Carnot cycle — full derivation, efficiency = 1 − T_C - T_H: Worked examples

1.7.21 · D3 · Physics › Thermodynamics › Carnot cycle — full derivation, efficiency = 1 − T_C - T_H

Yahan use kiye gaye har symbol ko parent note mein banaya gaya hai, lekin taaki tumhe kabhi wapas flip na karna pade, yahan sabhi symbols plain words mein hain — including do ( $Δ U$ aur $R$ , $e$ ) jinpe yeh examples zyada rely karte hain.

Recall Symbols, plain words mein

$T_{H}$ = hot reservoir (stove) ka temperature, kelvin mein.
$T_{C}$ = cold reservoir (ice) ka temperature, kelvin mein.
$Q_{H}$ = woh heat jo gas hot reservoir se absorb karti hai (joules mein ek positive number).
$Q_{C}$ = woh heat jo gas cold reservoir mein reject karti hai (joules mein ek positive number).
$W$ = ek poore cycle mein kiya gaya net useful work, $W = Q_{H} - Q_{C}$ .
$η$ (eta) = efficiency = "absorbed heat ka woh fraction jo work ban gaya" $= W / Q_{H}$ .
$Δ U$ = gas ki internal energy mein change — apne molecules ki random jiggling mein stored energy. Symbol $Δ$ (delta) ka matlab hai "final minus initial". Ek ideal gas ke liye $U$ sirf temperature par depend karta hai, isliye agar temperature apni starting value par wapas aa jaye, toh $Δ U = 0$ .
$R$ = universal gas constant, $R = 8.314 J K^{- 1} mol^{- 1}$ — woh fixed number jo ek mole gas ke liye $P V = R T$ mein pressure, volume aur temperature ko link karta hai.
$e$ = Euler's number $\approx 2.718$ , natural logarithm $ln$ ka base. Yeh woh special number hai jiske liye $ln e = 1$ hota hai; $e^{x}$ aur $ln x$ ek doosre ko undo karte hain.
Kelvin = absolute temperature scale jahan $0$ sabse thanda possible hai. Convert karo: $T (K) = T (^{\circ} C) + 273.15$ (arithmetic ke liye hum $273$ use karte hain).

Kyunki yahan reasoning pictures se chalti hai, teen figures examples ko anchor karti hain: engines ke liye ek energy-flow diagram, refrigerator ke liye mirrored flow, aur efficiency ko uski ceiling ki taraf chadh'te dikhata ek curve.

The scenario matrix

Yeh is topic ke produce kiye jaane wale sabhi case-classes ki poori list hai. Neeche har worked example us cell ke saath tagged hai jo woh fill karta hai.

#	Case class	Tricky kya hai	Example
A	Clean kelvin plug-in	kuch nahi — yeh baseline hai	Ex 1
B	Input Celsius mein	ratio banane se pehle convert karna zaroori	Ex 2
C	Find $Q_{C}$ , $W$ from $Q_{H}$	use karo $Q_{C} / Q_{H} = T_{C} / T_{H}$	Ex 3
D	Reverse direction (refrigerator)	efficiency ban jaati hai coefficient of performance	Ex 4
E	Degenerate: $T_{H} = T_{C}$	$η = 0$ , engine kuch nahi karta	Ex 5
F	Limiting: $T_{C} \to 0$	$η \to 1$ , kyun yeh unreachable hai	Ex 6
G	Real-world word problem	words ko $T_{H}, T_{C}, Q_{H}$ tak strip karo	Ex 7
H	Exam twist: "beat Carnot?" claim	Carnot's theorem trap pakad leta hai	Ex 8
I	Work back to a volume ratio	invert $Q_{H} = R T_{H} ln (V_{2} / V_{1})$	Ex 9
J	Sign edge: $T_{C} > T_{H}$	$η < 0$ signals a heat-pump regime	Ex 10

Neeche ka energy-flow picture Cells A–H ki backbone hai — ise saamne rakho.

Figure — Carnot cycle — full derivation, efficiency = 1 − T_C - T_H

Ab hum har cell fill karte hain.

The worked examples

Worked example Example 1 — Cell A: clean kelvin plug-in

Ek Carnot engine $T_{H} = 800 K$ aur $T_{C} = 320 K$ ke beech chalti hai. $η$ nikalo.

Forecast: Ratio $320/800$ yaani $0.4$ hai. Aage padhne se pehle andaza lagao ki $η$ $50%$ se upar hogi ya neeche.

