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

1.7.18 · D3 · Physics › Thermodynamics › Second law — Kelvin-Planck statement, Clausius statement

Yeh page ek decision gym hai. Parent note ne aapko dono statements di aur prove kiya ki woh equivalent hain. Yahan hum har tarah ki machine aap par throw karte hain — legal, illegal, degenerate, word-problem, exam-trap — aur har ek ko usi checklist se grind karte hain jab tak ki violation dhundhna reflex na ban jaaye.

Examples se pehle, humein ek shared toolkit chahiye. Neeche sab kuch sirf teen seedhe facts use karta hai, toh chaliye pehle unhe pin down karte hain — aur, kyunki unme se ek $U$ naam ki quantity name karta hai, hum usse bhi define karte hain.

Definition Sign convention: heat aur work directions kaise padhein

Is poori page mein hum $Q_{H}$ , $Q_{C}$ , $W$ ko magnitudes ki tarah likhte hain — hamesha positive numbers, sirf "kitne joules." Direction $+$ ya $-$ se nahi aata; yeh reservoir ke naam se aur jo machine humne draw ki us se aata hai:

$Q_{H}$ = Hot reservoir ke saath exchange hue joules. Engine ke liye yeh heat andar aa rahi hai (hot → machine); fridge ke liye yeh heat bahar ja rahi hai (machine → hot).
$Q_{C}$ = Cold reservoir ke saath exchange hue joules. Engine ke liye yeh heat bahar hai (machine → cold); fridge ke liye yeh heat andar hai (cold → machine).
$W$ = work ke joules. Engine work banata hai (out); fridge work kharcha karta hai (in). Toh jis moment aap "engine" ya "fridge" kehte hain, har arrow ki direction fix ho jaati hai aur aap sirf positive numbers handle karte hain. Aisa kyun? Yeh sign-juggling hatata hai: aap direction picture dekh kar decide karte hain, minus signs track karke nahi.

Formula Teen tools jo hum baar baar use karenge

Tool 1 — First Law over a cycle. First Law bookkeeping hai: heat in $-$ heat out $=$ work out $+$ rise in internal energy, yaani $Q_{H} - Q_{C} = W + Δ U$ . $Δ U$ kyun drop karein? Kyunki cycle mein $Δ U = 0$ (upar definition), isliye internal-energy term vanish ho jaata hai aur hum pure input-minus-output ke saath reh jaate hain: $W = Q_{H} - Q_{C} (First Law over a cycle)$ Neeche har jagah ise trust kyun kar sakte hain: is page ki har machine cycle mein chalti hai, isliye yeh bina exception ke hold karta hai.

Tool 2 — Engine efficiency (ek device jo work banata hai): $η = \frac{W}{Q _{H}} = 1 - \frac{Q _{C}}{Q _{H}}$ .

Tool 3 — Fridge/heat-pump COP (ek device jo work kharcha karke heat cold→hot move karta hai). "COP" Coefficient Of Performance ka short form hai — ise padhein "cold-side joules moved per joule of work paid": $COP = \frac{Q _{C}}{W}$

Yahan $Q_{H}$ = hot reservoir ke saath swap ki gayi heat, $Q_{C}$ = cold reservoir ke saath swap ki gayi heat, $W$ = net work, sab magnitudes (upar sign-convention box dekho).

Intuition Universal checklist (har claim par yeh run karo)

Pehle, step 3 mein hum jo metaphors use karte hain unke baare mein ek baat. Heat akeli chhodne par hamesha hot body se cold body ki taraf drift karti hai — jaise paani neeche ki taraf behta hai. Toh "downhill heat flow" ka matlab hai hot→cold (free, spontaneous direction, high temperature se low), aur "uphill heat flow" ka matlab hai cold→hot (low temperature se high — woh direction jo kabhi free mein nahi hota aur pump chahiye). Temperature height ka role play karta hai.

