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

WHAT we look at: one reservoir (the room), heat converted fully to work (turning the fan), no cold sink, no other change, done cyclically ("forever"). WHY that pins it down: a single-reservoir cyclic engine with $\eta =1$ is precisely the Kelvin–Planck forbidden machine. Answer: it violates the Kelvin–Planck statement.

WHAT we look at: heat travels cold→hot, $W=0$ (plugged into nothing), sole effect, cyclic. WHY: cold→hot as the only effect with zero work is exactly what Clausius forbids. Answer: it violates the Clausius statement.

WHAT: two reservoirs, $Q_C=250>0$, so $\eta <1$. WHY legal: neither law is broken — heat is only partly converted and heat is rejected to a genuine cold sink. Answer: Legal. (Its efficiency, computed later, is $\eta =1-\frac{250}{600}\approx 0.583$.)

Step 1 — WHY subtract: over a full cycle the working substance returns to its start, so its internal energy is unchanged — $\Delta U=0$ (defined in the toolbox above). The First Law then says whatever heat entered but did not leave must have left as work — so the work is the difference between heat in and heat out: $W=Q_H-Q_C=900-540=360\text{J}.$ Step 2 — WHY divide: efficiency asks "of the fuel I paid for ($Q_H$), what fraction came back as useful work?", so we divide the prize by the price: $\eta =\frac{W}{Q_H}=\frac{360}{900}=0.4=40\%.$ Check: $1-\frac{Q_C}{Q_H}=1-\frac{540}{900}=0.4$. ✔

Step 1 — WHY add: energy cannot vanish; everything the fridge takes in must come back out into the kitchen. It absorbs $Q_C$ from the cold food and the electrical work $W$, so both leave together as $Q_H$: $Q_H=Q_C+W=800+200=1000\text{J}.$ Step 2 — WHY COP not $\eta$: for a fridge the "prize" is heat removed, so we score $Q_C$ per unit work paid: ${\text{COP}}_{\text{fridge}}=\frac{Q_C}{W}=\frac{800}{200}=4.$ Answer: $Q_H=1000$ J, $\text{COP}=4$.

Step 1: $W=Q_H-Q_C=1000-0=1000$ J — First Law is happy. Step 2 — WHY it still fails: $Q_C=0$ from a single reservoir gives $\eta =1$. That is the exact machine Kelvin–Planck forbids. Answer: Impossible — violates Kelvin–Planck. First Law alone can't catch it; you need the Second.

Step 1 — WHY subtract for each: as in L2.1, over a cycle the work equals heat-in minus heat-rejected. So the engine that throws away less heat keeps more as work: $W_A=1000-400=600\text{J},W_B=1000-250=750\text{J}.$ Step 2 — efficiency each (prize over price): ${\eta }_A=\frac{600}{1000}=0.6,{\eta }_B=\frac{750}{1000}=\mathrm{0.75.}$ Step 3 — WHY compare rejected heat: with equal $Q_H$, the engine that wastes less heat wins on both counts. B rejects $150$ J less, so it delivers $150$ J more work and is $15$ percentage-points more efficient. Answer: B does $150$ J more work; ${\eta }_B-{\eta }_A=0.15$.

Step 1 — engine work: $W=1200-800=400$ J. Step 2 — feed it to the fridge: the fridge's work input is this same $W=400$ J. Step 3 — WHY COP inverts: $\text{COP}=\frac{Q_C^{\text{fridge}}}{W}$, so $Q_C^{\text{fridge}}=\text{COP}\times W=3\times 400=1200\text{J}.$ Answer: the fridge removes 1200 J from the cold room.

Step 1 — WHAT quantity is this really? $\frac{Q_H}{W}$ for a heat pump is its coefficient of performance ${\text{COP}}_{\text{pump}}$ (defined in the toolbox), not an engine efficiency $\eta =W/Q_H$. Step 2 — WHY no violation: a heat pump pays work $W$ and delivers more heat $Q_H=Q_C+W>W$, so $Q_H/W>1$ is normal and expected — you're getting free cold-reservoir heat along for the ride. It is not "work created from nothing." Step 3 — the actual engine efficiency (if you ran it as an engine) would be $\eta =W/Q_H=1/3\approx 33\%<100\%$. No law is threatened. Answer: nothing is broken. The student confused COP with efficiency.

