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

1.7.18 · D4 · Physics › Thermodynamics › Second law — Kelvin-Planck statement, Clausius statement

Intuition Is page ko kaise use karein

Ye 12 problems ek ladder climb karte hain: L1 Recognition (spot karo ki kaun sa law apply hota hai) → L2 Application ( $W = Q_{H} - Q_{C}$ mein plug karo) → L3 Analysis (machines compare karo, signs chase karo) → L4 Synthesis (do machines ko couple karo, ek contradiction build karo) → L5 Mastery (apna khud ka paradox banao aur usse toddo). Har solution ek collapsible callout mein chhupi hai taaki pehle tum khud test kar sako. Saare numbers machine-verified hain.

Shuru karne se pehle, char tools jo kabhi nahi bhoolne chahiye:

Engine (ek device jo ek cycle mein chalta hai): $W = Q_{H} - Q_{C}$ aur $η = \frac{W}{Q _{H}} = 1 - \frac{Q _{C}}{Q _{H}}$ .
Refrigerator: $Q_{H} = Q_{C} + W$ aur $COP_{fridge} = \frac{Q _{C}}{W}$ (prize = heat jo cold se remove ki gayi hai).
Heat pump (same hardware, alag goal): $COP_{pump} = \frac{Q _{H}}{W}$ (prize = heat jo hot ko deliver ki gayi hai). Ye ratio almost hamesha $1$ se bada hota hai — ye normal hai, aur ye koi efficiency NAHI hai.
Is page par sign convention: har $Q$ ek magnitude hai — joules ki ek positive number — aur iska direction (kisi reservoir mein/bahar, hot→cold ya cold→hot) hamesha words mein bataya jaata hai, kabhi sign mein chhupaaya nahi jaata. Jab hum kisi reservoir ka net change talliy karte hain toh hum ek signed sum likh sakte hain (jaise $+ 300 - 300$ ), lekin ye un directions ki bookkeeping hai jo hum ne words mein pehle hi naam de diye hain, na ki heat se attached koi sign.

Definition Do symbols jinpar hum baar baar lean karenge

$Δ U$ (internal energy mein change). $U$ woh total energy hai jo working substance (gas) ke andar stored hai — uske molecules ki kinetic aur potential energy. $Δ U$ (padho "delta U") ka matlab hai $U_{end} - U_{start}$ , yaani kisi process ke dauran us stored energy mein change. Ek cycle mein substance apni bilkul starting state par wapas aata hai, isliye $Δ U = 0$ : gas na toh stored energy mein richer bachi hai na poorer. Yahi ek fact ( $Δ U = 0$ ek cycle mein) First Law ko woh clean bookkeeping rule $W = Q_{H} - Q_{C}$ banata hai jo hum neeche har jagah use karte hain. Dekho First law of thermodynamics.
Reversible engine. Ek reversible engine ek idealised machine hai jiska har step usi states se bina kisi waste ke (koi friction nahi, koi sudden temperature jumps nahi, koi turbulence nahi) ulta chalaaya ja sakta tha. Iska consequence, Carnot's theorem mein prove kiya gaya, ye hai ki absolute temperatures $T_{H}$ (hot) aur $T_{C}$ (cold) par do fixed reservoirs ke beech ye maximum possible efficiency — Carnot bound — hit karta hai: $η_{m a x} = 1 - \frac{T _{C}}{T _{H}} < 1 (T in kelvin) .$ Inhi do reservoirs ke beech koi bhi real (irreversible) engine isse match nahi kar sakta, chhodne ki baat to door. Reversibility "best-case" idealisation hai; real machines isse kam rehti hain (dekho Reversible and irreversible processes).

Parent: Second Law — Kelvin–Planck & Clausius.

Level 1 — Recognition

L1.1 — Law ka naam batao

Problem. Ek gadget claim karta hai ki wo ek akele garm room mein baithkar, quietly ek fan chala sakta hai, sirf usi ek room ki air se heat suck karke, forever, kuch aur hue bina. Second Law ka kaun sa statement ye todata hai?

