Heat engines — efficiency η = 1 − Q_C - Q_H: Worked examples

1.7.19 · D3 · Physics › Thermodynamics › Heat engines — efficiency η = 1 − Q_C - Q_H

Numbers touch karne se pehle, plain words mein chaar reminders taaki koi bhi symbol mystery na rahe:

Definition Letters, words mein

$Q_{H}$ = woh heat (energy) jo hot side se pour hoti hai, hamesha positive joules ka count.
$Q_{C}$ = woh heat jo cold side mein dump hoti hai, yeh bhi positive joules ka count.
$W$ = woh useful work jo engine tumhe deta hai, joules mein.
$Δ U$ = gas ki internal energy mein change — abhi working substance ke andar kitni energy store hai versus start mein. (Symbol $Δ$ ka matlab hai "change in"; $U$ stored internal energy hai.) Ek complete cycle mein gas exactly apni starting state par wapas aati hai, isliye yeh stored energy unchanged rehti hai: $Δ U = 0$ .
$T_{H}$ , $T_{C}$ = hot aur cold reservoirs ki absolute (kelvin) temperatures. "Absolute" ka matlab hai true zero, $0 K$ se measure kiya gaya; kabhi Celsius nahi, kyunki hum temperatures divide karenge aur sirf kelvin ratios physically meaningful hain. Sabhi heats aur work yahan magnitudes (sizes) hain — hum kabhi negative $Q_{H}$ nahi likhte; direction already letter mein baki hui hai.

Intuition Har example ke peeche ek picture (neeche dekho)

Kisi bhi arithmetic se pehle, yeh apni aankh mein jalao: heat hot se cold ki taraf "girta" hai jaise paani ek wheel ke neeche, aur engine beech mein work $W$ skim kar leta hai. Figure flow dikhata hai — ek mota yellow pipe $Q_{H}$ hot block se neeche aata hai, ek patla blue pipe $Q_{C}$ cold block ki taraf jaata hai (patla pipe figure par hi "smaller" label kiya gaya hai), aur pink arrow $W$ side se tap hota hai. Har example sirf yahi picture hai alag-alag numbers ke saath pipes par.

Figure description (alt-text): A vertical chalkboard sketch. A pale-yellow HOT reservoir block sits at the top, a blue COLD reservoir block at the bottom, and a chalk-outlined ENGINE box in the middle. A thick yellow downward arrow labelled "Q_H (heat in, big)" enters the engine from the hot block; a thinner blue downward arrow labelled "Q_C (dumped, smaller)" leaves the engine into the cold block; a pink rightward arrow labelled "W = Q_H − Q_C (work out)" leaves the engine's side. The relation "eta = W / Q_H" is written beside the engine.

The scenario matrix

Har heat-engine question in cells mein se ek (ya blend) hota hai. Har row ek "type of shot" hai; last column us example ka naam deta hai jo use hit karta hai. (Yaad rahe: $η$ matlab hai $W / Q_{H}$ — useful work out over heat paid in.)

Cell	Kya diya hai / kya weird hai	Kya find karna hai	Example
A. Forward basics	$Q_{H}, Q_{C}$ diye hain	$W$ , $η = W / Q_{H}$	Ex 1
B. Reverse solve	$η$ aur $W$ diye hain	$Q_{H}$ , $Q_{C}$	Ex 2
C. Carnot ceiling check	$T_{H}, T_{C}$ , claimed $η = W / Q_{H}$	Kya yeh legal hai?	Ex 3
D. Degenerate: $Q_{C} = 0$	zero waste heat	Kya yeh possible hai?	Ex 4
E. Degenerate: $Q_{C} = Q_{H}$	sari heat dump	$η$ ki value	Ex 4
F. Limit: $T_{C} \to 0$ or $T_{H} \to \infty$	extreme temperatures	limiting $η$	Ex 5
G. Unit trap: Celsius diya	°C temperatures	correct Carnot $η$	Ex 6
H. Real-world word problem	power (watts), time, fuel	efficiency, dumped heat rate	Ex 7
I. Exam twist: two engines in series	ek ka waste doosre ko feed karta hai	combined $η$	Ex 8
J. PV-loop area	net work as loop area	cycle sketch se $η$	Ex 9

Hum inhe order mein karte hain; har ek apne cell ke saath label kiya gaya hai.

Worked examples

Cell A — Forward basics

Worked example Example 1 (Cell A)

Ek engine har cycle mein $Q_{H} = 800 J$ leta hai aur $Q_{C} = 600 J$ reject karta hai. $W$ aur $η$ find karo.

