit. Eliminating V with pV = nRT and using Cₚ = Cᵥ + R gives the equivalent ΔS = nCₚ ln(T₂/T₁) − nR ln(p₂/p₁), useful when pressures are what you measured. Two limiting cases check the result. Isothermal: the first term vanishes and ΔS = nR ln(V₂/V₁), which is Qᵣₑᵥ/T for the heat nRT ln(V₂/V₁) absorbed. Reversible adiabatic: ΔS = 0, so nCᵥ ln(T₂/T₁) = −nR ln(V₂/V₁), which rearranges to T V^(R/Cᵥ) = constant, and since R/Cᵥ = γ − 1 that is the familiar TV^(γ−1) = constant. Isentropic and adiabatic coincide here only because the path was reversible.

Open a partition and let 1.00 mol of ideal gas rush from V into an evacuated 2V. No heat crosses the insulated wall, Q = 0; the gas pushes on nothing, W = 0; so ΔU = 0 and, for an ideal gas whose internal energy depends only on temperature, T is unchanged. Now the temptation: Q = 0, so ∫dQ/T = 0, so ΔS = 0. That integral was taken along the real path, which is not reversible, and the definition does not permit it. The surrogate joining the same end states, same T and doubled V, is a reversible isothermal expansion, which absorbs Qᵣₑᵥ = nRT ln 2 while doing equal work, and ΔS = Qᵣₑᵥ/T = nR ln 2 = 8.314 × 0.693 = 5.76 J K⁻¹. Or read it off the two-term formula: the T term is zero and the V term is nR ln 2. The surroundings meanwhile exchanged nothing, so their change is zero and the whole 5.76 J K⁻¹ is generated. Boltzmann's count agrees: each molecule has twice the volume available, the multiplicity multiplies by 2^N, and k ln 2^N = Nk ln 2 = nR ln 2. The lost opportunity is the isothermal work nRT ln 2 the gas could have delivered and did not; at 300 K that is 1.73 kJ.

Let Q = 1000 J flow from a reservoir at Tₕ = 400 K to one at T_c = 300 K. The hot reservoir loses Q/Tₕ = 2.50 J K⁻¹; the cold one gains Q/T_c = 3.33 J K⁻¹; the sum S_gen = Q(1/T_c − 1/Tₕ) = 0.83 J K⁻¹ is positive because the same joules are worth more entropy at the lower temperature. The formula carries its own limiting case: with Tₕ = T_c + ΔT, S_gen ≈ Q ΔT/T², which vanishes as the gap closes, so conduction across a vanishing difference is the reversible heat transfer the surrogate paths assume. The generated entropy is also a price tag. A Carnot engine between the same reservoirs would have delivered W = Q(1 − T_c/Tₕ) = 250 J from the same 1000 J, and T_c S_gen = 300 × 0.833 = 250 J: lost work equals the cold temperature times the entropy generated. When the two bodies are finite rather than reservoirs, the logarithms return. Two equal bodies of heat capacity C at Tₕ and T_c meet at (Tₕ + T_c)/2, and S_gen = C ln[(Tₕ + T_c)²/(4 Tₕ T_c)], non-negative because the arithmetic mean beats the geometric mean, and zero only when the temperatures were already equal.