Recall Solution

KYA dekh rahe hain: ek reservoir (room), heat poori tarah work mein convert ho rahi hai (fan ghuma raha hai), koi cold sink nahi, koi aur change nahi, cyclically ("forever") ho raha hai. ISLIYE ye pin down hota hai: ek single-reservoir cyclic engine jisme $η = 1$ ho, exactly wahi Kelvin–Planck forbidden machine hai. Answer: ye Kelvin–Planck statement violate karta hai.

L1.2 — Law ka naam batao

Problem. Ek sealed box, kisi cheez se plugged in nahi, ek ice cube se ek garm mug mein $50$ J heat transfer karta hai, cycle ke baad cycle, as its only effect. Kaun sa statement ye todata hai?

Recall Solution

KYA dekh rahe hain: heat cold→hot travel karti hai, $W = 0$ (kisi se plugged in nahi), sole effect, cyclic. ISLIYE: zero work ke saath cold→hot as the only effect exactly wahi hai jo Clausius forbid karta hai. Answer: ye Clausius statement violate karta hai.

L1.3 — Legal hai ya illegal?

Problem. Ek engine ek hot reservoir se $Q_{H} = 600$ J absorb karta hai, ek cold reservoir ko $Q_{C} = 250$ J reject karta hai, aur baaki work ke roop mein karta hai. Legal hai ya nahi? (Classify karne ke liye koi number-crunching zaroori nahi.)

Recall Solution

KYA hai: do reservoirs, $Q_{C} = 250 > 0$ , isliye $η < 1$ . ISLIYE legal: koi bhi law break nahi hua — heat sirf partially convert hui hai aur heat ek genuine cold sink ko reject ki gayi hai. Answer: Legal. (Iska efficiency, baad mein compute kiya, $η = 1 - \frac{250}{600} \approx 0.583$ hai.)

Level 2 — Application

L2.1 — Missing quantity fill karo

Problem. Ek engine har cycle mein $Q_{H} = 900$ J leta hai aur $Q_{C} = 540$ J reject karta hai. $W$ aur $η$ find karo.

Recall Solution

Step 1 — ISLIYE subtract karte hain: ek poore cycle mein working substance apne start par wapas aata hai, isliye uski internal energy unchanged hai — $Δ U = 0$ (toolbox mein defined). First Law phir kehti hai jo bhi heat enter hui lekin nahi nikli, woh work ke roop mein nikli — isliye work, heat in aur heat out ka difference hai: $W = Q_{H} - Q_{C} = 900 - 540 = 360 J .$ Step 2 — ISLIYE divide karte hain: efficiency puchti hai "jo fuel maine pay kiya ( $Q_{H}$ ), uska kitna fraction useful work ke roop mein wapas aaya?", isliye hum prize ko price se divide karte hain: $η = \frac{W}{Q _{H}} = \frac{360}{900} = 0.4 = 40%.$ Check: $1 - \frac{Q _{C}}{Q _{H}} = 1 - \frac{540}{900} = 0.4$ . ✔

L2.2 — Refrigerator bookkeeping

Problem. Ek fridge $W = 200$ J electrical work use karke andar se $Q_{C} = 800$ J remove karta hai. Kitchen mein dump hua heat $Q_{H}$ aur COP find karo.

Recall Solution

Step 1 — ISLIYE add karte hain: energy gayab nahi ho sakti; fridge jo bhi leta hai woh sab kitchen mein wapas aana chahiye. Wo $Q_{C}$ cold food se aur electrical work $W$ absorb karta hai, isliye dono milkar $Q_{H}$ ke roop mein bahar aate hain: $Q_{H} = Q_{C} + W = 800 + 200 = 1000 J .$ Step 2 — ISLIYE COP, $η$ nahi: fridge ke liye "prize" removed heat hai, isliye hum pay ki gayi work per $Q_{C}$ score karte hain: $COP_{fridge} = \frac{Q _{C}}{W} = \frac{800}{200} = 4.$ Answer: $Q_{H} = 1000$ J, $COP = 4$ .