Forecast: Guess karo — kya yeh acha engine hai ya bekar? (Yeh jo khata hai uska zyada dump karta hai, toh low $η$ expect karo.)

Step 1. $W = Q_{H} - Q_{C} = 800 - 600 = 200 J$ . Yeh step kyun? Ek full cycle mein gas apni start par wapas aati hai, isliye uski stored internal energy unchanged hai, $Δ U = 0$ (upar define kiya gaya "change in internal energy"). First Law of Thermodynamics ke by, stored energy mein koi net change nahi, toh net heat ka har joule work ban jaata hai.

Step 2. $η = \frac{W}{Q _{H}} = \frac{200}{800} = 0.25 = 25%$ . Yeh step kyun? Efficiency hai jo milta hai ( $W$ ) divided by jo pay karte hain ( $Q_{H}$ ).

Verify: Second form $1 - Q_{C} / Q_{H} = 1 - 600/800 = 1 - 0.75 = 0.25$ ✓. Units: J/J dimensionless hai, fraction ke liye correct. Jaise forecast kiya, yeh bekar engine hai.

Cell B — Reverse solve

Cell C — Carnot ceiling check

Worked example Example 3 (Cell C)

Ek salesman ka engine $T_{H} = 500 K$ aur $T_{C} = 350 K$ (dono kelvin, jaise upar definitions mein flag kiya) ke beech run karta hai aur $η = 0.32$ claim karta hai (yaad karo $η = W / Q_{H}$ ). Legal hai ya fraud?

Forecast: Guess karo kya $0.32$ reversible ceiling ke neeche fit ho sakta hai.

Step 1. Best possible: $η_{Carnot} = 1 - \frac{T _{C}}{T _{H}} = 1 - \frac{350}{500} = 1 - 0.70 = 0.30$ . Yeh step kyun? Carnot Cycle in dono temperatures ke beech kisi bhi engine ke liye absolute ceiling deta hai. Hum kelvin use karte hain taaki ratio $T_{C} / T_{H}$ meaningful ho.

Step 2. Compare karo: claimed $0.32 > 0.30 = η_{Carnot}$ — impossible. Yeh step kyun? Reversible bound beat karna Second Law of Thermodynamics violate karega.

Verify: $η_{Carnot} = 0.30$ ; claim isse $0.02$ se exceed karta hai, isliye yeh fraudulent hai. (Agar unhone $0.29$ claim kiya hota, toh legal hota, halanki reversible ke impressively close.)

Cells D & E — Degenerate inputs

Worked example Example 4 (Cells D and E)

Do extreme "engines" examine karo: (a) ek jo kuch bhi dump nahi karta, $Q_{C} = 0$ ; (b) ek jo sab kuch dump karta hai, $Q_{C} = Q_{H}$ . $η$ find karo aur bolo kya dono exist kar sakte hain.

Forecast: Ek dream machine hai, ek useless. Guess karo kiska $η$ 1 hai aur kiska 0.

Step 1 — case (a), $Q_{C} = 0$ . $η = 1 - \frac{0}{Q _{H}} = 1 = 100%$ . Yeh step kyun? Degenerate value ko seedha $η = 1 - Q_{C} / Q_{H}$ mein plug karo.

Step 2 — kya (a) real hai? Nahi. $Q_{C} = 0$ matlab heat poori tarah work mein convert ho gayi kisi aur effect ke bina — Second Law of Thermodynamics ke Kelvin–Planck form se forbidden. Yeh unreachable ceiling $η = 1$ hai.

Step 3 — case (b), $Q_{C} = Q_{H}$ . $η = 1 - \frac{Q _{H}}{Q _{H}} = 1 - 1 = 0 = 0%$ . Yeh step kyun? Same formula, doosra extreme.

Step 4 — (b) ka matlab. $η = 0$ ke saath, $W = Q_{H} - Q_{C} = 0$ : heat andar aati hai aur seedha bahar jaati hai koi work nahi karte. Physically allowed (yeh sirf heat leak hai), lekin yeh engine nahi hai — yeh ek conductor hai.

Verify: $W_{(a)} = Q_{H} - 0 = Q_{H}$ (sab work, forbidden); $W_{(b)} = Q_{H} - Q_{H} = 0$ (koi work nahi). Poora allowed range $0 \leq η < 1$ hai, aur yeh dono cases exactly uske endpoints par hain. ✓

Cell F — Limiting behaviour

Worked example Example 5 (Cell F)

Ek Carnot engine $T_{H} = 400 K$ fixed rakhta hai. (a) Cold reservoir ko absolute zero ki taraf push karo, $T_{C} \to 0 K$ . (b) Bajaaye $T_{C} = 300 K$ fix karo aur $T_{H} \to \infty$ push karo. $η$ har mein kya approach karta hai? (Sab temperatures kelvin.)