Step 1 — cancel the cold side (WHY we matched $300$ J): direction tally on the cold reservoir — the engine adds $300$ J (heat flowing into cold) while the magic device removes $300$ J (heat flowing out of cold). Writing those named directions as a signed sum: $\underset{\text{into cold}}{\underbrace{+300}}\underset{\text{out of cold}}{\underbrace{-300}}=0\text{J}.$ The cold reservoir is left exactly as it was — as if it never existed. Step 2 — net on the hot side: engine removes $900$ J from hot; magic device returns $300$ J to hot. Net heat removed from hot: $900-300=600\text{J}.$ Step 3 — the work is unchanged: the magic device consumed no work, so it neither adds to nor subtracts from the work ledger. The combo's work output is therefore just the original engine's output, surviving untouched — nothing new is derived here, we merely re-read $W=Q_H-Q_C=900-300=600\text{J}$ from the engine we already had. Conclusion: the combo takes $600$ J from a single (hot) reservoir and turns it fully into $600$ J of work — a Kelvin–Planck violation. ∎ In the figure this is exactly the two equal-and-opposite cyan arrows on the cold bar cancelling, leaving only the hot bar drained and the amber work arrow out.

Step 1 — fridge intake (a): $Q_C=\text{COP}\times W=4\times 500=2000$ J from the cold reservoir. Step 2 — fridge output (b): $Q_H^{\text{fridge}}=Q_C+W=2000+500=2500$ J dumped into hot. Step 3 — net on the hot reservoir (c): the magic engine took $500$ J from hot; the fridge dumped $2500$ J into hot: $\text{net into hot}=2500-500=2000\text{J}.$ Step 4 — WHY this is Clausius-breaking: the combo pulled $2000$ J out of cold and pushed exactly $2000$ J into hot, with no external work (the work came internally from the magic engine) and no other change. Conclusion: heat moved cold→hot as the sole effect — a Clausius violation. ∎ Note $Q_C=2000$ J out of cold matches the $2000$ J into hot: the books balance perfectly.

Step 1 — Engine 1 output: $W_1=Q_H-Q_C=1000-200=800$ J of work (its claimed efficiency ${\eta }_1=800/1000=0.8$). Step 2 — drive the reverse machine: Engine 2 run backward needs $600$ J to move $600$ J cold→hot and dump $1200$ J into hot. Feed it $600$ J from Engine 1's $800$ J, leaving $200$ J of leftover work in your pocket. Step 3 — account each reservoir (directions named, then summed).

• Cold: Engine 1 dumps $+200$ J into cold; reverse machine sucks $600$ J out of cold. Net on cold $=200-600=-400$ J (cold loses $400$ J).
• Hot: Engine 1 draws $1000$ J out of hot; reverse machine dumps $1200$ J into hot. Net on hot $=1200-1000=+200$ J (hot gains $200$ J). Step 4 — the leftover work: $W_{\text{net}}=800-600=200$ J still comes out. Step 5 — read the paradox: the combo pulls $400$ J from cold, pushes $200$ J into hot, and hands you $200$ J of free work — energy still balances ($400=200+200$ ✔). But it does all this with no external input: heat flowed cold→hot AND we harvested work, from nothing. That is simultaneously a Clausius and a Kelvin–Planck violation. Step 6 — WHY the reversibility of Engine 2 is the crux: because Engine 2 is reversible, running it backward returns exactly the energies it would produce running forward — no waste is hidden. So the only way the combo can misbehave is if Engine 1 out-performs that reversible benchmark. Here Engine 1's ${\eta }_1=0.8$ exceeds what any reversible engine between these reservoirs could reach. Carnot's theorem says the reversible engine is the champion — no engine can beat it — so an ${\eta }_1$ above the reversible value is precisely what breaks the Second Law. The reversible Engine 2 is the "measuring stick"; Engine 1 is illegal because it beats the stick. Answer: the paradox is $-400$ J from cold, $+200$ J to hot, $+200$ J free work — impossible; it exists only if Engine 1 beats the reversible (Carnot) benchmark, which no engine can.