Formula likho: $η = 1 - \frac{T _{C}}{T _{H}}$ . Yeh step kyun? Yahi ek formula hai jo tab kaam karta hai jab dono temperatures pehle se kelvin mein hon — koi conversion nahi, koi heats nahi chahiye.
Substitute karo: $η = 1 - \frac{320}{800} = 1 - 0.4 = 0.6$ . Yeh step kyun? $T_{C} / T_{H}$ ek pure ratio hai; units cancel ho jaate hain, isliye answer dimensionless hai.

$η = 0.6 = 60%$

Verify: $0.6 < 1$ ✓ (ek engine kabhi $1$ exceed nahi kar sakti). Flow figure (s01) mein, efficiency green $W$ arrow ki width divided by incoming red $Q_{H}$ arrow hai — yahan $W$ $60%$ leta hai, jo hamare number se match karta hai.

Worked example Example 2 — Cell B: input in Celsius (the ratio trap)

Ek power plant $327^{\circ} C$ wale boiler aur $27^{\circ} C$ wale cooling tower ke beech kaam karta hai. $η$ nikalo.

Forecast: Jo student convert karna bhool jaata hai woh compute karta hai $1 - 27/327 \approx 0.92$ . Kya tum uss par trust karoge? Pehle asli answer ka andaza lagao.

Dono ko kelvin mein convert karo: $T_{H} = 327 + 273 = 600 K$ , $T_{C} = 27 + 273 = 300 K$ . Yeh step kyun? Efficiency ratio $T_{C} / T_{H}$ use karta hai, aur ratio depend karta hai ki zero kahan hai. Celsius zero ko freezing water par rakhta hai; kelvin zero ko absolute cold par. Sirf kelvin ratio physical hai.
Ab plug in karo: $η = 1 - \frac{300}{600} = 1 - 0.5 = 0.5$ . Yeh step kyun? Same formula jaise Ex 1 mein, lekin conversion ke baad hi trustworthy hai.

$η = 0.5 = 50%$

Verify: Celsius shortcut ne $0.92$ diya — $0.50$ se bilkul alag. Yeh gap exactly wahi kelvin mistake hai jo parent note mein warn ki gayi thi. Asli answer $50%$ kaafi lower aur correct hai.

Worked example Example 3 — Cell C: find

$Q_{C}$ and $W$ from $Q_{H}$ Ex 2 ki engine ( $η = 0.5$ ) har cycle mein $Q_{H} = 4000 J$ absorb karti hai. Work $W$ aur rejected heat $Q_{C}$ nikalo.

Forecast: $4000$ ka aadha work ban jaata hai — lekin doosra aadha kahan jaata hai? $Q_{C}$ ka andaza lagao.

Work: $W = η Q_{H} = 0.5 \times 4000 = 2000 J$ . Yeh step kyun? Definition se $η = W / Q_{H}$ , toh $W = η Q_{H}$ . Yeh absorbed heat ka woh fraction hai jo useful ban gaya.
First law se rejected heat: $Q_{C} = Q_{H} - W = 4000 - 2000 = 2000 J$ . Yeh step kyun? Ek poore cycle mein gas apne starting temperature par wapas aa jaati hai, toh uski internal energy bhi wapas aa jaati hai: $Δ U = 0$ (upar define kiya — final minus initial internal energy). First law $Q_{net} = Δ U + W$ tab ban jaata hai $Q_{H} - Q_{C} = W$ , toh jo bhi heat work nahi bani woh $Q_{C}$ ke roop mein baahar chali jaati hai.

$W = 2000 J, Q_{C} = 2000 J$

Verify: Reversible identity se cross-check karo: $Q_{C} = Q_{H} \frac{T _{C}}{T _{H}} = 4000 \times \frac{300}{600} = 2000 J$ ✓. Flow figure (s01) mein red $Q_{H}$ arrow exactly aadha-aadha split hota hai: aadha green $W$ ke roop mein exit karta hai, aadha blue $Q_{C}$ ke roop mein neeche jaata hai. Dono routes agree karte hain, jo first law aur reversible relation $Q_{H} / T_{H} = Q_{C} / T_{C}$ ki consistency confirm karta hai.

