The Second Law, Entropy & Heat Engines

You’ll be able to

State the second law of thermodynamics in terms of entropy
Compute the efficiency of a heat engine from heat and work
Calculate the maximum (Carnot) efficiency set by reservoir temperatures

The direction of time: entropy

The first law says energy is conserved, but it never says which way processes run. That job belongs to the second law: the total entropy of an isolated system — a measure of disorder, or how spread out the energy is — never decreases. Heat flows hot → cold, gases mix, and organized energy dissipates, all spontaneously. The reverse never happens on its own. Entropy is the arrow that points thermodynamics forward in time.

How a heat engine works

A heat engine exploits the natural flow of heat to do useful work. It draws heat Q_H from a hot reservoir, converts part of it into work W, and must dump the leftover Q_C into a cold reservoir. By energy conservation, W = Q_H − Q_C. The second law forbids dumping nothing — some waste heat is unavoidable, which is exactly why no engine is ever 100% efficient.

Thermal efficiency

e = W / Q_H = 1 − (Q_C / Q_H)

Efficiency is the fraction of input heat turned into work. It is always between 0 and 1 (0–100%).

Worked example

A heat engine absorbs 1000 J from a hot reservoir each cycle and exhausts 600 J to a cold reservoir. Find the work output and the efficiency.

1.Work per cycle: W = Q_H − Q_C = 1000 − 600 = 400 J.
2.Efficiency: e = W / Q_H = 400 / 1000 = 0.40.
3.Express as a percentage: 0.40 × 100% = 40%.

Answer: W = 400 J and e = 0.40 (40%)

Carnot (maximum) efficiency

e_c = 1 − (T_C / T_H)

The best efficiency any engine can reach between two reservoirs. T_C and T_H must be in kelvin. No real engine beats this ideal limit.

Checkpoint

A heat engine absorbs 800 J of heat and produces 200 J of work each cycle. What is its efficiency?

Watch out

Carnot efficiency uses only the reservoir temperatures, and they must be in kelvin. Plugging in Celsius values inflates the answer dramatically, because the offset zero of the Celsius scale distorts the ratio T_C/T_H.

Checkpoint

What is the maximum possible efficiency of an engine operating between reservoirs at 27 °C and 327 °C?

Checkpoint

During a spontaneous, irreversible process inside an isolated system, the total entropy of the system must:

On the exam

Compare the two efficiencies: e = 1 − Q_C/Q_H uses the actual heat flows, while e_c = 1 − T_C/T_H is the temperature-set ceiling. A real engine’s efficiency is always less than its Carnot limit — if a problem’s engine beats Carnot, you have an error.

Answer the 3 checkpoints as you read.