Reservoirs count karo. Ek ya do?
Work $W$ dhundho. Kya yeh andar ja raha hai (fridge) ya bahar (engine)? Kya yeh zero hai?
Heat ki direction dhundho. Hot→cold (downhill, free) ya cold→hot (uphill, costs)?
Matching statement apply karo. Single reservoir + $η = 1$ ⟹ Kelvin–Planck verdict. Cold→hot with $W = 0$ ⟹ Clausius verdict.
Numbers sanity-check karo: $0 \leq η < 1$ , sab magnitudes $\geq 0$ , energy balances.

Scenario matrix

Har second-law question inhi cells mein se ek hai. Ex# column batata hai kaunsa worked example us cell ko nail karta hai.

Cell	Ise kya special banata hai	Expected verdict	Ex#
A Single reservoir, $Q_{C} = 0$	Degenerate "perfect engine"	Illegal (K–P)	1
B Two reservoirs, $Q_{C} > 0$ , $η < 1$	Ordinary honest engine	Legal	2
C Cold→hot, $W = 0$	Degenerate "free fridge"	Illegal (Clausius)	3
D Cold→hot, $W > 0$	Ordinary honest fridge	Legal	4
E $η$ best possible se upar	Carnot limit se zyada	Illegal (implied)	5
F $η < 0$ ya $Q_{C} < 0$ ya $W > Q_{H}$	Numbers First Law hi tod dete hain	Illegal (First Law)	6
G Real-world word problem	Ocean/space engine, hidden reservoir	Reason it out	7
H Equivalence twist	Do devices combine karo, leak dhundho	Illegal (composite)	8
I Limiting case $Q_{C} \to 0$ ya $T_{C} \to 0$	Boundary par kya hota hai	Approaches limit	9

Cell E aur I ko hum Carnot engine and Carnot theorem ke saath light touch se cover karte hain, jo efficiency par ceiling set karta hai — parent note aapko batata hai $η < 1$ ; Carnot batata hai kitna kam.

Example 1 — Cell A: perfect engine

Worked example Claim: ek engine

ek reservoir se $Q_{H} = 600$ J leta hai aur per cycle $W = 600$ J output karta hai. Forecast: Kya yeh legal hai? Padhne se pehle guess karo — kya numbers mein waste heat ke liye room hai?

Reservoirs count karo. Ek (sirf ek hot source named hai). Yeh step kyun? Kelvin–Planck specifically single-reservoir engines ke baare mein hai.
First Law: $Q_{C} = Q_{H} - W = 600 - 600 = 0$ J. Kyun? Cycle mein $W = Q_{H} - Q_{C}$ , isliye $Q_{C}$ force ho jaata hai jab $Q_{H}, W$ fix hoon — hum ise choose nahi kar sakte.
Efficiency: $η = W / Q_{H} = 600/600 = 1$ . Kyun decisive? Ek reservoir se $η = 1$ bilkul wohi machine hai jo Kelvin–Planck forbid karta hai.
Verdict: Illegal — Kelvin–Planck violation. ❌

Verify: $Q_{C} = 0$ ka matlab zero heat rejected; ek real cycle mein kuch heat reject honi hi chahiye (gas ko reset karna padta hai), toh $Q_{C} = 0$ tell-tale hai. Units check: J − J = J. ✓

Example 2 — Cell B: honest engine

Example 3 — Cell C: free fridge

Example 4 — Cell D: honest fridge

Example 5 — Cell E: Carnot ceiling ko beat karna

Worked example Ek engine

$T_{H} = 500$ K aur $T_{C} = 300$ K ke beech run karta hai aur $η = 0.55$ claim karta hai. Forecast: Yeh heat reject karta hai, $η < 1$ , toh Kelvin–Planck khush hai. Lekin kya koi tighter limit hai?