Forecast: Dono ko efficiency upar push karni chahiye — lekin kitne number tak?

Step 1 — case (a). $η = 1 - \frac{T _{C}}{T _{H}} \to 1 - \frac{0}{400} = 1$ . Yeh step kyun? Jaise $T_{C} \to 0$ hota hai, wasted-fraction term $T_{C} / T_{H} \to 0$ , toh $η$ apni ceiling tak climb karta hai.

Step 2 — case (a) ke liye sample point. $T_{C} = 40 K$ par: $η = 1 - 40/400 = 0.90 = 90%$ . Colder cold ⇒ higher $η$ . Yeh step kyun? Ek finite numeric point trend visible banata hai: hum $T_{C}$ shrink karte hain aur $η$ ko ceiling ki taraf rise karte dekhte hain, toh hum trust kar sakte hain ki limit sirf algebra nahi hai.

Step 3 — kya case-(a) ceiling reachable hai? Nahi. Exactly $T_{C} = 0 K$ reach karna thermodynamics ke third law se forbidden hai, isliye $η = 1$ approached hota hai lekin kabhi mila nahi. Yeh step kyun? Ek limit batata hai hum kahan ja rahe hain, yeh nahi ki hum wahan pahunch sakte hain ya nahi; hume alag se physics state karni hai jo endpoint block karta hai, exactly jaise Second Law ne $Q_{C} = 0$ ko Example 4 mein block kiya.

Step 4 — case (b). $η = 1 - \frac{300}{T _{H}}$ . Jaise $T_{H} \to \infty$ , fraction $300/ T_{H} \to 0$ , toh $η \to 1$ . Sample: $T_{H} = 3000 K$ par, $η = 1 - 300/3000 = 0.90 = 90%$ . Yeh step kyun? Yeh mirror-image limit hai — cold side chill karne ki bajaye hot side heat karte hain, aur wahi wasted-fraction term vanish ho jaata hai, confirming "bada temperature gap ⇒ higher $η$ ."

Verify: Dono limits $η \to 1$ dete hain lekin kabhi $= 1$ nahi; sample points $0.90$ aur $0.90$ trend confirm karte hain "bigger temperature gap ⇒ higher efficiency" parent ke water-wheel picture se. ✓

Cell G — The Celsius trap

Worked example Example 6 (Cell G)

Ek engine $T_{H} = 227^{\circ} C$ aur $T_{C} = 27^{\circ} C$ ke beech kaam karta hai. Maximum efficiency find karo. (Units dhyan se dekho!)

Forecast: Ek student jo convert karna bhool jaata hai woh $1 - 27/227 \approx 0.88$ compute karega. Guess karo kya true answer isse upar hai ya neeche.

Step 1. Kelvin mein convert karo: $T (K) = T (^{\circ} C) + 273.15$ . $T_{H} = 227 + 273.15 = 500.15 K$ , $T_{C} = 27 + 273.15 = 300.15 K$ . Yeh step kyun? Carnot formula temperatures divide karta hai, aur sirf absolute temperature ratio ko physically meaningful banata hai. Celsius ka arbitrary zero hota hai, isliye °C ke ratios nonsense hain.

Step 2. $η_{Carnot} = 1 - \frac{300.15}{500.15} = 1 - 0.60012 = 0.39988 \approx 0.40 = 40%$ . Yeh step kyun? Ceiling formula ko correct units ke saath apply karo.

Verify: Galat Celsius answer hoga $1 - 27/227 = 0.881$ — bahut zyada optimistic. True $\approx 40%$ bahut kam hai, trap confirm karta hai. (Whole degrees tak round karne par, $T = 500, 300$ exactly $0.40$ deta hai.) ✓

Cell H — Real-world word problem

Worked example Example 7 (Cell H)

Ek power plant efficiency $η = 0.40$ par $P_{W} = 300 MW$ electrical power output karta hai. (a) Woh apne fuel se kitni rate par heat absorb karta hai? (b) Woh river mein waste heat kitni rate par dump karta hai? (c) $1$ ghante mein woh kitna fuel-heat use karta hai?

Forecast: 40% efficiency ke saath, guess karo river-dumping rate 300 MW ke useful output se badi hai ya choti.

