Heat engines — efficiency η = 1 − Q_C - Q_H: Question bank

1.7.19 · D5 · Physics › Thermodynamics › Heat engines — efficiency η = 1 − Q_C - Q_H

Symbols ka reminder (sab parent mein defined hain): $Q_{H}$ = hot reservoir se li gayi heat (ek positive magnitude), $Q_{C}$ = cold reservoir mein dump ki gayi heat (positive magnitude), $W$ = engine dwara kiya gaya useful work, $η = W / Q_{H} = 1 - Q_{C} / Q_{H}$ = kisi bhi engine ki efficiency, $T_{H}, T_{C}$ = reservoir temperatures kelvin mein.

Figure — Heat engines — efficiency η = 1 − Q_C - Q_H

True or false — justify karo

Ek heat engine ek single non-cyclic process mein absorbed heat ka 100% work mein convert kar sakta hai.

True — ek one-shot isothermal expansion gas ko constant temperature par rakhta hai (isliye uski internal energy unchanged hoti hai), aur First Law se saari absorbed heat work ban jaati hai; lekin gas ab ek different state mein expanded reh jaati hai, reset nahi hoti, isliye ye ek single process hai, repeatable cycle nahi. Kelvin–Planck ban sirf ek repeating cycle par apply hota hai.

Ek complete cycle mein working gas ki internal energy change zero hoti hai.

True — internal energy ek state function hai, aur ek cycle exactly wahin khatam hoti hai jahan shuru hui thi, isliye

Δ U_{cycle} = 0

chahe koi bhi path liya ho.

Agar

Q_{C} = 0

ho toh engine ka

η = 1

hoga, aur ye clever engineering se achieve kiya ja sakta hai.

False — algebraically

η = 1

ke liye

Q_{C} = 0

chahiye, lekin Second Law (Kelvin–Planck) cyclic heat ko fully work mein convert karna forbid karta hai; koi engineering isse nahi hata sakti, ye ek law of nature hai.

T_{H}

aur

T_{C}

dono ko double karne par Carnot efficiency unchanged rehti hai.

True —

η_{C} = 1 - T_{C} / T_{H}

sirf ratio par depend karta hai; dono temperatures ko same factor se scale karne par

T_{C} / T_{H}

fixed rehta hai.

Ek real engine Carnot efficiency exceed kar sakta hai agar use bahut achhi tarah lubricate kiya jaaye.

False — Carnot kisi bhi engine ke liye ceiling hai (reversible ones ise reach karte hain, irreversible ones neeche rehte hain); friction sirf tumhe bound se aur neeche le jaata hai, kabhi upar nahi.

Q_{C} / Q_{H}

quantity engine ki efficiency ke barabar hoti hai.

False —

Q_{C} / Q_{H}

waste hua fraction hai; efficiency

1 - Q_{C} / Q_{H}

hai. Dono ka sum exactly 1 hota hai.

Equal temperature ke reservoirs ke beech (

T_{H} = T_{C}

) chalne wala engine phir bhi net work kar sakta hai.

False —

η_{C} = 1 - T_{C} / T_{H} = 0

jab

T_{H} = T_{C}

; koi temperature "drop" nahi jis se heat gire, isliye koi net work extract nahi hota.

Cold reservoir ko 1 K thanda karna hot reservoir ko 1 K garam karne se zyada efficiency badhata hai.

True jab $T_{C} < T_{H}$ ho —

η_{C} = 1 - T_{C} / T_{H}

lo aur ek variable ek baar mein change karo.

T_{C}

d T

se kam karne par

η_{C}

mein

d T / T_{H}

ki badhot hoti hai.

T_{H}

d T

se badhane par

η_{C}

mein

T_{C} d T / T_{H}^{2} = (T_{C} / T_{H}) (d T / T_{H})

ki badhot hoti hai. Kyunki

T_{C} / T_{H} < 1

hai, cold-side change exactly

T_{C} / T_{H}

factor se jeetta hai.

PV diagram par cycle se enclosed area engine dwara kiya gaya net work

W

ke barabar hota hai.

True — dekho PV diagrams and Work; loop ka enclosed area

\oint P d V = W_{net}

hota hai, clockwise loop ke liye positive.

Error dhundho

η = 1 - Q_{C} / Q_{H} = 1 - 273/373 = 0.27

ek engine ke liye jisme

T_{H} = 100° C, T_{C} = 0° C

hai."

Formula heats ko temperatures ke saath mix up kar raha hai — temperature form

η_{C} = 1 - T_{C} / T_{H}

hai; numbers kelvin sahi use karte hain lekin symbols

Q

galat hain.

"Kyunki heat absorb hoti hai, cycle mein gas ki internal energy badhti hai, isliye

Δ U > 0

Heat ek loop mein absorb bhi hoti hai aur reject bhi; gas garm hoti hai phir wapas apni starting state par cool hoti hai, isliye

Δ U_{cycle} = 0

hota hai net heat inflow ke bawajood.

"Engine 40% efficient hai, isliye ye sirf 40% heat waste karta hai."

Ulta hai — 40% efficient matlab 40% work banta hai aur 60%

Q_{C} / Q_{H} = 1 - η = 0.60

ke roop mein dump hota hai.

T_{H} = 500° C, T_{C} = 20° C

ke liye Carnot efficiency

1 - 20/500 = 0.96

hai."

Celsius yahan illegal hai; pehle convert karo:

1 - 293/773 \approx 0.62

. °C use karne se answer bahut inflated ho jaata hai.

"Kyunki

W = Q_{H} - Q_{C}

hai, zyada heat dump karne wala engine zyada work karta hai."

Fixed

Q_{H}

ke liye, bada

Q_{C}

matlab kam work; statement relationship ko invert karta hai —

Q_{C}

badhne par

W

ghatata hai.