Pehle, ceiling kahan se aati hai? Hum yahan heavy machinery ke bina argument sketch kar sakte hain (full proof Carnot engine and Carnot theorem mein hai). Socho best conceivable engine — ek itna gentle ki woh bina waste ke fridge ki tarah backwards bhi chal sake (ek reversible wala, dekho Reversible and irreversible processes). Us ideal engine ke liye jo heats woh swap karta hai woh dono reservoirs ke absolute temperatures ke same ratio mein nikalni hain: $\frac{Q _{C}}{Q _{H}} = \frac{T _{C}}{T _{H}}$ . Ise $η = 1 - Q_{C} / Q_{H}$ mein plug karo aur ceiling milti hai $η_{m a x} = 1 - T_{C} / T_{H}$ .

Yeh ceiling kyun hai aur sirf ek possible value kyun nahi? Maan lo koi engine ise beat kare. Ideal reversible engine ko backwards fridge ki tarah chalao aur beater ki work se feed karo. Parent note ka equivalence trick phir single-reservoir $η = 1$ device banata hai — Kelvin–Planck violation. Toh kuch bhi $η_{m a x}$ exceed nahi kar sakta. Figure neeche dekho.

Best possible efficiency (Carnot bound): $η_{m a x} = 1 - \frac{T _{C}}{T _{H}} = 1 - \frac{300}{500} = 0.40$ . Kelvin kyun? Ratio absolute temperatures ka hai.
Compare karo: claimed $0.55 > 0.40 = η_{m a x}$ . Kyun decisive? Reversible bound exceed karne se Kelvin–Planck violator ban sakta hai.
Verdict: Illegal — Carnot limit beat karta hai. ❌

Verify: temperatures kelvin mein honi chahiye (absolute temperature ke ratios). $300/500 = 0.6$ , isliye $η_{m a x} = 0.4$ . Claimed $0.55$ fail karta hai. ✓

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

Figure s01 — Efficiency ceiling. Horizontal axis $T_{C} / T_{H}$ hai (0 se 1 tak); vertical axis efficiency $η$ hai. Ek magenta line $η_{m a x} = 1 - T_{C} / T_{H}$ trace karti hai: jab cold side absolute zero par hoti hai toh $η = 1$ se shuru hoti hai aur jab dono temperatures match karte hain toh $η = 0$ par aa jaati hai. Line ke upar ka poora region violet shaded hai aur stamped hai "FORBIDDEN (beats Carnot)"; neeche ka region peach hai aur stamped hai "allowed engines". Orange dot Example 5 mark karta hai: $T_{C} / T_{H} = 0.6$ , line par $η_{m a x} = 0.4$ par, aur claimed $η = 0.55$ forbidden zone mein red X ki tarah plot kiya gaya hai.

Kya dhundhen: magenta line ek hard fence hai. Har real engine ise par ya neeche rehna chahiye. Example 5 ka claim (red X) fence ke upar float karta hai — instant disqualification, koi arithmetic nahi chahiye ek baar aap ise dekh sako. Notice karo ki line sirf $η = 1$ ko far left par touch karti hai, jahan $T_{C} / T_{H} = 0$ (absolute-zero cold reservoir) — isliye full efficiency ek limit hai jise aap approach karte hain, kabhi reach nahi karte (Example 9).

Example 6 — Cell F: numbers jo First Law tod dete hain

Worked example Teen engine claims. Kaunse numbers

Second Law tak pahunche bina hi impossible hain? (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: Inme se kaunse Kelvin–Planck se bhi simple reasons ki wajah se nahi ho sakte?

