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

1.7.18 · D5 · Physics › Thermodynamics › Second law — Kelvin-Planck statement, Clausius statement

Definition Is POORE page par use hone wala sign convention

Har equation ko unambiguous rakhne ke liye hum heat aur work ko magnitudes with an arrow ke roop mein track karte hain, na ki signed numbers ke roop mein:

$Q_{H} > 0$ = heat ki woh amount jo hot reservoir se nikalta hai aur machine mein jaata hai.
$Q_{C} > 0$ = heat ki woh amount jo machine se nikalta hai aur cold reservoir mein jaata hai.
$W > 0$ = net work jo machine bahar deliver karta hai (refrigerator ke liye, $W > 0$ woh work hai jo machine mein deliver hota hai).
Kyunki ek complete cycle machine ko uske start par wapas laata hai, $Δ U = 0$ , aur First Law ek pure bookkeeping ban jaata hai: $heat in = heat out + work out$ , yaani $Q_{H} = Q_{C} + W$ . Rearrange karne par, $W = Q_{H} - Q_{C}$ .

$Q_{H} - W = Q_{C}$ ko aise padho: "jo bhi heat andar aayi lekin work nahi bani, use cold side mein dump karna padega." Koi hidden minus signs nahi hain kahin bhi.

Intuition Quick vocabulary refresh

Reservoir = ek itna bada body ki heat add ya remove karne par uska temperature kabhi nahi badlata.
Cycle = machine apni exact starting state par wapas aati hai, isliye ek loop mein $Δ U = 0$ hota hai.
Kelvin–Planck (K–P): koi bhi cyclic engine ek reservoir ki heat ko poori tarah work mein nahi badal sakta (iske liye $Q_{C} = 0$ chahiye, jo forbidden hai).
Clausius: koi bhi cyclic device heat ko cold→hot bina kuch aur hue nahi move kar sakti (iske liye $W = 0$ chahiye, jo forbidden hai).
Carnot engine (neeche kuch answers mein use hua hai): do reservoirs ke beech sabse zyada efficient possible engine; iski ceiling efficiency $η_{m a x} = 1 - \frac{T _{C}}{T _{H}}$ hai, jo kisi bhi real cold temperature $T_{C} > 0$ ke liye $1$ se kam hai. Ye inline fact yaad rakh taaki page chhodni na pade.

Worked example Do-question heuristic, ek concrete cycle par apply kiya gaya

Almost har trap do sawaal puchne se dissolve ho jaata hai: "Kya ye truly ek cycle hai?" aur "Hidden work / rejected heat kahan hai?" Chaliye dekhte hain ki ye ek real claim par kaise kaam karte hain. Claim: "Ek isothermal expansion $Q = 600$ J leta hai aur $W = 600$ J work output karta hai — 100% conversion, toh K–P khatam." Sawaal 1 — kya ye ek cycle hai? Nahi. Gas expanded end hoti hai (bada volume), apne start par wapas nahi. Toh $Δ U = 0$ sirf us ek stroke ke liye sach tha, ek loop ke liye nahi. Sawaal 2 — hidden rejected heat kahan hai? Gas ko reuse karne ke liye use compress karna padega. Maan lo return strokes mein $W_{comp} = 250$ J work andar daalni padti hai aur $Q_{C} = 250$ J cold side mein dump hota hai. Ab closed loop ka hisaab karo: $W_{net} = 600 - 250 = 350 J, Q_{C} = 250 J > 0, η = \frac{350}{600} \approx 0.583 < 1.$ Jaise hi loop close hota hai, $Q_{C} > 0$ appear ho jaata hai, $η$ 1 se neeche aa jaata hai, aur K–P safe hai. "100%" half cycle dekhne ka illusion tha. Upar ki figure dekho: missing $Q_{C}$ arrow hamesha dikhta hai.

True or false — justify

The First Law already forbids a cup of coffee heating itself by cooling the room.

False — woh process energy ko perfectly conserve karti hai (heat sirf room se coffee mein move hoti hai, total unchanged), isliye First Law usse theek maanta hai; sirf Second Law spontaneous cold→hot direction ko forbid karta hai, isliye hum ise kabhi nahi dekhte.

A single isothermal expansion converts 100% of absorbed heat into work, so Kelvin–Planck is violated.

False — ek expansion cycle nahi hai: gas expanded reh jaati hai, isliye

Δ U = 0

sirf us stroke ke liye sach tha. Loop close karo aur compression heat reject karta hai (

Q_{C} > 0

), jisse

η

1 se neeche aa jaata hai — bilkul wahi 600 J → 350 J worked example jaise.