L2.3 — Inventor ka claim, test kiya gaya

Problem. Ek inventor ka cyclic engine ek hot reservoir se $Q_{H} = 1000$ J draw karta hai, $Q_{C} = 0$ reject karta hai, aur $W = 1000$ J deliver karta hai. Kya ye possible hai?

Recall Solution

Step 1: $W = Q_{H} - Q_{C} = 1000 - 0 = 1000$ J — First Law khush hai. Step 2 — ISLIYE phir bhi fail hota hai: single reservoir se $Q_{C} = 0$ se $η = 1$ milti hai. Exactly yehi machine hai jo Kelvin–Planck forbid karta hai. Answer: Impossible — Kelvin–Planck violate karta hai. First Law akela isse pakad nahi sakta; tumhe Second ki zaroorat hai.

Level 3 — Analysis

L3.1 — Do engines, same fuel

Problem. Engine A: $Q_{H} = 1000$ J, $Q_{C} = 400$ J. Engine B: $Q_{H} = 1000$ J, $Q_{C} = 250$ J. Dono cyclic hain aur same do reservoirs use karte hain. Kaun zyada work karta hai, kaun zyada efficient hai, aur kitne se?

Recall Solution

Step 1 — ISLIYE har ek ke liye subtract karte hain: jaise L2.1 mein, ek cycle mein work equals heat-in minus heat-rejected. Isliye jo engine kam heat fenkta hai woh zyada work ke roop mein rakhta hai: $W_{A} = 1000 - 400 = 600 J, W_{B} = 1000 - 250 = 750 J .$ Step 2 — har ek ki efficiency (prize over price): $η_{A} = \frac{600}{1000} = 0.6, η_{B} = \frac{750}{1000} = 0.75.$ Step 3 — ISLIYE rejected heat compare karte hain: equal $Q_{H}$ ke saath, jo engine kam waste karta hai woh dono counts par jeetta hai. B, $150$ J kam reject karta hai, isliye $150$ J zyada work deliver karta hai aur $15$ percentage-points zyada efficient hai. Answer: B, $150$ J zyada work karta hai; $η_{B} - η_{A} = 0.15$ .

L3.2 — Engine ko fridge se chain karna

Problem. Ek engine ek hot reservoir se $Q_{H} = 1200$ J leta hai aur $Q_{C} = 800$ J reject karta hai. Uska poora work output ek fridge drive karta hai jiska $COP = 3$ hai. Fridge cold room se kitni heat pull karta hai?

Recall Solution

Step 1 — engine work: $W = 1200 - 800 = 400$ J. Step 2 — fridge ko feed karo: fridge ka work input yahi $W = 400$ J hai. Step 3 — ISLIYE COP invert hota hai: $COP = \frac{Q _{C}^{fridge}}{W}$ , isliye $Q_{C}^{fridge} = COP \times W = 3 \times 400 = 1200 J .$ Answer: fridge cold room se 1200 J remove karta hai.

L3.3 — "Efficiency = 120%" ka claim kahan se aata hai?

Problem. Ek student heat pump ki performance $\frac{Q _{H}}{W} = \frac{1200}{400} = 3$ compute karta hai aur report karta hai "efficiency $= 300%$ , jo $100%$ limit beat karta hai, isliye Second Law break ho gayi." Kuch galat hai kya?

