Second law — Kelvin-Planck statement, Clausius statement: Worked examples

This page is a decision gym. The parent note gave you the two statements and proved them equivalent. Here we throw every kind of machine at you — legal, illegal, degenerate, word-problem, exam-trap — and grind each one through the same checklist until spotting a violation becomes reflex.

Before the examples, we need one shared toolkit. Everything below uses only three plain facts, so let us pin them down first — and, because one of them names a quantity called $U$ , we define that too.

Definition Sign convention: how to read heat and work directions

All through this page we write $Q_{H}$ , $Q_{C}$ , $W$ as magnitudes — always positive numbers, just "how many joules." The direction is not carried by a $+$ or $-$ ; it is carried by which reservoir the letter names and which machine we drew:

$Q_{H}$ = joules exchanged with the Hot reservoir. For an engine this is heat flowing in (hot → machine); for a fridge it is heat flowing out (machine → hot).
$Q_{C}$ = joules exchanged with the Cold reservoir. For an engine this is heat out (machine → cold); for a fridge it is heat in (cold → machine).
$W$ = joules of work. An engine makes work (out); a fridge spends work (in). So the moment you say the word "engine" or "fridge," every arrow's direction is fixed and you only ever handle positive numbers. Why do it this way? It removes sign-juggling: you decide direction by reading the picture, not by tracking minus signs.

Formula The three tools we will reuse

Tool 1 — First Law over a cycle. The First Law is bookkeeping: heat in $-$ heat out $=$ work out $+$ rise in internal energy, i.e. $Q_{H} - Q_{C} = W + Δ U$ . Why drop $Δ U$ ? Because over a cycle $Δ U = 0$ (definition above), so the internal-energy term vanishes and we are left with pure input-minus-output: $W = Q_{H} - Q_{C} (First Law over a cycle)$ Why we can trust this everywhere below: every machine on this page runs in a cycle, so this holds without exception.

Tool 2 — Engine efficiency (a device that makes work): $η = \frac{W}{Q _{H}} = 1 - \frac{Q _{C}}{Q _{H}}$ .

Tool 3 — Fridge/heat-pump COP (a device that spends work to move heat cold→hot). "COP" is short for Coefficient Of Performance — read it as "cold-side joules moved per joule of work paid": $COP = \frac{Q _{C}}{W}$

Here $Q_{H}$ = heat swapped with the hot reservoir, $Q_{C}$ = heat swapped with the cold reservoir, $W$ = net work, all magnitudes (see the sign-convention box above).

Intuition The universal checklist (run this on EVERY claim)

First, one word about the metaphors we use in step 3. Heat left alone always drifts from the hot body to the cold body — like water running downhill. So "downhill heat flow" means hot→cold (the free, spontaneous direction, high temperature to low), and "uphill heat flow" means cold→hot (low temperature to high — the direction that never happens for free and needs a pump). Temperature plays the role of height.

Count the reservoirs. One or two?
Find the work $W$ . Is it going in (fridge) or out (engine)? Is it zero?
Find the direction of heat. Hot→cold (downhill, free) or cold→hot (uphill, costs)?
Apply the matching statement. Single reservoir + $η = 1$ ⟹ Kelvin–Planck verdict. Cold→hot with $W = 0$ ⟹ Clausius verdict.
Sanity-check numbers: $0 \leq η < 1$ , all magnitudes $\geq 0$ , energy balances.

The scenario matrix

Every second-law question is one of these cells. The Ex# column tells you which worked example nails that cell.

Cell	What makes it special	Verdict expected	Ex#
A Single reservoir, $Q_{C} = 0$	The degenerate "perfect engine"	Illegal (K–P)	1
B Two reservoirs, $Q_{C} > 0$ , $η < 1$	The ordinary honest engine	Legal	2
C Cold→hot, $W = 0$	The degenerate "free fridge"	Illegal (Clausius)	3
D Cold→hot, $W > 0$	The ordinary honest fridge	Legal	4
E $η$ above the best possible	Exceeds Carnot limit	Illegal (implied)	5
F $η < 0$ or $Q_{C} < 0$ or $W > Q_{H}$	Numbers break First Law itself	Illegal (First Law)	6
G Real-world word problem	Ocean/space engine, hidden reservoir	Reason it out	7
H Equivalence twist	Combine two devices, find the leak	Illegal (composite)	8
I Limiting case $Q_{C} \to 0$ or $T_{C} \to 0$	What happens at the boundary	Approaches limit	9

We cover cell E and I with a light touch on Carnot engine and Carnot theorem, which sets the ceiling on efficiency — the parent note tells you $η < 1$ ; Carnot tells you how much less.