Worked example Example 4 — Cell D: run it backwards (refrigerator)

Ab usi machine ko $T_{C} = 300 K$ (fridge ke andar) aur $T_{H} = 600 K$ (kitchen air) ke beech ek refrigerator ke roop mein ulta drive karo. Agar yeh kitchen mein $Q_{H} = 4000 J$ dump karta hai, toh coefficient of performance (COP) aur work input nikalo.

Forecast: Ek fridge engine ke opposite kaam karta hai — tum work pay karte ho taaki cold side se heat pump kar sako. Kya tumhare hisaab se COP (benefit per work spent) $1$ se zyada hoga ya kam?

Reversed flow (s01 se arrows flip kiye) yahan dikhaya gaya hai — ab work andar jaata hai aur heat upar pump hoti hai:

Reversed Carnot cycle ke liye heats ab bhi $\frac{Q _{H}}{T _{H}} = \frac{Q _{C}}{T _{C}}$ follow karti hain, toh $Q_{C} = Q_{H} \frac{T _{C}}{T _{H}} = 4000 \times \frac{300}{600} = 2000 J$ . Yeh step kyun? Reversibility ka matlab hai ki same temperature–heat relation hold karti hai chahe hum cycle kisi bhi direction mein chalayein; sirf arrows flip hote hain (s01 ko s02 se compare karo).
Tumhe jo work supply karni hogi: $W = Q_{H} - Q_{C} = 4000 - 2000 = 2000 J$ . Yeh step kyun? First law phir se: kitchen heat $Q_{H}$ barabar hai andar se kheinchi gayi heat plus tumhari pay ki gayi work ke.
Fridge COP $= \frac{Q _{C}}{W} = \frac{2000}{2000} = 1.0$ . Yeh step kyun? Fridge ke liye "benefit" cold space se remove ki gayi heat $Q_{C}$ hai, aur "cost" work $W$ hai.
Ise temperature-only formula mein convert karo. Shuru karo $COP = \frac{Q _{C}}{W} = \frac{Q _{C}}{Q _{H} - Q _{C}}$ se. Top aur bottom ko $Q_{H}$ se divide karo: $COP = \frac{Q _{C} / Q _{H}}{1 - Q _{C} / Q _{H}}$ . Ab $Q_{C} / Q_{H} = T_{C} / T_{H}$ use karo: $COP = \frac{T _{C} / T _{H}}{1 - T _{C} / T _{H}}$ . Top aur bottom ko $T_{H}$ se multiply karo: $COP = \frac{T _{C}}{T _{H} - T _{C}} = \frac{300}{600 - 300} = 1.0$ . Yeh step kyun? Humne temperature formula derive kiya rather than quote kiya — $Q_{H}$ se divide karke humne reversible ratio $T_{C} / T_{H}$ swap in kiya, aur $T_{H}$ se multiply karke fractions clear kiye.

$Q_{C} = 2000 J, W = 2000 J, COP = 1.0$

Verify: Dono routes ( $Q_{C} / W$ aur $T_{C} / (T_{H} - T_{C})$ ) $1.0$ dete hain ✓. Dhyan do ki COP $1 - T_{C} / T_{H}$ nahi hai — ek refrigerator ek alag question hai ("heat moved per work") rather than engine ("work per heat absorbed").

Worked example Example 5 — Cell E: degenerate case

$T_{H} = T_{C}$ Ek Carnot engine kaun si efficiency reach karta hai agar hot aur cold reservoirs same temperature par hon, jaise dono $400 K$ par?

Forecast: Bilkul bhi temperature gap nahi. Kya koi work nikal sakta hai?

Plug in karo: $η = 1 - \frac{T _{C}}{T _{H}} = 1 - \frac{400}{400} = 1 - 1 = 0$ . Yeh step kyun? Jab $T_{C} = T_{H}$ hota hai toh ratio exactly $1$ hota hai, jo poore expression ko khatam kar deta hai.
Interpret karo: $η = 0$ ka matlab hai $W = η Q_{H} = 0$ — engine koi net work produce nahi karta. Yeh step kyun? Engine ka poora kaam temperature difference ko exploit karna hai. Bina kisi gap ke, heat ke paas koi "downhill" direction nahi hai jahan woh work karte hue flow kare.

$η = 0$

Verify: Second law ke anusaar, tum ek single reservoir se ek cycle mein net work extract nahi kar sakte (woh second kind ki perpetual motion machine hogi). $η = 0$ exactly boundary par yahi law hai. Flow figure (s01) mein green $W$ arrow zero width tak shrink ho jaati. ✓

Worked example Example 6 — Cell F: limiting case

$T_{C} \to 0$ Maano cold reservoir ko absolute zero ki taraf cool kiya ja sake, $T_{C} \to 0 K$ , jabki $T_{H} = 500 K$ fixed rahe. $η$ kya approach karta hai, aur hum ise kabhi kyun reach nahi kar sakte?

