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

1.7.21 · D4 · Physics › Thermodynamics › Carnot cycle — full derivation, efficiency = 1 − T_C - T_H

Level 1 — Recognition

Goal: dono master formulas mein numbers plug karo. Koi rearranging nahi, units ke alawa koi trap nahi.

Recall Solution 1.1

KYA / KYUN: Dono temperatures pehle se kelvin mein hain (unke saath K unit hai), toh hum unhe seedha $η = 1 - T_{C} / T_{H}$ mein daal dete hain. Koi conversion nahi, koi rearranging nahi. $η = 1 - \frac{300}{500} = 1 - 0.6 = 0.4 = 40%$ Iska matlab: hot reservoir se li gayi heat ka $40%$ useful work banta hai; baaki $60%$ cold reservoir mein dump ho jaata hai.

Recall Solution 1.2

KYA / KYUN: Hume $η$ aur $T_{C}$ pata hai, $T_{H}$ chahiye. $η = 1 - T_{C} / T_{H}$ se start karo aur $T_{H}$ isolate karo — yeh pure algebra hai, ek variable. $0.25 = 1 - \frac{300}{T _{H}} \Rightarrow \frac{300}{T _{H}} = 0.75 \Rightarrow T_{H} = \frac{300}{0.75} = 400 K$

Recall Solution 1.3

KYA / KYUN: Efficiency defined hai as $η = W / Q_{H}$ , toh work ke liye multiply karo. Phir ek cycle mein first law use karo ( $Δ U = 0 \Rightarrow W = Q_{H} - Q_{C}$ ) taaki rejected heat mile. $W = η Q_{H} = 0.4 \times 800 = 320 J$ $Q_{C} = Q_{H} - W = 800 - 320 = 480 J$ Reservoir identity se cross-check: $Q_{C} = Q_{H} (T_{C} / T_{H})$ . Kyunki $η = 0.4$ hai, $T_{C} / T_{H} = 0.6$ , toh $Q_{C} = 800 \times 0.6 = 480 J$ ✓.

Level 2 — Application

Goal: units convert karo, dono master formulas combine karo, refrigerator ko ulta chalao.

Recall Solution 2.1

KYA / KYUN: Temperatures Celsius mein hain, lekin $η$ ek ratio use karta hai, jo sirf kelvin scale par meaningful hai (zero = koi thermal energy nahi). Toh pehle convert karo: $T (K) = T (^{\circ} C) + 273$ . $T_{H} = 327 + 273 = 600 K, T_{C} = 27 + 273 = 300 K$ $η = 1 - \frac{300}{600} = 0.5 = 50%$

Recall Solution 2.2

KYA / KYUN: First law absorbed heat deta hai: $Q_{H} = W + Q_{C}$ . Phir uski definition se efficiency. Phir $T_{C} / T_{H} = Q_{C} / Q_{H}$ (reversible identity) use karke $T_{H}$ nikalo. $Q_{H} = W + Q_{C} = 600 + 900 = 1500 J$ $η = \frac{W}{Q _{H}} = \frac{600}{1500} = 0.4 = 40%$ $\frac{Q _{C}}{Q _{H}} = \frac{T _{C}}{T _{H}}$ se: $\frac{900}{1500} = \frac{270}{T _{H}} \Rightarrow 0.6 = \frac{270}{T _{H}} \Rightarrow T_{H} = \frac{270}{0.6} = 450 K$ .

Recall Solution 2.3

KYA / KYUN: Ek refrigerator ulta Carnot cycle hai: hum work $W$ dete hain taaki $Q_{C}$ heat cold side se pump ho sake. COP measure karta hai "har unit work par kitni heat move hoti hai". Humein COP ka formula diya gaya hai, toh usse compute karo, phir $W$ ke liye solve karo. $COP = \frac{T _{C}}{T _{H} - T _{C}} = \frac{250}{300 - 250} = \frac{250}{50} = 5$ $W = \frac{Q _{C}}{COP} = \frac{500}{5} = 100 J$ Iska matlab: bijli ka har joule 5 J heat move karta hai — yeh "1 J work se 1 J heat" se kahin behtar hai, kyunki hum heat move kar rahe hain, create nahi kar rahe.

Level 3 — Analysis

Goal: reason karo ki $η$ changes ke response mein kaise badalta hai, aur hidden cycle data reconstruct karo.

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

Recall Solution 3.1

KYA / KYUN: Dono options ke liye $η$ compute karo aur compare karo. Yeh ek asymmetry expose karta hai: kyunki $T_{C}$ $T_{C} / T_{H}$ ke numerator mein hai aur $T_{H}$ denominator mein, woh $η$ ko equally affect nahi karte. Baseline: $η_{0} = 1 - 300/500 = 0.400$ .