Example 1 — Cell A: the perfect engine

Worked example Claim: an engine takes

$Q_{H} = 600$ J from one reservoir and outputs $W = 600$ J per cycle. Forecast: Is this legal? Guess before reading — do the numbers even leave room for waste heat?

Count reservoirs. One (only a hot source named). Why this step? Kelvin–Planck is specifically about single-reservoir engines.
First Law: $Q_{C} = Q_{H} - W = 600 - 600 = 0$ J. Why? Over a cycle $W = Q_{H} - Q_{C}$ , so $Q_{C}$ is forced once $Q_{H}, W$ are fixed — we don't get to choose it.
Efficiency: $η = W / Q_{H} = 600/600 = 1$ . Why decisive? $η = 1$ from one reservoir is exactly the machine Kelvin–Planck forbids.
Verdict: Illegal — Kelvin–Planck violation. ❌

Verify: $Q_{C} = 0$ means zero heat rejected; a real cycle must reject some heat (the gas has to be reset), so $Q_{C} = 0$ is the tell-tale. Units check: J − J = J. ✓

Example 2 — Cell B: the honest engine

Example 3 — Cell C: the free fridge

Example 4 — Cell D: the honest fridge

Example 5 — Cell E: beating the Carnot ceiling

Worked example An engine runs between

$T_{H} = 500$ K and $T_{C} = 300$ K and claims $η = 0.55$ . Forecast: It rejects heat, $η < 1$ , so Kelvin–Planck is happy. But is there a tighter limit?

First, where does the ceiling come from? We can sketch the argument here without heavy machinery (the full proof lives in Carnot engine and Carnot theorem). Imagine the best conceivable engine — one so gentle it could run backwards as a fridge with no waste (a reversible one, see Reversible and irreversible processes). For that ideal engine the heats it swaps turn out to be in the same ratio as the absolute temperatures of the two reservoirs: $\frac{Q _{C}}{Q _{H}} = \frac{T _{C}}{T _{H}}$ . Plug that into $η = 1 - Q_{C} / Q_{H}$ and you get the ceiling $η_{m a x} = 1 - T_{C} / T_{H}$ .

Why is it a ceiling and not just one possible value? Suppose some engine beat it. Run the ideal reversible engine backwards as a fridge and feed it with the beater's work. The parent note's equivalence trick then builds a single-reservoir $η = 1$ device — a Kelvin–Planck violation. So nothing may exceed $η_{m a x}$ . See the figure below.

Best possible efficiency (Carnot bound): $η_{m a x} = 1 - \frac{T _{C}}{T _{H}} = 1 - \frac{300}{500} = 0.40$ . Why kelvin? The ratio is of absolute temperatures.
Compare: claimed $0.55 > 0.40 = η_{m a x}$ . Why decisive? Exceeding the reversible bound would let you build a Kelvin–Planck violator.
Verdict: Illegal — beats the Carnot limit. ❌

Verify: temperatures must be in kelvin (ratios of absolute temperature). $300/500 = 0.6$ , so $η_{m a x} = 0.4$ . Claimed $0.55$ fails. ✓

Figure — Second law — Kelvin-Planck statement, Clausius statement

Figure s01 — Efficiency ceiling. The horizontal axis is $T_{C} / T_{H}$ (from 0 to 1); the vertical axis is efficiency $η$ . A magenta line traces $η_{m a x} = 1 - T_{C} / T_{H}$ : it starts at $η = 1$ when the cold side is at absolute zero and falls to $η = 0$ when the two temperatures match. The whole region above the line is shaded violet and stamped "FORBIDDEN (beats Carnot)"; the region below is peach and stamped "allowed engines". The orange dot marks Example 5: $T_{C} / T_{H} = 0.6$ , sitting on the line at $η_{m a x} = 0.4$ , with the claimed $η = 0.55$ plotted as a red X in the forbidden zone.

