Topic 5

The second law & entropy

The first law says energy is never lost. The second law says that is not enough — energy also has a quality, and every real process degrades it. That is why you cannot un-scramble an egg, why no engine is 100% efficient, and why time appears to run one direction.

Three statements, one law

The second law gets phrased three different ways depending on what is being discussed. They are logically equivalent — each implies the other two.

Form	Statement	Rules out
Clausius	Heat never flows spontaneously from a colder body to a hotter one.	A refrigerator that needs no power.
Kelvin–Planck	No cyclic engine can convert heat entirely into work. Some heat must always be dumped to a colder reservoir.	A ship that sails by cooling the ocean.
Entropy	Δ`S`_universe ≥ 0 for every process; = 0 only for a reversible one.	Any process that would make the universe more ordered overall.

Notice what is not forbidden

A refrigerator does move heat from cold to hot — it just cannot do so spontaneously; it must be paid for with work. And entropy can decrease locally (your freezer, a growing crystal, a living organism) as long as it increases more somewhere else. The second law is a statement about the total, not about every corner.

Heat engines

A heat engine runs a working substance around a cycle, taking heat Q_H from a hot reservoir, converting part of it to work W, and dumping the rest as Q_C into a cold reservoir. Because it is a cycle, ΔU = 0 and the first law gives an exact energy balance:

|Q_H| = |W| + |Q_C| Everything that goes in comes out, either as useful work or as waste heat. The second law is the extra constraint that Q_C can never be zero.

Arrow widths are drawn to scale for a fairly typical engine: of the heat taken in, only a slice becomes work. Widen the W arrow all the way and you would have to shrink Q_C to nothing — which the second law forbids.

Efficiency and the Carnot limit

Efficiency is “what you want divided by what you paid for”:

e = |W_net||Q_H| = 1 − |Q_C||Q_H| Dimensionless, between 0 and 1. Only heat absorbed goes in the denominator — never the total heat that moved.

Now the ceiling. No engine operating between two reservoirs can beat a fully reversible one, and every reversible engine between the same two reservoirs has the same efficiency:

e_Carnot = 1 − T_CT_H Kelvin only. This is a theoretical maximum: it assumes no friction, no turbulence, and infinitely slow operation. Real engines fall well short.

How this gets tested

A problem gives an engine’s reservoir temperatures and its claimed performance, then asks whether the claim is possible. Compute e_Carnot; if the claimed e exceeds it, the engine is impossible. Also expect: “how could this engine be improved?” — answer, raise T_H or lower T_C, since the ratio is all that matters.

Two consequences that are worth internalizing rather than deriving:

100% efficiency requires T_C = 0 K, which is unreachable. That is the Kelvin–Planck statement written as an equation.
Efficiency depends on the ratio, not the difference. An engine running between 600 K and 300 K has the same Carnot limit (50%) as one between 400 K and 200 K, even though the second has half the temperature difference.

Every one of these sits below its Carnot limit — and the gap is where the engineering happens.
Engine	Typical real efficiency
Gasoline car engine	~25%
Diesel engine	~35–40%
Coal-fired power plant	~35–40%
Combined-cycle gas turbine	~60%

Refrigerators and heat pumps

Run the engine diagram backwards. Now work goes in, and heat is pulled from the cold side and dumped on the hot side. On a PV diagram this is the counterclockwise cycle.

Note that Q_H is bigger than Q_C: the work you paid for also ends up as heat in the room. This is why leaving the fridge door open warms the kitchen rather than cooling it.

Entropy

Entropy S measures how spread out energy is — equivalently, how many microscopic arrangements are consistent with what you can see macroscopically. For heat transferred reversibly at a temperature that does not appreciably change:

ΔS = QT Units J/K. T in kelvin. Positive when heat flows in, negative when it flows out.

The T in the denominator is the interesting part. The same joule of heat produces a bigger entropy change when it lands somewhere cold. That single fact is why heat flows hot → cold and never the reverse:

ΔS_univ = −QT_H + QT_C > 0 when T_H > T_C The hot object loses a little entropy; the cold object gains more. Run it backwards and the total would be negative — forbidden.

Entropy is not “messiness”

The disorder metaphor is a rough hint, not a definition, and it misleads as often as it helps. Entropy counts microstates — how many ways the molecules can be arranged while still looking the same from outside. A gas filling a room has astronomically more arrangements available than the same gas in one corner, which is the entire reason it spreads.