$T_{H}$ ko $550$ tak badhao: $η = 1 - 300/550 = 1 - 0.5455 = 0.4545$ . Gain $\approx 0.0545$ .
$T_{C}$ ko $250$ tak ghataao: $η = 1 - 250/500 = 1 - 0.5 = 0.500$ . Gain $= 0.100$ . Conclusion: yahan cold side ghataana jeetta hai (gain $0.100$ vs $0.055$ ). Figure mein red curve dekho: is operating point ke paas, efficiency $T_{C}$ ke against $T_{H}$ ke against zyada steep hai.

Recall Solution 3.2

KYA / KYUN: $T_{H}$ par isothermal expansion ke dauran, $Δ U = 0$ toh saari absorbed heat work ban jaati hai: $Q_{H} = R T_{H} ln (V_{2} / V_{1})$ — log $\int P d V = \int R T_{H} d V / V$ integrate karne se aata hai. Phir adiabats force karte hain $V_{3} / V_{4} = V_{2} / V_{1}$ , toh $Q_{C} = R T_{C} ln (V_{2} / V_{1})$ same log ke saath. $Q_{H} = R T_{H} ln 2 = 8.314 \times 400 \times 0.6931 = 2304.9 J$ $Q_{C} = R T_{C} ln 2 = 8.314 \times 300 \times 0.6931 = 1728.7 J$ $W = Q_{H} - Q_{C} = 2304.9 - 1728.7 = 576.2 J$ Cross-check: $η = W / Q_{H} = 576.2/2304.9 = 0.250 = 1 - 300/400$ ✓.

Recall Solution 3.3

KYA / KYUN: "Same efficiency" ka matlab hai $1 - T_{m} /600 = 1 - 300/ T_{m}$ , yaani dono temperature ratios equal hain: $T_{m} /600 = 300/ T_{m}$ . Yeh $T_{m}$ ke liye ek single equation hai; cross-multiply karne par geometric mean milta hai. $T_{m}^{2} = 600 \times 300 = 180000 \Rightarrow T_{m} = 180000 \approx 424.3 K$ Iska matlab: equal-efficiency staging middle temperature ko geometric mean $T_{H} T_{C}$ par rakhta hai, arithmetic mean $450 K$ par nahi.

Level 4 — Synthesis

Goal: first-law bookkeeping, entropy, aur impossible-engine reasoning chain karo.

Recall Solution 4.1

KYA / KYUN: Ek reservoir ke liye jo fixed temperature $T$ par heat $Q$ deta/leta hai, uska entropy change hai $Δ S = Q / T$ (reservoir ek temperature par rehta hai). Hot reservoir $Q_{H}$ khota hai (toh $Δ S < 0$ ), cold reservoir $Q_{C}$ paata hai. Pehle $Q_{C}$ nikalo: reversible identity $Q_{C} = Q_{H} (T_{C} / T_{H}) = 1000 \times 250/500 = 500 J$ . $Δ S_{hot} = - \frac{Q _{H}}{T _{H}} = - \frac{1000}{500} = - 2.00 J/K$ $Δ S_{cold} = + \frac{Q _{C}}{T _{C}} = + \frac{500}{250} = + 2.00 J/K$ $Δ S_{total} = - 2.00 + 2.00 = 0 J/K$ Conclusion: total entropy change exactly zero hai, jo ek reversible cycle ki signature hai — $\oint d Q_{rev} / T = 0$ ke consistent. Gas bhi apne starting point par wapas aa jaata hai, toh $Δ S_{gas} = 0$ bhi.

Recall Solution 4.2

KYA / KYUN: Carnot value woh ceiling hai jo in reservoirs ke beech koi bhi engine reach kar sakti hai. Claimed efficiency aur maximum compute karo, phir entropy se confirm karo (second law ke zariye ultimate arbiter). Efficiency check: claimed $η = W / Q_{H} = 700/1000 = 0.70$ . Carnot max $= 1 - 300/600 = 0.50$ . Kyunki $0.70 > 0.50$ , yeh ceiling exceed karta hai — impossible. Entropy check: $Q_{C} = Q_{H} - W = 1000 - 700 = 300 J$ . $Δ S_{total} = - \frac{Q _{H}}{T _{H}} + \frac{Q _{C}}{T _{C}} = - \frac{1000}{600} + \frac{300}{300} = - 1.667 + 1.000 = - 0.667 J/K$ Ek negative total entropy change second law violate karta hai (universe ki entropy kabhi nahi gir sakti). Verdict: impossible.