Step 1 — heat-in rate. Power sirf energy per second hai, toh wahi algebra jo Example 2 mein hai rates par kaam karta hai: $\dot{Q}_{H} = \frac{P _{W}}{η} = \frac{300}{0.40} = 750 MW$ . Yeh step kyun? $η = W / Q_{H}$ mein har quantity ko cycle time se divide karne par joules watts ban jaate hain; ratio $η$ unchanged rehta hai. ( $\dot{Q}$ mein dot ka matlab sirf "per second" hai.)

Step 2 — waste-heat rate. $\dot{Q}_{C} = \dot{Q}_{H} - P_{W} = 750 - 300 = 450 MW$ . Yeh step kyun? Energy conservation rates par apply hota hai: jo work nahi bana woh dump hota hai, per second.

Step 3 — ek ghante mein fuel-heat. Pehle prefix aur time facts yaad karo: $1 MW = 1 MJ per second$ (ek megawatt har second ek megajoule hai), aur $1 h = 3600 s$ . Energy $=$ power $\times$ time, toh $Q_{H} = 750 MW \times 3600 s = 750 \frac{MJ}{s} \times 3600 s = 2700000 MJ = 2.7 \times 1 0^{6} MJ .$ Kyunki $1 MJ = 1 0^{6} J$ , ek baar aur convert karte hain: $2.7 \times 1 0^{6} MJ = 2.7 \times 1 0^{6} \times 1 0^{6} J = 2.7 \times 1 0^{12} J$ . Yeh step kyun? $MW = MJ / s$ likhne se seconds cleanly cancel ho jaate hain, aur $1 MJ = 1 0^{6} J$ batana dikhata hai same answer ke dono forms exactly kaise relate karte hain.

Verify: $\dot{Q}_{C} = 450 MW > P_{W} = 300 MW$ — plant electricity ke taur par deliver karne se zyada river mein dump karta hai, parent note ki warning se match karta hai. Rate check: $300/750 = 0.40 = η$ ✓. Energy check: $750 \times 3600 = 2700000 MJ$ ✓.

Cell I — Exam twist: two engines in series

Worked example Example 8 (Cell I)

Engine 1 $600 K$ aur $400 K$ ke beech run karta hai; uska waste heat engine 2 ko feed karta hai, jo $400 K$ aur $300 K$ ke beech run karta hai. Dono ideal (Carnot) hain. Engine 1 $Q_{H} = 1200 J$ absorb karta hai. Total work aur combined efficiency (work out ÷ original heat in) find karo.

Forecast: Guess karo kya do engines stack karna ek single engine ko beat karta hai jo directly $600 K$ se $300 K$ tak run karta hai.

Step 1 — engine 1. $η_{1} = 1 - 400/600 = 1/3$ . Toh $W_{1} = η_{1} Q_{H} = \frac{1}{3} (1200) = 400 J$ . Yeh step kyun? Har engine individually Carnot ceiling follow karta hai.

Step 2 — heat passed on. Engine 1 $Q_{C, 1} = Q_{H} - W_{1} = 1200 - 400 = 800 J$ $400 K$ reservoir mein dump karta hai, jo engine 2 ka input hai. Yeh step kyun? "Waste feeds the next engine" ka matlab hai $Q_{H, 2} = Q_{C, 1}$ .

Step 3 — engine 2. $η_{2} = 1 - 300/400 = 1/4$ . Toh $W_{2} = η_{2} \cdot 800 = \frac{1}{4} (800) = 200 J$ . Yeh step kyun? Same Carnot rule engine 2 ki temperatures ke saath.

Step 4 — totals. $W_{total} = W_{1} + W_{2} = 400 + 200 = 600 J$ . Combined efficiency $η_{tot} = \frac{W _{total}}{Q _{H}} = \frac{600}{1200} = 0.50$ . Yeh step kyun? Combined efficiency original heat input $Q_{H} = 1200 J$ use karta hai.

Verify: $600 K$ se seedha $300 K$ tak ek single Carnot engine $1 - 300/600 = 0.50$ deta hai — identical. Series mein do ideal engines ek bade ideal engine ke equal hain poore temperature drop par. ✓ Algebraic cross-check, har factor clearly likhke: $η_{tot} = η_{1} + η_{2} \cdot \frac{Q _{C, 1}}{Q _{H}}$ . Yahan $\frac{Q _{C, 1}}{Q _{H}} = \frac{800}{1200} = \frac{2}{3}$ (yeh $2/3$ exactly woh fraction hai original heat ka jo engine 1 se survive karke engine 2 mein jaata hai). Toh $η_{tot} = \frac{1}{3} + \frac{1}{4} \cdot \frac{2}{3} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$ ✓.