Forecast: Jaise cold side thandi hoti jaati hai, ratio $T_{C} / T_{H}$ shrink hota hai. $η$ ki ceiling kya hai?

Limit lo: jaise $T_{C} \to 0$ , $\frac{T _{C}}{T _{H}} \to 0$ , toh $η = 1 - \frac{T _{C}}{T _{H}} \to 1 - 0 = 1$ . Yeh step kyun? $T_{H}$ fixed aur finite hai, toh sirf ratio ka numerator zero ki taraf shrink hota hai.
Neeche sanity table banao ( $T_{H} = 500$ fix rakhke) aur dekho kaise $η$ $1$ ki taraf creep karta hai lekin kabhi touch nahi karta jabtak $T_{C} > 0$ hai. Yeh step kyun? Ek limit par tab zyada trust hota hai jab tum discrete values ko approach karte dekho.

$T_{C}$ (K)	$T_{C} / T_{H}$	$η = 1 - T_{C} / T_{H}$
250	0.50	0.50
50	0.10	0.90
5	0.01	0.99
$\to 0$	$\to 0$	$\to 1$

Yahi climb yahan ek curve ke roop mein draw ki gayi hai — notice karo ki yeh dashed $η = 1$ ceiling ke against press karti hai lekin kabhi reach nahi karti:

$η \to 1 as T_{C} \to 0, but η < 1 always$

Verify: $η = 1$ ke liye exactly $T_{C} = 0 K$ chahiye hoga, jo parent ke " $η < 1$ always" se consistent hai. ✓

Recall Absolute zero kyun unreachable hai? (third-law footnote)

Cooling kaam karti hai kisi substance se disorder (entropy) remove karke — lekin cooling ka har step sirf ek fraction hi extract kar sakta hai jo bacha hai. Jaise tum $0 K$ ki taraf approach karte ho, woh leftover entropy jo tumhe abhi bhi remove karni hai woh shrink hoti jaati hai, lekin saath hi koi bhi single step jo extract kar sakta hai woh bhi shrink hoti jaati hai, toh tumhe aur zyada steps chahiye aur kabhi pahuncho hi nahi. Intuition: yeh aise hai jaise kisi wall tak pahunchne ki koshish kar rahe ho hamesha adha raasta chalke. Yeh thermodynamics ka third law hai, aur isliye $η = 1$ ek aisa limit hai jo tum approach karte ho lekin kabhi achieve nahi kar sakte.

Worked example Example 7 — Cell G: real-world word problem

Ek geothermal plant $200^{\circ} C$ par steam tap karta hai aur waste heat $20^{\circ} C$ wali river mein reject karta hai. Yeh har second steam se $6 \times 1 0^{5} J$ heat absorb karta hai. Maximum possible power output (work per second) kya hai?

Forecast: Temperatures moderate hain — kya tumhare hisaab se aise real plant ke liye high ya low efficiency expect karoge?

Words ko numbers tak strip karo, kelvin mein convert karo: $T_{H} = 200 + 273 = 473 K$ , $T_{C} = 20 + 273 = 293 K$ . Yeh step kyun? "Steam" aur "river" bas hot aur cold reservoirs hain. Baki sab decoration hai; physics ko sirf $T_{H}$ , $T_{C}$ , aur $Q_{H}$ chahiye.
Maximum efficiency (Carnot, kisi bhi real engine ke liye upper bound): $η = 1 - \frac{293}{473}$ . Compute karo: $\frac{293}{473} = 0.6194$ , toh $η = 1 - 0.6194 = 0.3806 \approx 0.381$ . Yeh step kyun? "Maximum possible" signal hai Carnot use karne ka — in reservoirs ke beech koi bhi real engine ise beat nahi kar sakta.
Power = work per second $= η \times (heat absorbed per second) = 0.3806 \times 6 \times 1 0^{5} \approx 2.28 \times 1 0^{5} W$ . Yeh step kyun? Kyunki $η = W / Q_{H}$ per cycle hold karta hai, yeh per second bhi hold karta hai jab hum rates use karte hain: $\dot{W} = η \dot{Q}_{H}$ .