Recall Solution

Step 1 — YE quantity ACTUALLY kya hai? Heat pump ke liye $\frac{Q _{H}}{W}$ uska coefficient of performance $COP_{pump}$ hai (toolbox mein defined), engine efficiency $η = W / Q_{H}$ nahi. Step 2 — ISLIYE koi violation nahi: ek heat pump work $W$ pay karta hai aur zyada heat $Q_{H} = Q_{C} + W > W$ deliver karta hai, isliye $Q_{H} / W > 1$ normal aur expected hai — tum free mein cold-reservoir heat saath mein le ja rahe ho. Ye "kuch se kuch create hona" nahi hai. Step 3 — actual engine efficiency (agar isse engine ke roop mein run karo) hogi $η = W / Q_{H} = 1/3 \approx 33% < 100%$ . Koi law threatened nahi hai. Answer: kuch bhi break nahi hua. Student ne COP ko efficiency se confuse kar diya.

Level 4 — Synthesis

Agle problems tumse kehte hain ki equivalence proof khud build karo. Figure har joule track karta hai.

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

Definition Figure legend — diagram se pehle ye padho

Poora page ek fixed colour code use karta hai, jo blueprint figure se match karta hai:

Amber bar / amber arrows = HOT reservoir aur koi bhi heat ya work hot side se associated hai (work $W$ amber draw hai kyunki ye "valuable" output hai).
Cyan bar / cyan arrows = COLD reservoir aur cold side se/ko flow hone wali koi bhi heat.
Arrow direction = energy flow ki direction; har arrow par number = joules mein uska magnitude.

Figure kya dikhata hai. Ek lamba amber HOT reservoir bar upar across chalta hai, ek lamba cyan COLD reservoir bar neeche across, ek technical grid par. Unke beech do boxes hain. Left par, ek white box labelled ENGINE: ek amber $900$ J arrow HOT se neeche iske andar point karta hai, ek cyan $300$ J arrow isse neeche COLD mein point karta hai, aur ek horizontal amber arrow $W = 600$ J uski side se nikalta hai. Right par, ek cyan box labelled MAGIC C-viol (Clausius-violator): ek cyan $300$ J arrow COLD se upar iske andar point karta hai, aur ek cyan $300$ J arrow isse upar HOT mein point karta hai, koi work arrow bilkul nahi. Cold bar par do $300$ J arrows equal aur opposite hain, isliye cold net zero par aata hai; sirf hot bar se heat drain hoti hai aur sirf work bahar aata hai.

L4.1 — Clausius-violator ⟹ Kelvin–Planck-violator

Problem. Ek magic device zero work use karke $Q = 300$ J cold→hot move karta hai. Tumhare paas ek ordinary engine bhi hai (same do reservoirs) jo hot se $Q_{H} = 900$ J leta hai aur cold ko $Q_{C} = 300$ J reject karta hai. Inhe combine karo. Dikhao ki cold reservoir untouched rehta hai, aur compute karo ki combo hot reservoir se akele kitna net work produce karta hai.

Recall Solution

Step 1 — cold side cancel karo (ISLIYE humne $300$ J match kiya): cold reservoir par direction tally — engine $300$ J add karta hai (heat cold mein flow karti hai) jabki magic device $300$ J remove karta hai (heat cold se bahar flow karti hai). Un named directions ko signed sum ke roop mein likhte hain: $cold mein + 300 cold se bahar - 300 = 0 J .$ Cold reservoir bilkul waisa hi raha jaise tha — jaise kabhi exist hi nahi kiya. Step 2 — hot side par net: engine hot se $900$ J remove karta hai; magic device hot ko $300$ J return karta hai. Hot se remove hua net heat: $900 - 300 = 600 J .$ Step 3 — work unchanged hai: magic device ne koi work consume nahi kiya, isliye ye work ledger mein na kuch add karta hai na subtract. Combo ka work output isliye sirf original engine ka output hai, jो unchanged bacha hai — yahan kuch naya derive nahi ho raha, hum sirf $W = Q_{H} - Q_{C} = 900 - 300 = 600 J$ jo engine se already tha woh re-read kar rahe hain. Conclusion: combo ek single (hot) reservoir se $600$ J leta hai aur use poora $600$ J work mein convert karta hai — ek Kelvin–Planck violation. ∎ Figure mein ye exactly do equal-aur-opposite cyan arrows hain cold bar par jo cancel ho jaate hain, sirf hot bar drain hota hai aur amber work arrow bahar jaata hai.

