Entropy is a state function; heat is a transfer

For an infinitesimal reversible heat transfer at absolute temperature T,

To find the entropy difference between equilibrium states, integrate along a reversible path connecting them, even if the actual process was irreversible. One must not replace this with “actual heat divided by temperature” for every process.

For a closed system, entropy can enter with heat and can be produced internally by irreversibility. For an isolated system, the second law says ΔS≥0. A subsystem can lose entropy if enough is exported to its surroundings. This is why a refrigerator can cool its contents without violating the law.

A simple heat transfer exposes the inequality

Transfer 100 J from a very large 400 K reservoir to a very large 300 K reservoir. Their temperatures remain essentially fixed. The hot reservoir loses 100/400=0.250 J/K; the cold one gains 100/300=0.333 J/K. Total entropy rises by 0.0833 J/K.

The reverse transfer without some other change would decrease total entropy. Energy conservation alone would allow it; the second law excludes it. Transfer across a finite temperature difference is irreversible. “Reversible” is a limiting process that can be undone without a net change in system and surroundings, not merely something that runs backward mechanically.

For the insulated ideal-gas free expansion discussed in the first-law entry, actual Q=W=0, but entropy rises. Calculate it using a hypothetical reversible isothermal path between the same states:

For one mole doubling its volume this is 5.76 J/K. No heat secretly entered during free expansion; the reversible path is a calculation device for a state difference.

Derive the heat-engine ceiling

A cyclic engine absorbs heat magnitude QH from a hot reservoir at TH, rejects Q꜀ to a cold reservoir at T꜀, and delivers W=QH-Q꜀. Define efficiency

After one cycle the engine itself has no net entropy change. The two reservoirs have

Therefore

The equality is the Carnot limit, attained only by a reversible engine between those reservoirs. A Carnot cycle uses two reversible isothermal legs to exchange heat and two reversible adiabatic legs to change temperature without exchanging heat. The result is independent of a particular working gas or machine design.

An engine between 600 and 300 K cannot exceed 50% efficiency. If it absorbs 1000 J per cycle and produces 300 J of work, it rejects 700 J; reservoir entropy rises by -1000/600+700/300=0.667 J/K. Its 30% efficiency is thermodynamically allowed, although that alone does not prove a proposed mechanism works.

What the ceiling does and does not say

Rejected heat is not necessarily evidence of sloppy engineering. Some rejection is required for any ordinary two-reservoir cyclic heat engine. Friction, finite temperature gradients, mixing, and other irreversible processes further reduce useful output.

A car's combustion gas changes temperature continuously, exchanges matter, and loses energy through exhaust, cooling, and friction. It is misleading to insert one peak flame temperature and ambient temperature into Carnot's equation and call the result its actual efficiency. The equation provides a bounding idealization, not an engine-performance specification.

A refrigerator uses work to move heat from cold to hot. Its coefficient of performance is COP=Q꜀/W, which can exceed one because it moves thermal energy rather than converting work into an equal amount of cooling energy. The reversible bound is T(cold)/(T(hot) − T(cold)). Here T(cold) and T(hot) are the absolute temperatures of the two reservoirs. This is not efficiency above 100%.

Takeaway: The first law tracks how much energy exists; the second law constrains which transfers can occur and how much work a thermal gradient can yield. Energy is conserved while its capacity to produce useful work can diminish.