The first law says energy is conserved; the second law says which way it can flow and how much of it can become work. Heat flows on its own only from hot to cold, no engine can turn all the heat it receives into work, and every real process generates entropy. Entropy generation, S_gen ≥ 0, is the quantitative test.
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Two statements of the second law
The Kelvin–Planck statement: no cyclic device can receive heat from a single reservoir and produce an equal amount of work. Every heat engine must reject some heat to a colder sink. The Clausius statement: heat cannot flow from a colder body to a hotter one without work input. The two statements are equivalent; breaking one would break the other.
The best any engine can do between a source at T_H and a sink at T_L is the reversible, or Carnot, efficiency η = 1 − T_L/T_H, with both temperatures absolute. Real engines fall below it because of friction, heat transfer across finite temperature differences, and mixing.
Entropy
Entropy S is a property, like volume or internal energy. For a reversible process its change is dS = δQ/T. Because it is a property, you can find ΔS between two states along any convenient reversible path, even if the real process was irreversible. For a solid or liquid with constant specific heat, that gives ΔS = m·c·ln(T₂/T₁).
The entropy balance for a closed system is ΔS = Σ(Q/T_b) + S_gen, where T_b is the boundary temperature where each heat transfer crosses. S_gen is zero for a reversible process, positive for a real one and never negative. A claimed process that needs S_gen < 0 is impossible.
Entropy is the tool behind the turbine and pump efficiencies used in the Rankine cycle. It builds on the first law for closed systems, which you need for the energy side of every problem.
Key formulas
Formulas used on this page, with the textbook each is cited from
Formula
Expression
Where
Cited from
Carnot (reversible) efficiency
η_rev = 1 − T_L / T_H
T in kelvin
Çengel & Boles, Thermodynamics: An Engineering Approach
Entropy generation for A. The cycle returns to its start, so S_gen = Q_L/T_L − Q_H/T_H, with Q_L = 400 − 180 = 220 kW.S_gen,A = 220 / 300 − 400 / 650 = 0.733 − 0.615 = +0.118 kW/K
Same test for B, with Q_L = 400 − 240 = 160 kW.S_gen,B = 160 / 300 − 400 / 650 = 0.533 − 0.615 = −0.082 kW/K
Answer: The best possible efficiency is 53.8%. Claim A (45%) is possible, with positive entropy generation. Claim B (60%) beats the Carnot limit and would need negative entropy generation, so it is impossible. Both tests always agree.
Example: Entropy generated by quenching a hot casting
Given
A 4 kg aluminium casting at 300 °C is dropped into 40 kg of water at 20 °C in an insulated tank.
Take c = 0.90 kJ/(kg·K) for aluminium and 4.18 kJ/(kg·K) for water, typical values treated as constant. Evaporation is neglected.
Find
The final temperature.
The entropy change of each and the entropy generated.
Solution
First law: the tank is insulated and rigid, so heat lost by the casting equals heat gained by the water.T_f = (4 × 0.90 × 300 + 40 × 4.18 × 20) / (4 × 0.90 + 40 × 4.18) = 25.9 °C = 299.05 K
Entropy change of the casting, with absolute temperatures.ΔS_Al = 4 × 0.90 × ln(299.05 / 573.15) = −2.342 kJ/K
Entropy change of the water.ΔS_w = 40 × 4.18 × ln(299.05 / 293.15) = +3.333 kJ/K
No heat crosses the tank boundary, so the entropy generated is the total change.S_gen = −2.342 + 3.333 = +0.991 kJ/K
Answer: Everything settles at about 25.9 °C. The casting loses 2.34 kJ/K of entropy and the water gains 3.33 kJ/K, so the quench generates about 0.99 kJ/K. It is positive, as it must be: heat flowed across a large temperature difference.
Common mistakes
Using °C in efficiency or entropy formulas. T_L/T_H and ln(T₂/T₁) need absolute temperatures in kelvin.
Thinking entropy can never decrease. The entropy of a system can fall if it rejects heat, as the casting does. Only the entropy generated, S_gen, can never be negative.
Comparing a real engine with 100%. Judge efficiency against the Carnot limit for the same temperatures, not against one.
Using the system temperature instead of the boundary temperature in Q/T. Each heat transfer is divided by the temperature where it crosses the boundary.
Forgetting that ΔS is path-independent. Entropy is a property. Compute ΔS from the end states along any reversible path.
In industry
Frequently asked questions
What is entropy in simple terms?
A property that measures how spread out energy is among the possible states of a system. In engineering it is mostly used as a bookkeeping tool: any process that generates entropy has lost some ability to do work.
Why can’t a heat engine be 100% efficient?
The Kelvin–Planck statement forbids it: some heat must be rejected to a colder sink. The Carnot efficiency 1 − T_L/T_H would only reach one at a sink of absolute zero.
Is a refrigerator a violation of the second law?
No. It moves heat from cold to hot, but only with work input, which the Clausius statement allows.
What does a reversible process mean in practice?
An ideal limit with no friction, no heat transfer across a finite temperature difference and no mixing. No real process is reversible, but slow, well-designed ones come close.