Cell J — Efficiency from a PV loop area

Worked example Example 9 (Cell J)

Ek gas neeche sketch kiya gaya rectangular cycle run karta hai PV diagram par, clockwise traverse kiya. Enclosed rectangle ki width $Δ V = 2 \times 1 0^{- 3} m^{3}$ aur height $Δ P = 1 \times 1 0^{5} Pa$ hai. Heat sirf top leg aur left-hand rising leg ke dauran absorb hoti hai, total $Q_{H} = 500 J$ . $W$ aur $η$ find karo.

Figure description (alt-text): A chalkboard PV diagram (pressure on the vertical axis, volume on the horizontal). A rectangle is drawn and its interior shaded pink. Small arrows show the cycle running clockwise: the top edge is traversed left-to-right (yellow, labelled "heat absorbed Q_H"), the right edge downward (blue), the bottom edge right-to-left (chalk, labelled "heat rejected Q_C"), and the left edge upward (blue). The shaded area is labelled "W = area = dP x dV = 200 J", with the width labelled "dV = 2e-3 m^3" and the height "dP = 1e5 Pa".

Forecast: Net work loop ki area hai. Guess karo roughly yeh 500 J ka kitna fraction hoga.

Step 1 — kyun woh legs $Q_{H}$ carry karte hain. Top aur left-rising legs par gas expand ho rahi hai aur/ya higher pressure ki taraf push ho rahi hai, toh uska temperature (hence internal energy) rise karta hai aur woh outward work karta hai — dono energy in maangti hain, isliye heat gas mein flow karti hai wahan ( $Q_{H}$ ). Right-falling aur bottom legs par gas contract aur cool hoti hai, isliye heat out flow karti hai ( $Q_{C}$ ). Isliye $Q_{H} = 500 J$ outward legs ko assign kiya gaya hai. Yeh step kyun? First Law $Q = Δ U + W_{by gas}$ batata hai heat exactly wahan enter karti hai jahan internal energy aur outward work dono energy demand karte hain; kaunse legs $Q_{H}$ hain yeh name karna guess nahi hai, yeh travel ki direction se follow karta hai.

Step 2 — net work = loop area. Rectangle ke liye, area $= Δ P \times Δ V = (1 \times 1 0^{5}) (2 \times 1 0^{- 3}) = 200 J$ . Yeh step kyun? PV diagram par, gas by kiya gaya work $\int P d V$ hai; clockwise closed loop par yeh integral enclosed area ke equal hota hai (clockwise loop positive net work deta hai — figure mein shaded pink region dekho). Units: $Pa \cdot m^{3} = J$ .

Step 3 — efficiency. $η = \frac{W}{Q _{H}} = \frac{200}{500} = 0.40 = 40%$ . Yeh step kyun? Same master definition; loop area hi useful work hai.

Step 4 — dumped heat. $Q_{C} = Q_{H} - W = 500 - 200 = 300 J$ . Yeh step kyun? Energy conservation accounts close karta hai.

Verify: $1 - Q_{C} / Q_{H} = 1 - 300/500 = 0.40$ ✓. Area $200 J$ nikla, exactly $500 J$ input ka $40%$ — forecast se match karta hai ki heat ka sirf kuch hissa loop ki area ban jaata hai. ✓

Active recall

Recall Kaunsa cell kaunsa hai? (answers cover karo)

"Dono heats diye, $η$ find karo" ::: Cell A (Ex 1).
" $η$ aur $W$ diye, heats find karo" ::: Cell B (Ex 2).
"Kya claimed efficiency legal hai?" ::: Cell C — Carnot se compare karo (Ex 3).
" $Q_{C} = 0$ se $η = ?$ aur kya yeh real hai?" ::: $η = 1$ , Second Law se forbidden (Ex 4).
"Poore temperature drop par series mein do engines" ::: us drop par ek Carnot engine ke equal hai (Ex 8).
"PV loop se net work" ::: clockwise loop ki enclosed area (Ex 9).

Connections

Parent topic (Hinglish) — yahan work kiye gaye master formulas.
First Law of Thermodynamics — $W = Q_{H} - Q_{C}$ ka source.
Second Law of Thermodynamics — Cells C aur D ko kill karta hai (koi $η \geq η_{Carnot}$ nahi, koi $η = 1$ nahi).
Carnot Cycle — Ex 3, 5, 6, 8 mein use ki gayi ceiling.
Entropy — woh deeper reason kyun ceiling exist karti hai.
Refrigerators and Heat Pumps — in engines mein se kisi ko bhi backwards run karo.
PV diagrams and Work — Ex 9 ka loop-area method.