$η \approx 38.1%, P_{m a x} \approx 2.28 \times 1 0^{5} W$

Verify: Units: $[dimensionless] \times [J/s] = [W]$ ✓. Aur $\approx 38%$ ek believable low-temperature-plant ceiling hai — real geothermal plants isse neeche chalte hain, around $10$ – $20%$ .

Worked example Example 8 — Cell H: exam twist ("I beat Carnot!")

Ek inventor claim karta hai ki uski engine $600 K$ source se $Q_{H} = 1000 J$ leti hai, sirf $Q_{C} = 400 J$ $400 K$ sink ko reject karti hai, aur $W = 600 J$ rakh leti hai. Kya yeh possible hai?

Forecast: Unki claimed efficiency $600/1000 = 60%$ hai. Decide karne se pehle compare karo ki in reservoirs ke beech Carnot kya allow karta hai.

In reservoirs ke liye Carnot ceiling: $η_{Carnot} = 1 - \frac{400}{600} = 1 - 0.6667 = 0.3333 = 33.3%$ . Yeh step kyun? Carnot's theorem kehta hai ki koi bhi engine do fixed reservoirs ke beech reversible efficiency $1 - T_{C} / T_{H}$ beat nahi kar sakta.
Inventor ki claimed efficiency: $η_{claim} = \frac{W}{Q _{H}} = \frac{600}{1000} = 0.6 = 60%$ . Yeh step kyun? Unke claim ko apni terms par compute karo taaki hum like ke saath like compare kar sakein.
Compare karo: $0.60 > 0.333$ — claim Carnot limit exceed karta hai, toh yeh impossible hai. Yeh step kyun? Carnot beat karna tumhe second kind ki perpetual motion machine banane deta, jo second law violate karta.

$Impossible: 60% > η_{Carnot} = 33.3%$

Verify (entropy route): Yahan kyun entropy criterion sahi test hai. Har reservoir ki entropy (usse mili heat) ke hisaab se change hoti hai $\div$ (uska temperature). Hot source $Q_{H}$ khota hai $T_{H}$ par, toh uski entropy $- Q_{H} / T_{H}$ change hoti hai. Cold sink $Q_{C}$ pata hai $T_{C}$ par, toh uski entropy $+ Q_{C} / T_{C}$ change hoti hai. Gas apne start par wapas aati hai, toh ek cycle mein uski khud ki entropy change zero hai. Second law demand karta hai ki universe ki total entropy kabhi decrease na ho: $Δ S_{univ} = \frac{Q _{C}}{T _{C}} - \frac{Q _{H}}{T _{H}} \geq 0.$ Claim test karo: $\frac{400}{400} - \frac{1000}{600} = 1 - 1.667 = - 0.667 < 0$ ✗. Negative total entropy change forbidden hai, jo confirm karta hai ki yeh engine exist nahi kar sakta. ✓

Worked example Example 9 — Cell I: work back to a volume ratio

Ek mole ideal gas isothermal expansion (step 1) $T_{H} = 500 K$ par undergo karta hai aur $Q_{H} = 2882 J$ absorb karta hai. $R = 8.314 J K^{- 1} mol^{- 1}$ lo (gas constant, upar define kiya). Volume ratio $V_{2} / V_{1}$ nikalo.

Forecast: Absorbed heat $ln (V_{2} / V_{1})$ ke saath badhti hai. Agar $Q_{H}$ $R T_{H}$ se thodi kam hai, toh kya tumhare hisaab se ratio $e$ (about $2.7$ ) ke paas hoga, ya kaafi bada?

Ek isothermal step ke dauran temperature constant hota hai, toh $Δ U = 0$ (internal energy sirf temperature par depend karti hai) aur saari absorbed heat expansion work ban jaati hai. $P = R T_{H} / V$ ko volume par integrate karne se log formula milta hai: $Q_{H} = R T_{H} ln \frac{V _{2}}{V _{1}} .$ Yeh step kyun? $\int d V / V = ln V$ , toh logarithm exactly wahi hai jo pop out hota hai jab pressure $1/ V$ ki tarah fall karta hai — isliye $ln$ , aur koi aur function nahi, yahan appear karta hai.
Ratio ke liye solve karo: $ln \frac{V _{2}}{V _{1}} = \frac{Q _{H}}{R T _{H}} = \frac{2882}{8.314 \times 500} = \frac{2882}{4157} = 0.6933$ . Yeh step kyun? Divide karne se log isolate hota hai; hum $R T_{H}$ se multiplication ko "undo" karte hain.
Log remove karne ke liye exponentiate karo: $\frac{V _{2}}{V _{1}} = e^{0.6933} = 2.000$ . Yeh step kyun? $ln$ aur $e$ inverse operations hain — $e^{l n x} = x$ (yahi $e$ ka poora point hai, upar define kiya) — jo logarithm ke andar trapped ratio ko release karta hai.