Your household fridge violates the Clausius statement because it moves heat from cold food to warm air.

False — Clausius cold→hot ko tab forbid karta hai jab ye sole effect ho, yaani

W = 0

ho. Tumhare fridge ka compressor electrical work khaata hai (

W > 0

, roughly wall se 100–200 W), aur woh non-zero

W

hi "other effect" hai jo ise legal rakhta hai.

The Second Law says heat can never be turned into work.

False — heat routinely work mein badla jaata hai (yahi toh har engine karta hai); law sirf ise entirely work mein badalne se rokta hai ek single reservoir se ek cycle mein, yaani ye

η < 1

cap karta hai, ye

W > 0

ban nahi karta.

If Kelvin–Planck were false, Clausius could still be true.

False — dono logically equivalent hain: K–P-violating engine ke free work ko ek ordinary fridge mein lagao aur tum heat cold→hot pump kar sakte ho bina net work ke, jisse Clausius toot jaata hai. Ek todo aur arrows doosra bhi todne dete hain.

An engine with efficiency exactly

η = 1

is allowed as long as it uses two reservoirs.

False —

η = 1

Q_{C} = 0

force hota hai, isliye cold-side arrow gayab ho jaata hai aur cold reservoir untouched rehta hai; machine effectively single reservoir se khaati hai aur K–P violate karti hai, chahe paper par kitne bhi reservoirs list karo.

A refrigerator can have a coefficient of performance greater than 1.

True —

COP = Q_{C} / W

aasaani se 3–5 ho sakta hai kyunki tum spend kiye gaye work se zyada heat move karte ho (1 J work spend karo, 4 J heat shift karo). Ye koi law nahi todta, kyunki

W > 0

genuinely pay kiya jaata hai — arrows sirf ek chota

W

aur bada

Q_{C}

dikhate hain.

The word "cyclic" in both statements is just a technicality with no physical bite.

False — ye hi poora point hai: "cyclic" hatao aur dono statements false ho jaati hain, kyunki ek single non-cyclic stroke can heat ko fully work mein convert kar sakta hai ya heat ko bina kuch kiye uphill shift kar sakta hai. Bite yeh hai ki machine ko apne start par wapas aana padta hai, jo rejected-heat aur paid-work arrows ko wapas laane par majboor karta hai.

Spot the error

"My engine takes 500 J from one reservoir and outputs 500 J of work per cycle — perfectly legal since energy is conserved."

Error: convention se

Q_{C} = Q_{H} - W = 500 - 500 = 0

; ek single reservoir ke saath

Q_{C} = 0

ka matlab

η = 1

hai, exactly woh K–P-forbidden machine. Energy conservation zaroori hai legality ke liye lekin kaafi nahi.

"To prove K–P and Clausius equivalent, I show that violating K–P also violates K–P."

Error: equivalence ke liye cross implications chahiye — ek K–P violator ko dikhana hoga ki wo Clausius violator banata hai (aur vice versa), na ki apna hi statement phir se todta hai. Same-statement implication kuch prove nahi karta.

"A Clausius-violator moves

Q_{C}

cold→hot for free, so I pair it with an engine whose rejected heat is anything I like."

Error: tumhe engine ki rejected heat ko exactly magic device ke

Q_{C}

ke barabar tune karna hoga, taaki cold reservoir ka net change exactly zero ho aur cancel ho jaaye; warna combined machine ab bhi cold reservoir ko touch karti hai aur ek clean single-reservoir engine nahi hai.

"Since the fridge dumps

Q_{H} = Q_{C} + W

into the hot side, it creates energy."

Error: kuch create nahi hota — paid work

W

pulled heat

Q_{C}

mein add hota hai, aur unka sum hi dump hota hai. Ye sirf First Law arrows ka add up hai (

Q_{H} = Q_{C} + W

), energy kahi se nahi.

"The Carnot engine reaches

η = 1

, so it beats the Second Law."

Error: Carnot sabse zyada efficient allowed engine hai, jiska

η_{m a x} = 1 - T_{C} / T_{H}

hota hai, jo kisi bhi finite

T_{C} > 0

ke liye 1 se neeche rehta hai. Ye K–P ki set limit ko chhuta hai; kabhi cross nahi karta. (Poora treatment: Carnot engine and Carnot theorem.)

"Feeding an engine's full work output into a fridge always violates Clausius."

Error: sirf tab agar engine magic K–P violator ho (

W = Q_{H}

, koi heat reject nahi). Ek ordinary engine

Q_{C} > 0

reject karta hai, isliye combined device reservoirs par extra effect chhodta hai aur legal rehta hai.