(a) $Q_C\to 0$: then $W\to Q_H=1000$ J and $\eta \to 1$. This is the Kelvin–Planck ceiling — heat fully converted to work with nothing rejected. Forbidden to actually reach: you may approach it but never touch it, because a single cold-free cyclic engine at $\eta =1$ is exactly the K–P machine. (b) $Q_C\to Q_H=1000$ J: then $W\to 0$ and $\eta \to 0$. All the heat entering from hot passes straight through and leaves into cold, extracting no work. Allowed — this is just ordinary hot→cold heat flow dressed up as a (useless) "engine." Nothing forbids a machine that does nothing. (c) $W=0$ with two reservoirs: setting $W=0$ in $W=Q_H-Q_C$ forces $Q_C=Q_H=1000$ J — identical to case (b). It is plain spontaneous hot→cold conduction, which the Second Law positively permits (indeed insists on as the natural direction). Allowed. (d) $Q_H=0$: no heat is absorbed from hot, so $W=Q_H-Q_C=-Q_C\le 0$. A non-negative work output is impossible unless $Q_C=0$ too (then everything is zero — the machine simply idles). If instead $Q_C>0$ and work is supplied ($W<0$), the device is no longer an engine at all: it is absorbing work to push heat around, i.e. it has become a fridge/heat-pump. Allowed, but only as a relabelled machine — there is no engine here to violate anything. Big picture: across the whole range the only outlawed limit is (a), $Q_C\to 0$. Every efficiency from $\eta =0$ up to but not including $\eta =1$ is bounded above by the achievable Carnot ceiling ${\eta }_{\max }=1-T_C/T_H<1$ (toolbox). Kelvin–Planck bans only $\eta =1$; Carnot bans everything above ${\eta }_{\max }$, which is stricter still.

Step 1 — WHY use temperatures now: efficiency in joules ($\eta =1-Q_C/Q_H$) tells you what an engine did; the Carnot bound ${\eta }_{\max }=1-T_C/T_H$ tells you the best any engine could ever do between these two reservoirs. Absolute (kelvin) temperatures are what set that ceiling. (a) ${\eta }_{\max }=1-\frac{T_C}{T_H}=1-\frac{300}{600}=1-0.5=\mathrm{0.50.}$ So no engine here can exceed $50\%$. (b) $\eta =0.40<0.50$, so it is legal. Fraction of the ideal it achieves: $\frac{0.40}{0.50}=0.80$ — it reaches $80\%$ of the reversible ceiling (a good, but not perfect, real engine). (c) $\eta =0.55>0.50={\eta }_{\max }$. It claims to beat the reversible (Carnot) champion. By Carnot's theorem that is impossible — hooking it to a reversible engine run backward reproduces exactly the paradox of L5.1. Illegal, even though $0.55<1$ so bare Kelvin–Planck alone wouldn't catch it. Answer: (a) ${\eta }_{\max }=0.50$; (b) legal, $80\%$ of the ceiling; (c) illegal — exceeds the Carnot bound.

L2.1 $W=360$ J, $\eta =0.4$ ::: L2.2 $Q_H=1000$ J, COP$=4$ ::: L2.3 impossible ($\eta =1$) L3.1 $\Delta W=150$ J, $\Delta \eta =0.15$ ::: L3.2 fridge removes $1200$ J ::: L3.3 no violation ($\eta =1/3$) L4.1 $W=600$ J from hot alone ::: L4.2 $Q_C=2000$, dump $2500$, net $2000$ J into hot L5.1 cold $-400$, hot $+200$, free work $+200$ ::: L5.2 only $Q_C\to 0$ forbidden ::: L5.3 ${\eta }_{\max }=0.50$, real reaches $80\%$, $0.55$ illegal.