Question bank — Second law — Kelvin-Planck statement, Clausius statement

False — that process conserves energy perfectly (heat merely moves from room to coffee, total unchanged), so the First Law is happy with it; only the Second Law forbids the spontaneous cold→hot direction, which is why we never see it.

False — an expansion is not a cycle: the gas is left expanded, so $\Delta U=0$ held only for that stroke. Close the loop and the compression rejects heat ($Q_C>0$), dropping $\eta$ below 1 — exactly the 600 J → 350 J worked example above.

False — Clausius forbids cold→hot as the sole effect, i.e. with $W=0$. Your fridge's compressor eats electrical work ($W>0$, roughly 100–200 W from the wall), and that non-zero $W$ is the "other effect" that keeps it legal.

False — heat is routinely turned into work (that's what every engine does); the law only forbids turning it entirely into work in a cycle from a single reservoir, i.e. it caps $\eta <1$, it doesn't ban $W>0$.

False — the two are logically equivalent: hook a K–P-violating engine's free work into an ordinary fridge and you pump heat cold→hot with no net work, breaking Clausius. Break one and the arrows let you break the other.

False — $\eta =1$ forces $Q_C=0$, so the cold-side arrow vanishes and the cold reservoir is untouched; the machine effectively draws from a single reservoir and violates K–P regardless of how many reservoirs you list on paper.

True — $\text{COP}=Q_C/W$ can easily be 3–5 because you move far more heat than the work you spend (spend 1 J of work, shift 4 J of heat). This breaks no law, since $W>0$ is still genuinely paid — the arrows just show a small $W$ and a big $Q_C$.

False — it is the whole point: drop "cyclic" and both statements become false, because a single non-cyclic stroke can convert heat fully to work or shift heat uphill for free. The bite is that the machine must return to its start, which forces the rejected-heat and paid-work arrows to reappear.

Error: by the convention $Q_C=Q_H-W=500-500=0$; a single reservoir with $Q_C=0$ means $\eta =1$, exactly the K–P-forbidden machine. Energy conservation is necessary but not sufficient for legality.

Error: equivalence needs the cross implications — a K–P violator must be shown to build a Clausius violator (and vice versa), not to re-break its own statement. Same-statement implication proves nothing.

Error: you must tune the engine's rejected heat to equal the magic device's $Q_C$, so the cold reservoir's net change is exactly zero and cancels; otherwise the combined machine still touches the cold reservoir and isn't a clean single-reservoir engine.

Error: nothing is created — the paid work $W$ is added to the pulled heat $Q_C$, and the sum is what's dumped. This is just the First Law arrows adding up ($Q_H=Q_C+W$), not energy from nowhere.

Error: Carnot is the most efficient allowed engine, with ${\eta }_{\max }=1-T_C/T_H$, which stays below 1 for any finite $T_C>0$. It hugs the limit set by K–P; it never crosses it. (Fuller treatment: Carnot engine and Carnot theorem.)

Error: only if the engine is the magic K–P violator ($W=Q_H$, no heat rejected). An ordinary engine rejects $Q_C>0$, so the combined device leaves an extra effect on the reservoirs and stays legal.

Because both statements are precisely about what these two cyclic machines cannot do — the forbidden $Q_C=0$ engine and the forbidden $W=0$ fridge. Without the machines the law has nothing concrete to forbid.

Over a full loop the working substance returns to its start ($\Delta U=0$), which forces a compression stroke; that stroke rejects heat, making the net $Q_C$ positive — as the 250 J in the worked example showed.

Contradiction lets you assume a rule-breaking magic device exists and bolt an ordinary machine onto it to break the other rule, showing the two prohibitions stand or fall together. See Reversible and irreversible processes for why real machines never approach the magic ideal.

Because it singles out a preferred direction for spontaneous change (hot→cold, work-free processes forbidden in reverse), giving time a forward orientation absent from the time-symmetric First Law — see Arrow of time.

The First Law only checks that the arrows balance ($Q_H=Q_C+W$); it is direction-blind and permits the forbidden and allowed machines equally, so it cannot distinguish or link the two Second-Law prohibitions.

The uphill transfer lowers the reservoirs' total entropy, so it must be paid for by work that dumps at least that much entropy back; a free version would break the Clausius inequality.

No — the Carnot ceiling ${\eta }_{\max }=1-T_C/T_H$ would read 1 at $T_C=0$, but reaching $T_C=0$ is itself forbidden by the Third Law, so the loophole never physically opens.

With one effective reservoir there is no temperature drop to exploit; ${\eta }_{\max }=1-T_C/T_H=0$, so no net work can be extracted at all — the strongest possible statement of K–P.

Possibly legal against Clausius as literally worded (there is another effect), but that side effect must itself supply the needed entropy/work; a genuinely cost-free version remains impossible.

Then heat flows hot→cold spontaneously and needs no work — but that's ordinary downhill flow, not the forbidden uphill Clausius transfer; the labels "cold" and "hot" were just swapped.

No — as long as $Q_C>0$ strictly, $\eta <1$ and the law holds; K–P forbids only the exact limit $Q_C=0$, which is unreachable in a cycle.

Yes — extracting any net work from a lone reservoir per cycle means $Q_C=0$ (the cold arrow is gone) and $\eta =1$; the two descriptions are the identical forbidden machine.