Why questions

Why must we define "heat engine" and "refrigerator" before stating the Second Law?

Kyunki dono statements precisely inhi do cyclic machines ke baare mein hain ki ye kya nahi kar sakti — forbidden

Q_{C} = 0

engine aur forbidden

W = 0

fridge. Machines ke bina law ke paas forbid karne ke liye kuch concrete nahi.

Why does "in a cycle" rescue Kelvin–Planck from the isothermal-expansion counterexample?

Ek complete loop mein working substance apne start par wapas aati hai (

Δ U = 0

), jo ek compression stroke ko force karta hai; woh stroke heat reject karta hai, jisse net

Q_{C}

positive ho jaata hai — jaise worked example mein 250 J ne dikhaya.

Why does the equivalence proof use contradiction rather than a direct construction?

Contradiction se tum ek rule-breaking magic device ko assume karke uspar ek ordinary machine bolt kar sakte ho taaki doosra rule toot jaaye, ye dikhate hue ki dono prohibitions saath khade hain ya girte hain. Dekho Reversible and irreversible processes ki real machines magic ideal ke paas kabhi kyun nahi pahunchti.

Why is the Second Law called the "arrow of time"?

Kyunki ye spontaneous change ke liye ek preferred direction single out karta hai (hot→cold, work-free processes reverse mein forbidden), jo time ko ek forward orientation deta hai jo time-symmetric First Law mein absent hai — dekho Arrow of time.

Why can't you argue the two statements are equivalent using only the First law of thermodynamics?

First Law sirf check karta hai ki arrows balance ho rahe hain (

Q_{H} = Q_{C} + W

); ye direction-blind hai aur forbidden aur allowed machines ko equally permit karta hai, isliye ye do Second-Law prohibitions ko distinguish ya link nahi kar sakta.

Why does moving heat cold→hot always cost work in reality, connecting to Entropy and the Clausius inequality?

Uphill transfer reservoirs ki total entropy kum karta hai, isliye ise work se pay karna padta hai jo kam se kam utni entropy waapas dump kare; ek free version Clausius inequality todega.

Edge cases

If a cold reservoir is at absolute zero (

T_{C} = 0

), does

η = 1

become allowed?

Nahi — Carnot ceiling

η_{m a x} = 1 - T_{C} / T_{H}

T_{C} = 0

par 1 padega, lekin

T_{C} = 0

tak pahunchna khud Third Law se forbidden hai, isliye loophole physically kabhi nahi khulta.

What happens to the K–P statement if the two reservoirs are at the same temperature?

Ek effective reservoir ke saath exploit karne ke liye koi temperature drop nahi hai;

η_{m a x} = 1 - T_{C} / T_{H} = 0

, isliye koi net work extract nahi ho sakta — K–P ka sabse strong possible statement.

A device transfers heat cold→hot but also permanently magnetizes a bar of iron as a side effect. Legal?

Clausius literally worded ke against possibly legal hai (ek aur effect hai), lekin us side effect ko zaroori entropy/work supply karni chahiye; genuinely cost-free version phir bhi impossible rehta hai.

Can a refrigerator run with

W = 0

if the "cold" side is actually hotter than the "hot" side?

Tab heat hot→cold spontaneously flow karta hai aur koi work nahi chahiye — lekin ye ordinary downhill flow hai, forbidden uphill Clausius transfer nahi; sirf "cold" aur "hot" labels swap kiye gaye the.

Does an engine with

Q_{C} \to 0^{+}

(approaching but never reaching zero rejected heat) violate K–P?

Nahi — jab tak

Q_{C} > 0

strictly hai,

η < 1

hai aur law hold karta hai; K–P sirf exact limit

Q_{C} = 0

ko forbid karta hai, jo ek cycle mein unreachable hai.

Is a machine that creates net work each cycle from a single reservoir the same trap as one that reaches

η = 1

Haan — ek akele reservoir se per cycle koi bhi net work extract karna matlab

Q_{C} = 0

hai (cold arrow gone) aur

η = 1

; dono descriptions ek hi forbidden machine hain.

Recall Har trap ki ek-line summary

Almost saare traps do sawaalon mein dissolve ho jaate hain: "Kya ye truly ek cycle hai?" (agar nahi, toh missing rejected-heat arrow chhup raha hai) aur "Hidden work kahan hai?" (agar ek fridge free lagta hai, uska $W > 0$ dhundo). Energy-flow arrows banao aur koi bhi trap survive nahi karta.