L4.2 — Kelvin–Planck-violator ⟹ Clausius-violator

Problem. Ek magic K–P engine hot reservoir se $Q_{H} = 500$ J ko poora work $W = 500$ J mein convert karta hai (kuch reject nahi karta). Wo work ek ordinary fridge mein feed karo jiska $COP = 4$ hai. Find karo (a) fridge cold se kitna heat $Q_{C}$ pull karta hai, (b) fridge hot mein kitna heat dump karta hai, (c) hot ko net heat deliver hui, aur confirm karo ki ye ek work-free cold→hot transfer ke barabar hai.

Recall Solution

Step 1 — fridge intake (a): $Q_{C} = COP \times W = 4 \times 500 = 2000$ J cold reservoir se. Step 2 — fridge output (b): $Q_{H}^{fridge} = Q_{C} + W = 2000 + 500 = 2500$ J hot mein dump kiya. Step 3 — hot reservoir par net (c): magic engine ne hot se $500$ J liya; fridge ne hot mein $2500$ J dump kiya: $net hot mein = 2500 - 500 = 2000 J .$ Step 4 — ISLIYE ye Clausius-breaking hai: combo ne cold se $2000$ J nikala aur exactly $2000$ J hot mein push kiya, bina kisi external work ke (work internally magic engine se aaya) aur koi aur change nahi. Conclusion: heat cold→hot as the sole effect move hui — ek Clausius violation. ∎ Note karo ki cold se $Q_{C} = 2000$ J bahar aana hot mein $2000$ J se match karta hai: books perfectly balance hain.

Level 5 — Mastery

L5.1 — Apna khud ka paradox invent karo aur destroy karo

Problem. Tumhare paas same hot aur cold reservoirs ke beech DO engines hain. Engine 1 (ek claimed super-engine) hot se $Q_{H} = 1000$ J leta hai aur sirf $Q_{C} = 200$ J reject karta hai. Engine 2 ek reversible engine hai (toolbox mein defined) jo, jab backward fridge ke roop mein run kiya jata hai, cold se hot mein $600$ J move karne ke liye (aur $1200$ J hot mein dump karne ke liye) $W = 600$ J work chahiye. Engine 1 forward run karo aur uski work ka kuch hissa Engine 2 ko backward drive karne mein use karo. Dikhao ki agar aisa over-good Engine 1 exist karta, tum cold se hot heat pump kar sakte aur phir bhi free work pocket kar sakte — exact joules ke saath paradox pin down karo, aur explain karo ki Engine 2 ki reversible nature hi Engine 1 ko illegal kyun banati hai.

Recall Solution

Step 1 — Engine 1 output: $W_{1} = Q_{H} - Q_{C} = 1000 - 200 = 800$ J work (iska claimed efficiency $η_{1} = 800/1000 = 0.8$ ). Step 2 — reverse machine drive karo: Engine 2 backward run karne ke liye $600$ J chahiye taaki $600$ J cold→hot move ho sake aur $1200$ J hot mein dump ho sake. Engine 1 ke $800$ J mein se $600$ J feed karo, aur $200$ J leftover work apni pocket mein rakh lo. Step 3 — har reservoir account karo (directions naam diye, phir sum kiya).

