Properties, State and the First Law
System and surroundings, state properties, work and heat, the first law for closed and open systems, enthalpy and steady-flow devices.
System, surroundings and state
Thermodynamics starts by drawing a boundary. Everything inside is the system; everything outside is the surroundings. A closed system (a sealed cylinder of gas) exchanges energy but not mass; an open system or control volume (a turbine, a pump, a boiler) has mass flowing through it. An isolated system exchanges neither.
The condition of a substance is its state, described by properties: pressure p, temperature T, specific volume v (m³/kg), density ρ = 1 ÷ v, internal energy u, enthalpy h and entropy s (all per kilogram, in kJ/kg or kJ/kg·K). For a pure substance, two independent properties fix the state of a single phase; the rest follow from tables or equations. Properties are state functions: a change in a property depends only on the end states, not the path. Work and heat are not properties; they depend on the path.
Always use absolute pressure and temperature in thermodynamic equations: pabs = pgauge + 101.325 kPa, and T in kelvin (K = °C + 273.15).
Work and heat
Work is energy transferred by a force acting through a distance; for a gas expanding against a piston it is W = ∫ p dV, the area under the curve on a p–V diagram. Heat is energy transferred because of a temperature difference. Both are energy in transit: a system holds energy, never heat or work. Heat added to a substance with no phase change raises its temperature by Q = m c ΔT, where c is the specific heat (4.186 kJ/kg·K for liquid water, 1.005 for air at constant pressure, 0.47 for steel).
The first law for a closed system
Energy is conserved. For a closed system the change in stored energy equals the heat added minus the work done by the system:
Q − W = ΔU + ΔKE + ΔPE (usually just Q − W = m (u₂ − u₁))
Worked example: heating water in a tank
A 3 kW element heats 2 kg of water from 20 °C to 80 °C in a rigid, insulated tank. Neglect the tank’s own heat capacity.
- Energy needed Q = m c ΔT = 2 × 4.186 × 60 = 502.3 kJ.
- Time = Q ÷ power = 502.3 ÷ 3 = 167 s (about 2.8 minutes).
- In practice add the tank, losses and the heater’s efficiency: allow 10 to 30 percent more.
Open systems and enthalpy
In a flowing system the fluid also carries flow work p v across the boundary as it enters and leaves. Combining it with internal energy gives enthalpy, h = u + p v, the energy a flowing kilogram carries. The steady-flow energy equation (one inlet, one outlet) is:
Q̇ − Ẇ = ṁ [ (h₂ − h₁) + (V₂² − V₁²) ÷ 2 + g (z₂ − z₁) ]
For an ideal gas h changes with temperature only: Δh = cp ΔT. For water and steam, h comes from tables (see Steam, Water and Property Tables). This is the same energy balance as the pipe-flow Bernoulli equation of Fluid Mechanics and Pipe Flow with the heat and internal-energy terms added.
Steady-flow devices
| Device | What it does | Simplified energy balance |
|---|---|---|
| Heater / boiler / condenser | adds or removes heat; no work, small pressure change | Q̇ = ṁ (h₂ − h₁) |
| Turbine | extracts work from an expanding fluid; adiabatic | Ẇ = ṁ (h₁ − h₂) |
| Compressor / pump | adds work to raise pressure; adiabatic | Ẇ = ṁ (h₂ − h₁) |
| Nozzle | converts enthalpy into velocity | V₂ = √(2 (h₁ − h₂) + V₁²) |
| Throttle valve | drops pressure with no work or heat | h₂ = h₁ |
| Mixing chamber / heat exchanger | combines or exchanges between streams | Σ ṁ h in = Σ ṁ h out |
Worked example: four devices
- Air heater: 0.5 kg/s from 20 °C to 120 °C: Q̇ = 0.5 × 1.005 × 100 = 50.3 kW.
- Nozzle: an enthalpy drop of 50 kJ/kg from rest gives V = √(2 × 50 000) = 316.2 m/s.
- Compressor: 0.2 kg/s of air from 100 kPa and 300 K to 600 kPa (isentropic): T₂ = 300 × 60.2857 = 500.6 K, Ẇ = 0.2 × 1.005 × 200.6 = 40.3 kW.
- Throttle: saturated liquid at 1 MPa (h = 762.51) throttled to 100 kPa stays at h = 762.51 but now exceeds the liquid enthalpy at 100 kPa (417.51), so a fraction x = (762.51 − 417.51) ÷ 2 257.5 = 15.3 percent flashes to steam.
- Steam turbine (the Rankine example): h₁ = 3 213.6 and h₂ = 2 143.8 kJ/kg give 1 070 kJ/kg of shaft work (see Steam Power Cycles: Rankine).
Units and sign convention
Work is positive when the system does work on the surroundings (a turbine) and heat is positive when it enters the system. Some textbooks use the opposite sign for work, so check which one a source uses. Keep kJ and kW together (1 kW = 1 kJ/s), and write molar quantities in kmol and kJ/kmol. Use the Energy, Pressure and Temperature converters when mixing unit systems.
Related
- Next: Ideal Gases and Processes.
- Fluid Mechanics and Pipe Flow for the mechanical energy equation.
- Formula sheets for a printable thermodynamics sheet.
- Thermodynamics and Steam: the whole guide.
FAQ
What is the difference between internal energy and enthalpy?
Internal energy u is the energy stored in the molecules. Enthalpy h = u + p v adds the flow work needed to push the fluid across a boundary. Use h for flowing fluids and u for a fixed mass.
Why does a throttle keep enthalpy constant?
There is no work, no heat (it is short and fast) and the kinetic energy is about the same before and after, so h₁ = h₂ by the steady-flow equation. The temperature of a real gas or a liquid–vapor mixture can still change.
Is heat a property?
No. A system holds internal energy, not heat. Heat and work describe energy crossing the boundary and depend on the path.
Do I need absolute pressure?
Yes for gas laws and tables. Gauge pressure is only for differences from the atmosphere.
References
- Çengel, Y. A. and Boles, M. A., Thermodynamics: An Engineering Approach, McGraw-Hill (first law, steady-flow devices).
- Moran, M. J., et al., Fundamentals of Engineering Thermodynamics, Wiley.
- Sonntag, R. E., Borgnakke, C., Fundamentals of Thermodynamics, Wiley.
- NIST Chemistry WebBook and IAPWS-IF97 for water and steam properties.
Last reviewed: October 2026. Specific heats and table values are typical.