Refrigerators and heat pumps — COP: Exercises — Hinglish deep dive

1.7.20 · D4 · Physics › Thermodynamics › Refrigerators and heat pumps — COP

Yahan $Q_{C}$ woh heat hai jo cold reservoir se khींची gayi (magnitude, joules mein), $Q_{H}$ woh heat jo hot reservoir mein dump ki gayi, aur $W_{in}$ woh work jo pay kiya gaya (compressor ko electricity). Teeno positive numbers hain; directions pehle se hi formulas mein bake hain.

Sign-convention map (problems se pehle yeh padho)

Neeche har problem is baat par hinges karti hai ki $Q$ aur $W$ kis direction mein point karte hain. Neeche di gayi figure ek baar rule fix karti hai, ek picture ke roop mein: working substance mein jaane wale arrows positive inputs hain, bahar jaane wale arrows outputs hain, aur teeno ek cycle mein zero tak balance ho jaate hain.

Figure — Refrigerators and heat pumps — COP

Quantity	Direction (fridge / heat pump ke liye)	Yahan sign
$Q_{C}$	heat in, cold reservoir se khींची gayi	$+$
$W_{in}$	work in, mains/compressor se	$+$
$Q_{H}$	heat out, hot reservoir mein dump ki gayi	$+$ (substance se bahar jaati hai)
balance	$Q_{C} + W_{in} = Q_{H}$ ek poore cycle mein	$Δ U = 0$

Picture aur table ek hi baat keh rahe hain: do arrows in ( $Q_{C}$ , $W_{in}$ ), ek arrow out ( $Q_{H}$ ), aur woh same total mein sum hote hain.

Level 1 — Recognition

L1.1

Ek machine cold body se $Q_{C} = 300 J$ lete hai aur har cycle mein $W_{in} = 100 J$ work chahiye. Kaun si quantity "benefit" hai agar device ko fridge ki tarah use kiya jaaye, aur $COP_{r e f}$ kya hai?

Recall Solution

Kya / kyun. Ek fridge ko uski cooling se judge kiya jaata hai, isliye benefit hai $Q_{C} = 300 J$ ; cost hai work $W_{in} = 100 J$ . $COP_{r e f} = \frac{Q _{C}}{W _{in}} = \frac{300}{100} = 3.$ Matlab: har 1 J electricity andar se 3 J heat remove karti hai.

L1.2

L1.1 wali same machine. Scratch se recompute kiye bina, iska heat-pump COP batao.

Recall Solution

Shortcut kyun kaam karta hai. Dono COPs mein hamesha exactly 1 ka difference hota hai (same denominator $W_{in}$ , numerators mein $Q_{H} - Q_{C} = W_{in}$ ka difference hai). $COP_{h p} = COP_{r e f} + 1 = 3 + 1 = 4.$ Definition se check karo: $Q_{H} = Q_{C} + W_{in} = 300 + 100 = 400$ , isliye $COP_{h p} = Q_{H} / W_{in} = 400/100 = 4.$ ✓

L1.3

Ek fridge andar $T_{C} = 250 K$ aur room $T_{H} = 300 K$ ke beech run karti hai. Refrigerator ke roop mein iske maximum possible COP ke liye formula likhο (evaluate mat karo abhi), aur batao kaun sa temperature upar hoga.

Recall Solution

Refrigerator ke liye cold temperature upar hoti hai (cold benefit hai): $COP_{r e f}^{C a r n o t} = \frac{T _{C}}{T _{H} - T _{C}} = \frac{250}{300 - 250} .$ Denominator hai climb $T_{H} - T_{C}$ .

Level 2 — Application

L2.1

Ek refrigerator jiska $COP_{r e f} = 5$ hai, har cycle mein $W_{in} = 120 J$ work consume karta hai. $Q_{C}$ aur $Q_{H}$ nikalo.