What to look for: the magenta line is a hard fence. Every real engine must live on or below it. Example 5's claim (red X) floats above the fence — instant disqualification, no arithmetic needed once you can see it. Notice the line only touches $η = 1$ at the far left, where $T_{C} / T_{H} = 0$ (an absolute-zero cold reservoir) — which is why full efficiency is a limit you approach, never reach (Example 9).

Example 6 — Cell F: numbers that break the First Law

Worked example Three engine claims. Which numbers are impossible

before we even reach the Second Law? (a) $Q_{H} = 500$ J, $W = 600$ J. (b) $Q_{H} = 700$ J, $Q_{C} = - 50$ J. (c) $Q_{H} = 400$ J, $W = 400$ J, $Q_{C} = 100$ J.

Forecast: Which of these can't happen for reasons even simpler than Kelvin–Planck?

(a): $Q_{C} = Q_{H} - W = 500 - 600 = - 100$ J. Why bad? A negative rejected-heat magnitude means work exceeds input — energy created from nothing. First Law broken. ❌
(b): $Q_{C} = - 50$ J is a negative magnitude. Why bad? Magnitudes can't be negative; there's no physical "negative rejected heat" for an engine. First Law / definition broken. ❌
(c): Check consistency: $W$ should equal $Q_{H} - Q_{C} = 400 - 100 = 300$ J, but the claim says $W = 400$ J. Why bad? The three numbers don't balance — $400 \neq = 300$ . First Law broken. ❌
Verdict: all three fail the First Law; the Second Law never even gets a turn.

Verify: (a) $500 - 600 = - 100 < 0$ ; (b) $- 50 < 0$ ; (c) $400 - 100 = 300 \neq = 400$ . All inconsistent. ✓ Moral: the parent note's First Law is the first gate; Second Law is the second.

Example 7 — Cell G: the ocean-powered ship (word problem)

Worked example An inventor pitches a ship that "burns no fuel — it runs its engine purely by cooling the surrounding ocean, which is one big reservoir at

$15^{\circ} C$ , and turns that heat into propulsion." Forecast: Energy is conserved (the ocean loses exactly the work gained). So which law, if any, does it break?

Count reservoirs. Just the ocean — a single reservoir at one temperature. Why crucial? No colder body is available to dump waste heat into.
Trace the energy: ocean heat $\to$ work, $Q_{C} = 0$ , $η = 1$ . Why? If all extracted heat becomes propulsion with nothing rejected, that's a single-reservoir $η = 1$ engine.
Match the statement: exactly the Kelvin–Planck-forbidden machine. Why it fools people: it does obey the First Law — the ocean cools, so energy is conserved. That's why we need a second law to rule it out.
Verdict: Illegal — Kelvin–Planck. The ship would be a perpetual motion machine of the second kind. ❌

Verify: With one reservoir, $η = 1 - Q_{C} / Q_{H}$ can equal $1$ only if $Q_{C} = 0$ ; $Q_{C} = 0$ from a single source is banned. This is Cell A wearing a nautical costume — see Reversible and irreversible processes and Arrow of time for why nature enforces the direction.

Example 8 — Cell H: spot the leak in a composite device

Worked example Someone bolts two boxes together and claims the pair is legal:

Box 1 (engine): takes $Q_{H} = 1000$ J from the hot reservoir, outputs $W = 1000$ J, rejects $Q_{C} = 0$ .
Box 2 (fridge): fed that $W = 1000$ J, pulls $Q_{C} = 400$ J from the cold reservoir, dumps $1400$ J into hot.

"Each box has honest-looking numbers," they say. Is the combination legal? Forecast: Track the net heat at each reservoir — where does it end up?