$\frac{V _{2}}{V _{1}} = 2.0$

Verify: Plug back karo: $R T_{H} ln 2 = 8.314 \times 500 \times 0.6931 = 2882 J$ ✓ diya gaya $Q_{H}$ recover ho jaata hai.

Recall Compression step ko yeh ratio kyun share karna zaroori hai

Cycle ke do adiabatic legs $T V^{γ - 1} = const$ follow karte hain. Ise hot→cold leg ke liye likhte hain ( $T_{H} V_{2}^{γ - 1} = T_{C} V_{3}^{γ - 1}$ ) aur cold→hot leg ke liye ( $T_{H} V_{1}^{γ - 1} = T_{C} V_{4}^{γ - 1}$ ) aur ek ko doosre se divide karte hain toh temperatures aur exponent cancel ho jaate hain, jo $V_{2} / V_{1} = V_{3} / V_{4}$ force karta hai. Toh yahan $V_{3} / V_{4} = 2.0$ bhi hai — aur yahi equality hai jiske kaaran dono logs cancel ho jaate hain aur efficiency mein sirf $T_{C} / T_{H}$ bachta hai.

Worked example Example 10 — Cell J: sign edge

$T_{C} > T_{H}$ (negative $η$ ) Galti se koi reservoirs ko ulta label kar deta hai aur engine formula mein $T_{H} = 300 K$ , $T_{C} = 400 K$ plug karta hai. Kya nikalta hai, aur sign ka kya matlab hai?

Forecast: "Cold" reservoir ab "hot" wale se zyada hot hai. Kya tumhare hisaab mein $η$ $1$ se upar hogi, exactly $0$ , ya $0$ se neeche?

Formula blindly apply karo: $η = 1 - \frac{T _{C}}{T _{H}} = 1 - \frac{400}{300} = 1 - 1.3333 = - 0.3333$ . Yeh step kyun? Jab "cold" label wala reservoir actually hotter hota hai, toh ratio $T_{C} / T_{H}$ $1$ se zyada ho jaata hai, toh $1$ minus yeh negative ho jaata hai.
Sign ko physically read karo. Negative $η$ real engine efficiency nahi ho sakti (positive heat in ke liye negative work out nahi mil sakti). Yeh signal karta hai ki heat naturally hotter body ( $400 K$ ) se cooler ( $300 K$ ) ki taraf khud se flow karti hai — yahan koi engine nahi hai. Heat ko doosri taraf move karne ke liye tumhe work supply karni hogi: yeh heat-pump / refrigerator regime hai (Cell D), engine nahi. Yeh step kyun? Formula mein koi built-in referee nahi hai; sign hi referee hai. Negative result yeh maths ka tarika hai tumhe batane ka ki dono reservoirs swap hain aur tum reverse (heat-pump) situation mein ho.

$η = - 0.333 \Rightarrow not an engine; this is a heat-pump regime$

Verify: Labels ko physical roles par wapas swap karo ( $T_{H} = 400$ , $T_{C} = 300$ ) aur honest engine efficiency hai $1 - 300/400 = 0.25 > 0$ ✓ — ek sensible positive number. Negative value purely mislabelling ka symptom thi, aur uske sign ne correctly reversed regime flag kiya.

Recall Har example ne kaunsa cell fill kiya?

Cell A (clean) ::: Ex 1 Cell B (Celsius conversion) ::: Ex 2 Cell C (find $Q_{C}$ , $W$ ) ::: Ex 3 Cell D (refrigerator / COP) ::: Ex 4 Cell E (degenerate $T_{H} = T_{C}$ ) ::: Ex 5 Cell F (limit $T_{C} \to 0$ ) ::: Ex 6 Cell G (word problem) ::: Ex 7 Cell H (beat-Carnot twist) ::: Ex 8 Cell I (invert to volume ratio) ::: Ex 9 Cell J (sign edge $T_{C} > T_{H}$ ) ::: Ex 10