Recall Solution

Step 1 (cooling). $Q_{C} = COP_{r e f} \times W_{in} = 5 \times 120 = 600 J$ . Step 2 (energy balance, first law). $Q_{H} = Q_{C} + W_{in} = 600 + 120 = 720 J$ . Room ko 720 J milta hai jabki andar sirf 600 J kho ta hai — extra 120 J woh work hai jo ab room ko warm kar raha hai.

L2.2

Ek Carnot heat pump ek ghar ko $T_{H} = 293 K$ par rakha hai jabki bahar $T_{C} = 273 K$ hai. $COP_{h p}$ nikalo, aur batao ki 1 J electricity par yeh kitni heat deliver karta hai.

Recall Solution

Heat-on-top kyun. Heat pump ka benefit hai heating $Q_{H}$ , isliye $T_{H}$ upar jaata hai. $COP_{h p}^{C a r n o t} = \frac{T _{H}}{T _{H} - T _{C}} = \frac{293}{293 - 273} = \frac{293}{20} = 14.65.$ Isliye har 1 J electricity ideally ghar mein 14.65 J heat deliver karta hai — ek 1:1 electric heater se kaafi better.

L2.3

Ek inventor claim karta hai ki $T_{C} = 280 K$ aur $T_{H} = 310 K$ ke beech run hone wala fridge $COP_{r e f} = 12$ achieve karta hai. Kya second law yeh allow karta hai?

Recall Solution

Step 1 — ceiling. $COP_{r e f}^{C a r n o t} = \frac{280}{310 - 280} = \frac{280}{30} = 9.33.$ Step 2 — compare. Claimed $12 > 9.33 = COP_{r e f}^{C a r n o t}$ . Koi bhi real ya reversible device Carnot ko beat nahi kar sakta, isliye claim impossible hai — yeh universe mein negative entropy generate karega. Reject.

Level 3 — Analysis

L3.1

Ek Carnot fridge $T_{C} = 250 K$ hold karta hai jabki room $T_{H} = 300 K$ se $T_{H} = 310 K$ tak warm ho jaata hai. Pehle aur baad mein $COP_{r e f}^{C a r n o t}$ compute karo. Kya room temperature badhne se fayda hota hai ya nuksan, aur kyun?

Recall Solution

Pehle: $\frac{250}{300 - 250} = \frac{250}{50} = 5.0$ . Baad mein: $\frac{250}{310 - 250} = \frac{250}{60} \approx 4.17$ . Kyun girta hai. Denominator hai "hill height" $T_{H} - T_{C}$ . Hotter room hill ko steeper banata hai, isliye same heat pump karne mein zyaada work lagta hai — COP girta hai. Neeche ki plot yeh visible banati hai: dono COP curves neeche slide karte hain jaise horizontal axis par gap badhta hai, aur do plum dots exactly is problem ke before/after pair mark karte hain.

Neeche ki figure trace karti hai ki temperature gap badhne par COP kaise girta hai; do plum dots is problem ke before-and-after points hain (50 K aur 60 K ke gaps). Ise is tarah padho: reservoirs ko aur dur push karo aur leverage collapse ho jaata hai.

L3.2

Ek fridge apna darwaza khola hue sealed, insulated room mein rakhа hai. Samay ke saath, kya room thanda hoga, garam hoga, ya same rahega? $Q_{H}, Q_{C}, W_{in}$ se justify karo.

Recall Solution

Reasoning. Har cycle mein machine room-air se $Q_{C}$ pull karti hai (khule darwaze se) aur $Q_{H} = Q_{C} + W_{in}$ usi room mein wapas dump karti hai (coils ke through). Room mein har cycle mein net heat add hoti hai: $Q_{H} - Q_{C} = W_{in} > 0.$ $Q_{C}$ cancel ho jaata hai — woh room se gaya aur seedha wapas aa gaya. Jo bachta hai woh compressor work $W_{in}$ hai, jo sab heat mein convert ho jaata hai. Room garam hota hai, bilkul electrical power draw hone ki rate par.

L3.3

Do Carnot refrigerators same room $T_{H} = 300 K$ share karte hain. Fridge A $T_{C} = 290 K$ tak cool karta hai; Fridge B (ek deep freezer) $T_{C} = 250 K$ tak cool karta hai. Kiska COP higher hai, aur roughly kis factor se?