Why entropy increases

Nothing forces gas molecules to spread out. There is no repulsive law pushing them apart. Spreading is simply overwhelmingly more likely.

For just 12 molecules, the odds of finding them all on one side are 1 in 2¹² ≈ 4000. For a mole, the exponent becomes 6×10²³ — a probability so small that “never” is an entirely fair description. The second law is statistics, not a force.

Boltzmann made this exact: S = k_B ln Ω, where Ω is the number of microstates. You will not be asked to compute with this in AP Physics 2, but it is the reason the second law is true, and it explains why the law is statistical — entropy decreases are not strictly impossible, merely so improbable that they never happen.

Reversible vs. irreversible

A process is reversible if it could run backwards with no net change to the universe — which requires it to be quasi-static and frictionless. Nothing real qualifies. Every actual process (friction, mixing, free expansion, heat across a finite temperature difference) makes ΔS_univ > 0. Reversible processes are the idealized limiting case, and they set the Carnot bound.

Worked examples

1. Engine efficiency and waste heat

An engine absorbs 3600 J per cycle from a hot reservoir and expels 2800 J to a cold one. Find the work per cycle and the efficiency.

|W| = |Q_H| − |Q_C| = 3600 − 2800 = 800 J e = 800/3600 = 0.22 = 22%

This is the same engine as the cycle worked out on the PV diagrams page — there the 800 J came from the enclosed area of the loop, here from the energy balance. Same number, two routes.

2. Is this engine possible?

An inventor claims an engine operating between 550 K and 300 K with 50% efficiency. Evaluate the claim.

e_Carnot = 1 − 300550 = 0.4545 = 45.5%

The claimed 50% exceeds the Carnot limit, so the engine is impossible. No design refinement can help: the bound depends only on the reservoir temperatures, not on the working substance, the cycle shape, or the engineering.

Follow-up they like: what is the most work per cycle this engine could produce from 3600 J of input? (0.4545)(3600) = 1636 J.

3. Entropy change of melting ice

1.0 kg of ice melts at 0 °C. Find the entropy change of the ice.

The temperature is constant during melting, so ΔS = Q/T applies directly with Q = mL_f.

ΔS = (1.0)(3.34×10⁵)273 = +1.2×10³ J/K

Positive — liquid water has vastly more available arrangements than a rigid ice crystal. Note the kelvin conversion: using 0 in the denominator would divide by zero, which is a useful reminder that this formula is kelvin-only.

4. Why heat only flows one way

1200 J of heat flows from a 600 K reservoir to a 300 K reservoir. Find ΔS_universe. Then check the reverse.

ΔS = −1200600 + 1200300 = −2.0 + 4.0 = +2.0 J/K Allowed — the total went up.

Now reverse it, 1200 J from cold to hot:

ΔS = −1200300 + 1200600 = −4.0 + 2.0 = −2.0 J/K Forbidden — a spontaneous decrease in total entropy.

Both directions conserve energy perfectly. The first law cannot tell them apart; only the second law rules one out. That is the cleanest demonstration of why we need a second law at all.

5. Free expansion — entropy without heat

An insulated container is divided by a partition, gas on one side and vacuum on the other. The partition is removed. What are Q, W, ΔU, ΔT, and ΔS?

Q = 0 — the container is insulated.
W = 0 — the gas expands into vacuum, pushing against nothing.
ΔU = Q + W = 0, so ΔT = 0 for an ideal gas.
ΔS = nR ln(V_f/V_i) > 0.

The point: entropy increased even though no heat flowed. The formula ΔS = Q/T only applies to reversible transfers, and free expansion is violently irreversible — the gas will never spontaneously crowd back into half the box. For an irreversible process you compute ΔS along an imagined reversible path between the same two states, which is where the logarithm comes from.

This is also a good check on the difference between the two laws: the first law is entirely satisfied by the gas returning to one side. Only the second law says it will not.

Checklist

I can state the second law in all three forms and say what each one forbids.
I use only absorbed heat in the denominator of the efficiency formula.
I convert to kelvin before touching the Carnot formula.
I can test a claimed engine against the Carnot limit in one line.
I can explain, using Q/T, why heat flows hot to cold.
I know that local entropy decreases are fine as long as the universe’s total rises.