Cold: Engine 1 cold mein $+ 200$ J dump karta hai; reverse machine cold se $600$ J suck karti hai. Cold par net $= 200 - 600 = - 400$ J (cold $400$ J lose karti hai).
Hot: Engine 1 hot se $1000$ J draw karta hai; reverse machine hot mein $1200$ J dump karti hai. Hot par net $= 1200 - 1000 = + 200$ J (hot $200$ J gain karta hai). Step 4 — leftover work: $W_{net} = 800 - 600 = 200$ J phir bhi bahar aata hai. Step 5 — paradox padho: combo cold se $400$ J pull karta hai, hot mein $200$ J push karta hai, aur tumhe $200$ J free work deta hai — energy phir bhi balance karti hai ( $400 = 200 + 200$ ✔). Lekin ye sab bina kisi external input ke karta hai: heat cold→hot flow hui AUR humne work harvest ki, kuch se. Ye simultaneously ek Clausius aur ek Kelvin–Planck violation hai. Step 6 — ISLIYE Engine 2 ki reversibility crux hai: kyunki Engine 2 reversible hai, ise backward run karna exactly wohi energies return karta hai jo ye forward run karne par produce karta — koi waste hidden nahi. Isliye combo sirf tabhi misbehave kar sakta hai jab Engine 1 us reversible benchmark se out-perform kare. Yahan Engine 1 ki $η_{1} = 0.8$ in reservoirs ke beech kisi bhi reversible engine ki reach se exceed karti hai. Carnot's theorem kehta hai reversible engine champion hai — koi engine ise beat nahi kar sakta — isliye reversible value se upar $η_{1}$ exactly wahi hai jo Second Law todata hai. Reversible Engine 2 "measuring stick" hai; Engine 1 illegal hai kyunki wo stick ko beat karta hai. Answer: paradox hai cold se $- 400$ J, hot ko $+ 200$ J, free work $+ 200$ J — impossible; ye tab hi exist karta hai jab Engine 1, reversible (Carnot) benchmark ko beat kare, jo koi engine nahi kar sakta.

L5.2 — Degenerate & limiting cases

Problem. Fixed $Q_{H} = 1000$ J wale ek cyclic engine ke liye, describe karo ki har limiting case mein $η$ , $W$ , aur $Q_{C}$ ka kya hota hai, aur batao ki kaun sa (agar koi) Second Law allow karta hai: (a) $Q_{C} \to 0$ , (b) $Q_{C} \to Q_{H} = 1000$ J, (c) $W = 0$ do reservoirs ke saath, (d) $Q_{H} = 0$ .

Recall Solution

(a) $Q_{C} \to 0$ : tab $W \to Q_{H} = 1000$ J aur $η \to 1$ . Ye Kelvin–Planck ceiling hai — heat poori tarah work mein convert, kuch reject nahi. Actually pahunchne ke liye Forbidden: tum iske paas ja sakte ho lekin chhoo nahi sakte, kyunki $η = 1$ par ek single cold-free cyclic engine exactly K–P machine hai. (b) $Q_{C} \to Q_{H} = 1000$ J: tab $W \to 0$ aur $η \to 0$ . Hot se enter hua saara heat straight through jaata hai aur cold mein nikal jaata hai, koi work extract nahi. Allowed — ye sirf ordinary hot→cold heat flow hai jo ek (useless) "engine" ke roop mein disguise mein hai. Koi machine jo kuch nahi karti use koi cheez forbid nahi karti. (c) $W = 0$ do reservoirs ke saath: $W = Q_{H} - Q_{C}$ mein $W = 0$ set karna $Q_{C} = Q_{H} = 1000$ J force karta hai — case (b) se identical. Ye plain spontaneous hot→cold conduction hai, jise Second Law positively permit karta hai (actually natural direction ke roop mein insist karta hai). Allowed. (d) $Q_{H} = 0$ : hot se koi heat absorb nahi hoti, isliye $W = Q_{H} - Q_{C} = - Q_{C} \leq 0$ . Non-negative work output impossible hai jab tak $Q_{C} = 0$ bhi na ho (tab sab kuch zero hai — machine simply idle rehti hai). Agar instead $Q_{C} > 0$ aur work supplied hai ( $W < 0$ ), device ab engine rahi hi nahi: ye heat around push karne ke liye work absorb kar rahi hai, yaani ye fridge/heat-pump ban gayi. Allowed, lekin sirf relabelled machine ke roop mein — yahan kuch violate karne ke liye koi engine nahi hai. Big picture: poore range mein sirf ek hi outlawed limit hai (a), $Q_{C} \to 0$ . $η = 0$ se lekin $η = 1$ shaamil nahi tak har efficiency achievable Carnot ceiling $η_{m a x} = 1 - T_{C} / T_{H} < 1$ (toolbox) se bounded above hai. Kelvin–Planck sirf $η = 1$ ban karta hai; Carnot $η_{m a x}$ se upar sab kuch ban karta hai, jo stricter hai.