(a): $Q_{C} = Q_{H} - W = 500 - 600 = - 100$ J. Kyun bura? Negative rejected-heat magnitude ka matlab work input se zyada hai — energy kuch se create hoti hai. First Law broken. ❌
(b): $Q_{C} = - 50$ J negative magnitude hai. Kyun bura? Magnitudes negative nahi ho sakti; engine ke liye koi physical "negative rejected heat" nahi hoti. First Law / definition broken. ❌
(c): Consistency check karo: $W$ hona chahiye $Q_{H} - Q_{C} = 400 - 100 = 300$ J, lekin claim kehta hai $W = 400$ J. Kyun bura? Teen numbers balance nahi karte — $400 \neq = 300$ . First Law broken. ❌
Verdict: teeno First Law fail karte hain; Second Law ko kabhi chance hi nahi milta.

Verify: (a) $500 - 600 = - 100 < 0$ ; (b) $- 50 < 0$ ; (c) $400 - 100 = 300 \neq = 400$ . Sab inconsistent. ✓ Moral: parent note ka First Law pehla gate hai; Second Law doosra.

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

Worked example Ek inventor ek ship pitch karta hai jo "koi fuel nahi jalaati — woh apna engine purely surrounding ocean ko cool karke chalati hai, jo

$15^{\circ} C$ par ek bada reservoir hai, aur woh heat propulsion mein convert karta hai." Forecast: Energy conserved hai (ocean exactly utna lose karta hai jitna work gained hota hai). Toh kaunsa law, agar koi, yeh todta hai?

Reservoirs count karo. Sirf ocean — ek single reservoir ek temperature par. Kyun crucial? Waste heat dump karne ke liye koi colder body available nahi hai.
Energy trace karo: ocean heat $\to$ work, $Q_{C} = 0$ , $η = 1$ . Kyun? Agar sab extracted heat propulsion ban jaaye bina kuch reject kiye, toh yeh single-reservoir $η = 1$ engine hai.
Statement match karo: exactly Kelvin–Planck-forbidden machine. Kyun yeh log fool karta hai: yeh First Law toh obey karta hai — ocean thanda padta hai, isliye energy conserved hai. Isliye hume ise rule out karne ke liye second law chahiye.
Verdict: Illegal — Kelvin–Planck. Ship ek perpetual motion machine of the second kind hogi. ❌

Verify: Ek reservoir ke saath, $η = 1 - Q_{C} / Q_{H}$ tab $1$ ho sakta hai jab $Q_{C} = 0$ ; single source se $Q_{C} = 0$ banned hai. Yeh Cell A hai nautical costume mein — Reversible and irreversible processes aur Arrow of time dekho ki kyun nature direction enforce karta hai.

Example 8 — Cell H: composite device mein leak dhundho

Worked example Koi do boxes ko bolts se jodta hai aur claim karta hai ki pair legal hai:

Box 1 (engine): hot reservoir se $Q_{H} = 1000$ J leta hai, $W = 1000$ J output karta hai, $Q_{C} = 0$ reject karta hai.
Box 2 (fridge): woh $W = 1000$ J fed hai, cold reservoir se $Q_{C} = 400$ J pull karta hai, hot mein $1400$ J dump karta hai.

"Har box ke numbers honest lagte hain," woh kehte hain. Kya combination legal hai? Forecast: Har reservoir par net heat track karo — woh kahan end hoti hai?

Box 1 akele test karo: single-source engine with $Q_{C} = 0$ , $η = 1$ . Kyun? Pehle se hi Kelvin–Planck violation hai (Cell A) — toh composite doomed hai. Lekin composite ka signature bhi dekhte hain.
Cold reservoir par net: sirf Box 2 ise touch karta hai: $- 400$ J (cold se heat removed). Kyun track karo? Hum duniya par net effect hunt kar rahe hain.
Hot reservoir par net: $- 1000$ (Box 1 ko) $+ 1400$ (Box 2 se) $= + 400$ J gained. Kyun? Hot side ke saath exchange ki gayi har heat add karo.
Net effect padho: $400$ J cold→hot moved, work internally cancel ho jaata hai ( $1000$ made $= 1000$ spent), koi aur change nahi. Kyun decisive? Woh heat cold→hot hai sole effect ki tarah — Clausius violation.
Verdict: Illegal. Ek impossible box ( $Q_{C} = 0$ engine) Clausius-violating composite banata hai — exactly parent note ka equivalence argument, numbers ke saath run kiya gaya. ❌

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

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

Worked example Ek engine ki best-possible efficiency ka kya hota hai jab cold reservoir thanda aur thanda hota jaata hai,

$T_{C} \to 0$ K, aur $T_{H} = 400$ K fixed hai? Forecast: Kya $η_{m a x}$ kabhi actually $1$ reach kar sakta hai?

