5.3 Enthalpy, Entropy, and Heat Capacity

5.3 Enthalpy, Entropy, and Heat Capacity

Quick Answer: Cp and Cv convert temperature changes into sensible Δh and Δu; latent heat covers phase change at constant T. For ideal gases, Cp − Cv = R and both u and h depend only on T. Entropy generation flags irreversibility; isentropic means ΔS = 0, the reversible adiabatic ideal.

Section 5.1 classified laws and properties; Section 5.2 fixed PvT behavior. This section turns those ideas into the heat-capacity and entropy relations you use on every heater, cooler, flash feed preheat, and compressor efficiency item in Domain B.

Definitions: U, H, Cp, Cv

Internal energy U is the thermodynamic store associated with molecular modes (for ideal gases, mainly T). Enthalpy is defined as:

H = U + PV

so for changes:

ΔH = ΔU + Δ(PV)

For constant pressure processes with only PdV work, Q_p = ΔH for a closed system—the operational reason enthalpy is so useful. For steady-flow devices, energy balances are naturally written in h.

Heat capacities (molar forms shown):

Cv = (∂Ū/∂T)_v , Cp = (∂H̄/∂T)_p

Ideal Gas Relations (Memorize)

For an ideal gas:

• U = U(T) only → ΔU = ∫ Cv dT
• H = H(T) only → ΔH = ∫ Cp dT
• Cp − Cv = R (molar); cp − cv = R_specific (mass basis)
• γ = Cp/Cv > 1 (often ≈ 1.4 for diatomic gases near ambient T)

If Cp is constant:

Δh = Cp ΔT , Δu = Cv ΔT

If Cp = a + bT + …, integrate polynomial forms as given in the stem or data sheet.

Sensible vs Latent Enthalpy Changes

Sensible heat changes temperature without changing phase. Latent heat changes phase (or sometimes reaction progress) at essentially constant temperature for a pure substance at fixed P.

Heating liquid from T1 to boiling, vaporizing, then superheating vapor is a three-segment path:

1. Liquid sensible: ∫ Cp,liquid dT from T1 to T_boil(P)
2. Latent: ΔH_vap at T_boil
3. Vapor sensible: ∫ Cp,vapor dT from T_boil to T2

Exam stems that give only “heat 1 kg of water from 25 °C to steam at 150 °C at 1 atm” expect you to include boiling latent heat—not a single Cp water liquid over the whole range.

Reference states. Tabulated enthalpies (steam tables, process simulators) use a reference (e.g., liquid at 0.01 °C, or ideal gas at 25 °C). Only differences ΔH matter in balances; never mix two tables with incompatible references without converting.

Entropy for Ideal Gases

Entropy s is a state function. For an ideal gas between (T1, P1) and (T2, P2):

Δs = ∫_{T1}^{T2} Cp dT / T − R ln(P2/P1)

(molar form). Equivalent volume form:

Δs = ∫ Cv dT / T + R ln(v2/v1)

Constant-γ isentropic relations (ideal gas):

T2/T1 = (P2/P1)^{(γ−1)/γ} = (v1/v2)^{γ−1}

These define the isentropic path used in compressor/turbine efficiency:

η_isen,compressor ≈ (h2s − h1)/(h2,actual − h1)
η_isen,turbine ≈ (h1 − h2,actual)/(h1 − h2s)

where state 2s is the exit state at the real exit pressure with s2s = s1.

Entropy Generation as an Irreversibility Indicator

For any process:

ΔS_system = ∫ δQ/T |_boundary + S_gen

with S_gen ≥ 0. For an adiabatic process, ∫ δQ/T = 0, so ΔS_system = S_gen ≥ 0. Thus a real adiabatic compressor produces s2 > s1.

Lost work / availability concepts (exergy) extend S_gen into “how much shaft work you sacrificed.” Licensing MCQs more often ask qualitative direction (s increases, efficiency < 100%) than full exergy algebra.

Worked Example: Sensible + Latent Duty

Problem. A feed preheater raises 1000 kg/h of pure liquid organic (approximate cp,liquid = 2.2 kJ/(kg·K)) from 30 °C to its boiling point 80 °C at the heater pressure, then vaporizes 40% of the stream. Δĥ_vap = 380 kJ/kg at 80 °C. Neglect pressure-drop effects and assume constant cp. Find the heat duty.

Step 1 — Sensible duty (all mass heated to 80 °C).

Q_sensible = m cp ΔT = 1000 kg/h × 2.2 kJ/(kg·K) × (80 − 30) K = 110,000 kJ/h = 110 MJ/h

Step 2 — Latent duty (40% vaporized).

m_vap = 0.40 × 1000 = 400 kg/h
Q_latent = 400 × 380 = 152,000 kJ/h = 152 MJ/h

Step 3 — Total.

Q_total = 110 + 152 = 262 MJ/h (about 72.8 kW if continuous steady power equivalent: 262,000 kJ/h ÷ 3600 s/h ≈ 72.8 kJ/s).

Exam lesson: Partial vaporization still requires full-stream sensible heat to the bubble point (for a pure fluid at fixed P), then latent only on the vaporized fraction. Mixtures need bubble-point and composition-dependent latent duties—Chapter 6.

Worked Example: Isentropic vs Actual Compression (Conceptual Numbers)

Ideal gas, constant Cp = 1.0 kJ/(kg·K), γ = 1.4, inlet T1 = 300 K, P2/P1 = 2.0.

Isentropic exit temperature:

T2s = T1 (P2/P1)^{(γ−1)/γ} = 300 × 2^{0.4/1.4} = 300 × 2^{0.286} ≈ 300 × 1.22 ≈ 366 K

Δh_s = Cp(T2s − T1) ≈ 1.0 × 66 ≈ 66 kJ/kg

If actual T2 = 390 K, Δh_actual = 90 kJ/kg, and

η_isen ≈ 66/90 ≈ 73%

Actual compression is adiabatic but not isentropic: s2 > s1, and more shaft work is required than the isentropic ideal.

Linking Back to First and Second Laws

UPDA Chemical Exam Workflow

1. Identify phases and whether T changes, phase changes, or both.
2. Choose mass or molar basis consistent with the Cp data.
3. Ideal gas? Use Δh = ∫ Cp dT and Cp − Cv = R if Cv is needed.
4. For machines, construct the isentropic end state at real exit P, then apply efficiency if given.
5. State clearly that adiabatic ≠ isentropic unless reversibility is assumed.

Common Traps

• Using Cp in a constant-volume ΔU calculation without converting via Cv = Cp − R (ideal gas).
• Omitting latent heat when steam or refrigerant changes phase.
• Applying Δs = Cp ln(T2/T1) at constant pressure when the process is not isobaric without the −R ln(P2/P1) term (ideal gas).
• Setting S_gen = 0 for a real insulated compressor.
• Adding sensible heats with inconsistent units (°C differences are fine for ΔT, but absolute T is required in ln(T2/T1)).

With laws, EOS, and heat-capacity/entropy tools in place, Chapter 6 applies them to phase equilibria, bubble and dew points, flash, and chemical reaction equilibrium—still under Domain B of the UPDA/MMUP Chemical study map.