Test Box 1 alone: single-source engine with $Q_{C} = 0$ , $η = 1$ . Why? Already a Kelvin–Planck violation (Cell A) — so the composite is doomed. But let's see the composite's signature too.
Net at cold reservoir: only Box 2 touches it: $- 400$ J (heat removed from cold). Why track this? We're hunting the net effect on the world.
Net at hot reservoir: $- 1000$ (to Box 1) $+ 1400$ (from Box 2) $= + 400$ J gained. Why? Add every heat exchanged with the hot side.
Read the net effect: $400$ J moved cold→hot, work cancels internally ( $1000$ made $= 1000$ spent), no other change. Why decisive? That's heat cold→hot as the sole effect — a Clausius violation.
Verdict: Illegal. One impossible box ( $Q_{C} = 0$ engine) breeds a Clausius-violating composite — exactly the equivalence argument from the parent note, run with numbers. ❌

Verify: cold: $- 400$ ; hot: $- 1000 + 1400 = + 400$ ; work net $= 1000 - 1000 = 0$ . Net = $400$ J transferred cold→hot for free. ✓

Example 9 — Cell I: the limiting case $T_{C} \to 0$

Worked example What happens to the best-possible efficiency of an engine as the cold reservoir gets colder and colder,

$T_{C} \to 0$ K, with $T_{H} = 400$ K fixed? Forecast: Can $η_{m a x}$ ever actually reach $1$ ?

Write the bound: $η_{m a x} = 1 - \frac{T _{C}}{T _{H}}$ . Why? It's the ceiling derived in Example 5.
Evaluate at several $T_{C}$ : $T_{C} = 200 \Rightarrow η_{m a x} = 0.5$ ; $T_{C} = 40 \Rightarrow 0.90$ ; $T_{C} = 4 \Rightarrow 0.99$ . Why tabulate? To watch the trend, not just the endpoint.
Take the limit: as $T_{C} \to 0^{+}$ , $T_{C} / T_{H} \to 0$ , so $η_{m a x} \to 1$ . Why does this not violate K–P? Reaching $η = 1$ would need $T_{C} = 0$ exactly — an unattainable absolute-zero reservoir. So $η < 1$ always holds for real machines; we only approach $1$ .
Also check the other degenerate end ( $Q_{C} \to 0$ with two reservoirs): $η = 1 - Q_{C} / Q_{H} \to 1$ as $Q_{C} \to 0$ , but $Q_{C}$ can only approach zero, never hit it for a cyclic engine — the same boundary from the heat side.
Verdict: limiting behaviour is $η_{m a x} \to 1$ but never $= 1$ . The boundary is approached, never touched. ✓

Verify: $1 - 40/400 = 0.9$ ; $1 - 4/400 = 0.99$ ; $1 - 200/400 = 0.5$ . The sequence climbs toward but never equals $1$ . ✓ On the figure in Example 5, this is the magenta line creeping up toward the top-left corner without ever landing on $η = 1$ except at the single unreachable point $T_{C} / T_{H} = 0$ .

Recall Quick self-test across the matrix

Single reservoir, $η = 1$ — which law? ::: Kelvin–Planck (Cell A / Ex 1, 7). Cold→hot with $W = 0$ — which law? ::: Clausius (Cell C / Ex 3). Cold→hot with $W > 0$ — legal or not? ::: Legal, ordinary fridge (Cell D / Ex 4). Engine claims $η$ above $1 - T_{C} / T_{H}$ — verdict? ::: Illegal, beats Carnot bound (Cell E / Ex 5). $W > Q_{H}$ for an engine — which law fails first? ::: The First Law, before the Second even applies (Cell F / Ex 6). As $T_{C} \to 0$ , what does $η_{m a x}$ do? ::: Approaches 1 but never reaches it (Cell I / Ex 9).

Connected reading

Parent: Second law — Kelvin-Planck statement, Clausius statement (index 1.7.18)
The efficiency ceiling used in Ex 5 & 9: Carnot engine and Carnot theorem
Where the quantitative law lives: Entropy and the Clausius inequality
The gate before the Second Law (Ex 6): First law of thermodynamics
Why direction is baked in: Reversible and irreversible processes, Arrow of time