Bound likho: $η_{m a x} = 1 - \frac{T _{C}}{T _{H}}$ . Kyun? Yeh Example 5 mein derived ceiling hai.
Kai $T_{C}$ par evaluate karo: $T_{C} = 200 \Rightarrow η_{m a x} = 0.5$ ; $T_{C} = 40 \Rightarrow 0.90$ ; $T_{C} = 4 \Rightarrow 0.99$ . Kyun tabulate karo? Trend dekhne ke liye, sirf endpoint nahi.
Limit lo: jab $T_{C} \to 0^{+}$ , $T_{C} / T_{H} \to 0$ , toh $η_{m a x} \to 1$ . Yeh K–P violate kyun nahi karta? $η = 1$ reach karne ke liye $T_{C} = 0$ exactly chahiye hoga — ek unattainable absolute-zero reservoir. Toh real machines ke liye $η < 1$ hamesha hold karta hai; hum sirf $1$ approach karte hain.
Doosra degenerate end bhi check karo ( $Q_{C} \to 0$ do reservoirs ke saath): $η = 1 - Q_{C} / Q_{H} \to 1$ jab $Q_{C} \to 0$ , lekin $Q_{C}$ sirf zero approach kar sakta hai, cyclic engine ke liye kabhi hit nahi kar sakta — heat side se wahi boundary.
Verdict: limiting behaviour $η_{m a x} \to 1$ hai lekin kabhi $= 1$ nahi. Boundary approach ki jaati hai, kabhi touch nahi hoti. ✓

Verify: $1 - 40/400 = 0.9$ ; $1 - 4/400 = 0.99$ ; $1 - 200/400 = 0.5$ . Sequence $1$ ki taraf climb karti hai lekin kabhi equal nahi hoti. ✓ Example 5 ke figure par, yeh magenta line hai jo top-left corner ki taraf creep karti hai bina kabhi $η = 1$ par landing kiye, sirf us ek unreachable point $T_{C} / T_{H} = 0$ ko chhodkar.

Recall Matrix par quick self-test

Single reservoir, $η = 1$ — kaunsa law? ::: Kelvin–Planck (Cell A / Ex 1, 7). Cold→hot with $W = 0$ — kaunsa law? ::: Clausius (Cell C / Ex 3). Cold→hot with $W > 0$ — legal hai ya nahi? ::: Legal, ordinary fridge (Cell D / Ex 4). Engine $η$ above $1 - T_{C} / T_{H}$ claim kare — verdict? ::: Illegal, Carnot bound beat karta hai (Cell E / Ex 5). Engine ke liye $W > Q_{H}$ — pehle kaunsa law fail hota hai? ::: First Law, Second Law apply hone se pehle (Cell F / Ex 6). Jab $T_{C} \to 0$ , $η_{m a x}$ kya karta hai? ::: 1 approach karta hai lekin kabhi reach nahi karta (Cell I / Ex 9).

Connected reading

Parent: Second law — Kelvin-Planck statement, Clausius statement (index 1.7.18)
Ex 5 & 9 mein use ki gayi efficiency ceiling: Carnot engine and Carnot theorem
Quantitative law kahan rehta hai: Entropy and the Clausius inequality
Second Law se pehle ka gate (Ex 6): First law of thermodynamics
Direction baked in kyun hai: Reversible and irreversible processes, Arrow of time