L5.3 — Carnot ceiling: real engine kitna close ja sakta hai?

Problem. Ek reservoir pair hai $T_{H} = 600$ K (hot) aur $T_{C} = 300$ K (cold). (a) Inke beech kisi bhi engine ki maximum possible efficiency kya hai? (b) Inhi reservoirs par ek real engine $η = 0.40$ deliver karta hai. Kya ye legal hai, aur ye ideal ceiling ka kitna fraction reach karta hai? (c) Ek inventor same reservoirs par $η = 0.55$ claim karta hai. Legal?

Recall Solution

Step 1 — ISLIYE ab temperatures use karte hain: joules mein efficiency ( $η = 1 - Q_{C} / Q_{H}$ ) tumhe batati hai engine ne kya kiya; Carnot bound $η_{m a x} = 1 - T_{C} / T_{H}$ tumhe batata hai in do reservoirs ke beech koi bhi engine best mein kya kar sakta hai. Absolute (kelvin) temperatures hi woh ceiling set karte hain. (a) $η_{m a x} = 1 - \frac{T _{C}}{T _{H}} = 1 - \frac{300}{600} = 1 - 0.5 = 0.50.$ Isliye yahan koi engine $50%$ exceed nahi kar sakta. (b) $η = 0.40 < 0.50$ , isliye ye legal hai. Ideal ka fraction jo ye achieve karta hai: $\frac{0.40}{0.50} = 0.80$ — ye reversible ceiling ka $80%$ reach karta hai (ek achha, lekin perfect nahi, real engine). (c) $η = 0.55 > 0.50 = η_{m a x}$ . Ye reversible (Carnot) champion को beat karne ka claim karta hai. Carnot's theorem ke zariye ye impossible hai — ise backward run kiye gaye reversible engine se hook karna exactly L5.1 ka paradox reproduce karta hai. Illegal, even though $0.55 < 1$ isliye bare Kelvin–Planck akela ise nahi pakad paata. Answer: (a) $η_{m a x} = 0.50$ ; (b) legal, ceiling ka $80%$ ; (c) illegal — Carnot bound exceed karta hai.

Recall Quick self-check ledger (sab verified)

L2.1 $W = 360$ J, $η = 0.4$ ::: L2.2 $Q_{H} = 1000$ J, COP $= 4$ ::: L2.3 impossible ( $η = 1$ ) L3.1 $Δ W = 150$ J, $Δ η = 0.15$ ::: L3.2 fridge $1200$ J remove karta hai ::: L3.3 koi violation nahi ( $η = 1/3$ ) L4.1 $W = 600$ J hot se akele ::: L4.2 $Q_{C} = 2000$ , dump $2500$ , hot mein net $2000$ J L5.1 cold $- 400$ , hot $+ 200$ , free work $+ 200$ ::: L5.2 sirf $Q_{C} \to 0$ forbidden ::: L5.3 $η_{m a x} = 0.50$ , real $80%$ reach karta hai, $0.55$ illegal.

Connections

Parent: Kelvin–Planck & Clausius
In bounds ka implied ceiling: Carnot engine and Carnot theorem
Sharper, sign-tracked version: Entropy and the Clausius inequality
Energy accounting jis par hum rely kiye: First law of thermodynamics
"Cyclic + no other effect" kyun matter karta hai: Reversible and irreversible processes
Poore law ka deep meaning: Arrow of time