"Ek ideal reversible engine best hai kyunki ye sabse tez chalta hai."

Reversibility efficiency maximise karta hai, speed nahi; truly reversible processes infinitely slow (quasi-static) hote hain, isliye real trade-off efficiency versus power ka hai.

Why questions

Ek cyclic heat engine ko cold reservoir mein kuch heat kyun reject karni padti hai?

Ek cycle ko gas ko exact start state par wapas laana hota hai, isliye

Q_{H}

ke roop mein jo bhi energy daali gayi jo work ke roop mein nahi nikli wo heat ke roop mein nikelni chahiye — warna loop par

Δ U

zero nahi hoga. Kelvin–Planck statement ise sharpen karta hai: koi bhi cycle exist nahi kar sakta jiska only effect heat ko fully work mein turn karna ho, isliye

Q_{C} > 0

forced hai. (Upar energy-flow figure dekho.)

Carnot formula mein Kelvin (absolute) temperature kyun required hai?

Ratio

T_{C} / T_{H}

ko true thermodynamic temperature reflect karna chahiye; sirf ek absolute scale hi aise ratios ko physically meaningful banata hai. Celsius scale par freezing ke paas ratio meaningless hai (aur

T_{H}

se division blow up ho jaata hai jab

T_{H} \to 0° C

, jo physical zero nahi hai); true singular point

0

K, absolute zero hai.

Reservoirs ke beech bada temperature difference higher efficiency kyun deta hai?

Bada gap

1 - T_{C} / T_{H}

badhata hai; intuitively heat ek greater drop se "girती" hai, isliye uska zyada hissa work mein divert ho sakta hai — jaise ek taller waterfall wheel ko zyada tez spin karta hai.

Carnot engine maximum-efficiency benchmark kyun hai?

Ye fully reversible hai, zero entropy generate karta hai; koi bhi real irreversibility (friction, finite temperature differences) entropy create karta hai jiske liye extra dumped heat "pay" karni padti hai,

η

ghata deti hai. Dekho Entropy.

Hum engine ko perfectly insulate kyun nahi kar sakte taaki heat loss rok sakein aur

η = 1

reach kar sakein?

Rejected heat

Q_{C}

ek accidental "leak" nahi hai jise tum insulate kar sako — ye cycle complete karne ke liye Second Law dwara required ek thermodynamically mandatory dump hai.

Power plants abhi bhi "efficient" hone par bhi use se zyada energy kyun waste karte hain?

Typical

η \approx 0.4

matlab

Q_{C} / Q_{H} = 0.6

; aadhe se zyada input heat waste ke roop mein nikelni chahiye cycle constraint satisfy karne ke liye, chahe plant kitna bhi modern ho.

Efficiency

W / Q_{H}

ke roop mein define ki jaati hai,

W / (Q_{H} + Q_{C})

ke roop mein nahi, kyun?

Tum sirf us fuel ke liye "pay" karte ho jo

Q_{H}

supply karta hai; dumped

Q_{C}

ek loss hai, koi purchased input nahi, isliye honest "value for money" ratio

W / Q_{H}

hai.

Edge cases

Jab

Q_{C} = Q_{H}

ho toh efficiency kya hai?

η = 1 - Q_{C} / Q_{H} = 0

; saari absorbed heat dump ho jaati hai, koi work produce nahi hota — machine pointlessly cycle karti rehti hai.

Jab

Q_{C} \to 0

ho (forbidden limit) toh efficiency kya hai?

η \to 1

, theoretical maximum, lekin ye Kelvin–Planck violate karta hai; ye sirf conceptually approached limit hai, kabhi reach nahi hoti.

Jab

T_{C} \to 0

K (absolute zero cold reservoir) ho toh

η_{C}

ka kya hota hai?

η_{C} \to 1

; absolute zero par ek cold reservoir perfect conversion allow karega, lekin absolute zero unattainable hai (Third Law), isliye ye bhi sirf ek limit hai.

Kya efficiency kabhi negative ho sakti hai, aur iska matlab kya hoga?

Ek engine ke liye, nahi — negative

η

matlab

W < 0

(gas par kiya gaya work), jo ek refrigerator describe karta hai, ise efficiency se nahi COP se measure karte hain.

Agar engine heat absorb kare lekin cycle reverse mein run ho toh

W

kya hoga?

Loop reverse karne par

W < 0

ho jaata hai: tum work supply karte ho heat ko cold se hot tak pump karne ke liye — wo device ek heat pump/refrigerator hai, engine nahi.

Ek engine jisme

T_{H} = T_{C}

ho lekin genuinely different

Q_{H}

aur

Q_{C}

hon — possible hai?

Nahi — equal temperatures ke saath

η_{C} = 0

hai, isliye koi bhi real engine wahan

W \leq 0

rakhega; temperature difference ke bina net work extract nahi kar sakte.

Agar koi inventor apne reservoirs ke liye

η > η_{C}

claim kare, toh instantly kya galat hai?

Wo reversible bound beat karne ka claim kar raha hai, jo tumhe ek aisi machine banane deta jo universe ki entropy decrease kare — ye direct Second Law violation hai, isliye claim bina koi detail check kiye galat hai.

Connections

Parent: Heat engines — efficiency
First Law of Thermodynamics — $W = Q_{H} - Q_{C}$ ke peeche $Δ U = 0$ cycle argument.
Second Law of Thermodynamics — yahan har "impossible" verdict ka source.
Carnot Cycle — edge cases mein use ki gayi efficiency ceiling.
Entropy — kyun irreversibility efficiency cost karta hai.
Refrigerators and Heat Pumps — jahan negative $W$ / COP over le leta hai.
PV diagrams and Work — $W$ ki loop-area interpretation.