Laws of Thermodynamics

The zeroth, first, second and third laws define temperature, energy conservation, the direction of processes and absolute zero. Learn energy balances for closed and open (steady-flow) systems and what the second law forbids.

Four laws govern all energy conversion. The first says energy can't be created or destroyed. The second says it naturally degrades — heat flows from hot to cold, and no engine can turn all heat into work. Together they explain why power stations need cooling towers and cars waste most of their fuel.

Energy balances based on the first law are the everyday calculation of mechanical engineers: heaters, boilers, turbines, compressors, HVAC.

An analogy

First law: you can't get something for nothing (energy bank balance). Second law: you can't even break even (every transaction charges a fee in lost work quality).

Zeroth law

If A and B are each in thermal equilibrium with C, they are in equilibrium with each other. This is why a thermometer (C) can compare temperatures.

First law — energy conservation

Heat added minus work done by the system equals the change in stored energy. Sign convention here: Q in is +, W out is +. For a cycle, ΔU = 0 so net heat in = net work out.

Steady-flow devices

For turbines, compressors, pumps, nozzles, heat exchangers and throttles, mass flow in = out and the steady-flow energy equation applies. Turbines: w = h₁ − h₂. Throttling valves: h₁ = h₂. Nozzles: enthalpy drop becomes kinetic energy.

Second law statements

Kelvin–Planck: no cyclic device can produce work from a single heat reservoir (100 % efficient engines are impossible). Clausius: heat can't flow from cold to hot without work input (refrigerators need power). Both are equivalent.

Third law and exergy

The entropy of a perfect crystal at 0 K is zero, fixing absolute entropy values. Combining the first and second laws gives exergy — the maximum useful work obtainable relative to the environment. Exergy analysis pinpoints where real plants waste potential work (typically the boiler combustion).

Formulas

Worked example: Heating water in a kettle

Given: 2 kg water, 20 → 80 °C; c = 4.18 kJ/kg·K; Kettle 2 kW, no losses

Find: Heat and time.

Answer: 502 kJ, ≈ 4.2 min

Worked example: Steam turbine power

Given: ṁ = 10 kg/s; h₁ = 3232 kJ/kg, h₂ = 2193 kJ/kg; Adiabatic, negligible KE/PE

Find: Power output.

Answer: ≈ 10.4 MW

Common mistakes

Applications

Key takeaways

Going further