Recall Solution 4.3

KYA / KYUN: First law $Q_{C}$ deta hai. Real $η$ uski definition se; Carnot $η$ benchmark ke roop mein. Kyunki yeh engine irreversible hai, hum expect karte hain $Δ S_{total} > 0$ (strictly positive). $Q_{C} = Q_{H} - W = 1000 - 400 = 600 J$ $η_{real} = \frac{400}{1000} = 0.40, η_{Carnot} = 1 - \frac{300}{600} = 0.50$ Real Carnot se neeche hai ( $0.40 < 0.50$ ) — allowed. Entropy generated: $Δ S_{total} = - \frac{1000}{600} + \frac{600}{300} = - 1.667 + 2.000 = + 0.333 J/K$ Conclusion: positive entropy generation ek irreversible lekin legal engine confirm karta hai.

Level 5 — Mastery

Goal: theorem prove karo, aur ek abstract statement ke peeche ke numbers derive karo.

Recall Solution 5.1

KYA / KYUN: Ulta assume karo aur dikhao ki yeh second law todne deta hai. Yeh proof by contradiction hai — standard weapon jab koi quantity ceiling exceed karne ka claim kare. Setup (figure dekho): Suppose ek "super-engine" X ka $η_{X} > η_{C}$ hai. X ko forward chalao taaki work $W$ produce ho, hot reservoir se $Q_{H}^{X}$ absorb karte hue. Saara $W$ use karo ek Carnot engine ko ulta refrigerator ki tarah drive karne ke liye, jo cold se hot mein heat pump kare. Work match karo: reversed Carnot engine wahi $W$ consume karta hai. Usse choose karo taaki exactly $Q_{H}^{C}$ hot reservoir mein reject kare aur cold se $Q_{C}^{C}$ absorb kare, $W = Q_{H}^{C} - Q_{C}^{C}$ ke saath. X dwara hot reservoir se li gayi heat hai $Q_{H}^{X} = W / η_{X}$ . Fridge dwara hot reservoir mein wapas di gayi heat hai $Q_{H}^{C} = W / η_{C}$ . Kyunki $η_{X} > η_{C}$ , hume milta hai $Q_{H}^{X} = W / η_{X} < W / η_{C} = Q_{H}^{C}$ . Contradiction: hot reservoir $Q_{H}^{C}$ receive karta hai aur $Q_{H}^{X}$ khota hai, net gain $Q_{H}^{C} - Q_{H}^{X} > 0$ . First law se same net heat cold reservoir se draw hoti hai — koi external work nahi (dono engines ka work cancel ho gaya). Humne spontaneously cold se hot mein heat move ki, yeh second kind ki perpetual-motion machine hai. Yeh second law violate karta hai. Isliye $η_{X} \leq η_{C}$ . $■$

Recall Solution 5.2

KYA / KYUN: Abstract proof ko numbers mein badlo. Compute karo ki X hot side se kitni heat kheenchta hai aur reversed Carnot fridge kitni wapas karta hai; unka difference woh illegal free heat hai. $Q_{H}^{X} = \frac{W}{η _{X}} = \frac{300}{0.6} = 500 J (X dwara hot se draw ki gayi)$ $Q_{H}^{C} = \frac{W}{η _{C}} = \frac{300}{0.5} = 600 J (fridge dwara hot mein wapas ki gayi)$ $Hot reservoir mein net = Q_{H}^{C} - Q_{H}^{X} = 600 - 500 = 100 J > 0$ Conclusion: har cycle mein $100 J$ cold $\to$ hot flow karta hai zero net work ke saath — impossible. Positive number hi contradiction hai, confirm karta hai $η_{X} \leq 0.5$ .

Recall Solution 5.3

KYA / KYUN: Parent derivation mein working-substance-dependent logs cancel ho gaye, aur kisi bhi reversible engine ke liye $Q_{C} / Q_{H} = T_{C} / T_{H}$ mila. Toh reversible engines ke heat ratios ek substance-free thermometer dete hain — yahi Kelvin scale ki definition hai. $T_{H} = 273.16 K$ ko reference fix karke, kisi bhi doosre reservoir ki temperature hai: $T_{C} = T_{H} \cdot \frac{Q _{C}}{Q _{H}} = 273.16 \times 0.500 = 136.58 K$ Iska matlab: humne kabhi "molecular speed" measure nahi ki ya koi gas use nahi ki — sirf ek heat ratio. Isliye Kelvin scale ko absolute kaha jaata hai: ise koi material nahi chahiye.

Recall Jaane se pehle ek-line self-test

Master fact ::: $η = 1 - T_{C} / T_{H}$ aur $Q_{C} / Q_{H} = T_{C} / T_{H}$ , temperatures kelvin mein. Reversible signature ::: $Δ S_{total} = 0$ (irreversible $> 0$ , impossible $< 0$ ). Theorem ::: koi engine Carnot ko nahi peetta; proof ek super-engine ko reversed Carnot se couple karke second law todta hai.