Laws of Thermodynamics and Properties

FE Exam Weight: Thermodynamics and Heat Transfer is 9–14 questions (~10%) on the FE Other Disciplines module. Every equation below appears in the searchable NCEES FE Reference Handbook, so the skill being tested is finding the right relation and applying it with correct units — not memorization. The exam is open only to that Handbook plus an NCEES-approved calculator.

State, Properties, and Process

A thermodynamic property is any quantity whose value depends only on the current state, not on how the system got there: temperature T, pressure P, specific volume v, internal energy u, enthalpy h, and entropy s are all state (point) functions. Heat Q and work W are NOT properties — they are path-dependent; you cannot speak of "the heat in a system," only the heat that crosses a boundary during a process. Intensive properties (T, P, v, u, h, s) are independent of system size; extensive properties (V, U, H, S) scale with mass.

The state postulate says a simple compressible substance is fully fixed by two independent intensive properties — so knowing P and v (or T and x, the quality) lets you read every other property from a table.

Enthalpy h = u + Pv is defined for convenience: in any open (flow) device the combination u + Pv naturally appears because the flowing fluid carries both its internal energy and the flow work Pv needed to push it across the boundary. A classic FE trap is using temperatures in °C inside ratios or absolute-temperature formulas — entropy changes, the ideal-gas law, Carnot efficiency, and radiation ALL require kelvin (K = °C + 273.15).

The Four Laws

Zeroth Law: if A is in thermal equilibrium with C and B is in equilibrium with C, then A and B are in equilibrium — this is what makes temperature measurable and lets a thermometer work.

** For a closed system (fixed mass, no flow): Q − W = ΔU, or per unit mass q − w = Δu. With the standard sign convention, Q is positive when heat is added to the system and W is positive when the system does work on its surroundings.

For an open system at steady state, energy entering by mass and heat equals energy leaving by mass and shaft work: $\dot{Q} - \dot{W}_s = \dot{m}\Big[(h_2-h_1) + \frac{V_2^2 - V_1^2}{2} + g(z_2 - z_1)\Big]$ In most FE problems the kinetic and potential terms are negligible, leaving Q̇ − Ẇs = ṁ(h₂ − h₁) — note it is enthalpy, not internal energy, that appears in flow devices (turbines, compressors, nozzles, heat exchangers).

Second Law. The Kelvin–Planck statement: no cyclic device can convert heat entirely into work — some heat must be rejected to a cold reservoir. The Clausius statement: heat cannot move from cold to hot without work input. Quantitatively, for any process the entropy of the universe never decreases: $\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} \geq 0$ Equality holds only for a reversible (ideal) process; every real process is irreversible and generates entropy. Third Law: the entropy of a perfect crystal approaches zero as T → 0 K, setting an absolute entropy reference.

Specific Heats and Ideal-Gas Relations

The ideal-gas law is PV = mRT = nR̄T, where R is the specific gas constant (R = R̄/M, with universal R̄ = 8.314 kJ/(kmol·K)). For any ideal gas, internal energy and enthalpy depend on temperature ALONE:

Standard air values: cp = 1.005, cv = 0.718 kJ/(kg·K), R = 0.287 kJ/(kg·K), k = 1.4. Check: cp − cv = 1.005 − 0.718 = 0.287 = R. ✓

Reversible Processes for Ideal Gases

Many FE questions ask which property stays constant during a named process and how to compute the work:

Worked first-law example (closed system). A piston–cylinder holds 2 kg of air. During a process 150 kJ of heat is added and the air does 40 kJ of boundary work on the piston. Find the change in internal energy and, if cv = 0.718 kJ/(kg·K), the temperature rise.

Step 1 — First law: Q − W = ΔU → ΔU = 150 − 40 = 110 kJ (Q is +, work done BY the gas is +).

Step 2 — Temperature change: ΔU = m·cv·ΔT → ΔT = ΔU/(m·cv) = 110 / (2 × 0.718) = 110 / 1.436 = 76.6 K rise.

The most common error here is subtracting work the wrong way (using +W instead of −W) or applying ΔU = m·cp·ΔT — remember cv goes with internal energy, cp goes with enthalpy.

Worked isothermal-work example. 1 kg of air (R = 0.287) expands isothermally at 350 K from 100 kPa to 50 kPa. Since v₂/v₁ = P₁/P₂ = 2, W = mRT ln(v₂/v₁) = 1 × 0.287 × 350 × ln 2 = 100.45 × 0.693 = 69.6 kJ. Because T is constant, Δu = 0, so by the first law Q = W = 69.6 kJ — all the heat added leaves as work.

Exam tip: before plugging numbers, identify (1) closed vs. open system, (2) what is held constant, and (3) whether temperatures must be absolute. Doing those three checks first eliminates most thermodynamics mistakes on the FE.