Recall Solution

$A: \frac{290}{300 - 290} = \frac{290}{10} = 29, B: \frac{250}{300 - 250} = \frac{250}{50} = 5.$ Fridge A ka COP B ka $29/5 = 5.8 \times$ hai. Kyun. Colder targets ka matlab hai bada gap $T_{H} - T_{C}$ aur chota numerator — dono COP ko neeche push karte hain. Deep freezing expensive hota hai.

Level 4 — Synthesis

L4.1

Ek real fridge apne Carnot COP ka 40% achieve karta hai. Yeh $T_{C} = 275 K$ aur $T_{H} = 305 K$ ke beech run karta hai aur har cycle mein $Q_{C} = 900 J$ remove karna chahiye. Actual COP nikalo, phir $W_{in}$ jo ise draw karna hoga, phir $Q_{H}$ .

Recall Solution

Step 1 — Carnot ceiling. $COP_{r e f}^{C a r n o t} = \frac{275}{305 - 275} = \frac{275}{30} \approx 9.1 \overline{6}$ (bar ka matlab $9.1666 \dots$ , exactly $\frac{55}{6}$ ). Step 2 — real COP. $COP_{r e a l} = 0.40 \times 9.1 \overline{6} = 3.6 \overline{6}$ (yaani $3.6666 \dots = \frac{11}{3}$ ). Step 3 — work. $W_{in} = \frac{Q _{C}}{COP _{r e a l}} = \frac{900}{3.6 6} \approx 245.45 J$ . Step 4 — heat dumped. $Q_{H} = Q_{C} + W_{in} = 900 + 245.45 \approx 1145.45 J$ . Ideal machine ko sirf $900/9.1 \overline{6} \approx 98.2 J$ chahiye hota; real wale ko irreversibility ki wajah se $\approx 245$ J chahiye.

L4.2

Ek device sardi mein heat pump ke roop mein use hota hai aur, garmi mein, same unit ulta run karke fridge (air conditioner) ki tarah kaam karta hai. Sardi mein $T_{H} = 293 K$ , $T_{C} = 273 K$ . Iska Carnot heat-pump COP nikalo; phir matching Carnot refrigerator COP ko do independent tareekon se confirm karo.

Recall Solution

Heat-pump COP: $\frac{293}{293 - 273} = \frac{293}{20} = 14.65$ . Tarika 1 — 1 ghatao: $COP_{r e f} = COP_{h p} - 1 = 14.65 - 1 = 13.65$ . Tarika 2 — direct formula: $\frac{T _{C}}{T _{H} - T _{C}} = \frac{273}{20} = 13.65$ . ✓ Dono agree karte hain, confirm karte hain ki $COP_{h p} = COP_{r e f} + 1$ same $W_{in}$ par sirf algebra hai.

L4.3

Ek Carnot heat pump $T_{H} = 300 K$ par ek ghar mein har cycle mein $Q_{H} = 1200 J$ deliver karta hai, $T_{C} = 280 K$ par bahari hawa se heat draw karke. $W_{in}$ aur har cycle mein universe ka entropy change nikalo. Comment karo.

Recall Solution

Step 1 — COP. $COP_{h p} = \frac{300}{300 - 280} = \frac{300}{20} = 15$ . Step 2 — work. $W_{in} = \frac{Q _{H}}{COP _{h p}} = \frac{1200}{15} = 80 J$ . Step 3 — heats. $Q_{C} = Q_{H} - W_{in} = 1200 - 80 = 1120 J$ . Step 4 — entropy of universe. Ek reversible (Carnot) cycle ke liye, cold se pull ki gayi entropy hot mein dump ki gayi entropy ke barabar hoti hai: $Δ S = \frac{Q _{H}}{T _{H}} - \frac{Q _{C}}{T _{C}} = \frac{1200}{300} - \frac{1120}{280} = 4.0 - 4.0 = 0.$ Comment. $Δ S_{u ni v er se} = 0$ ek reversible machine ki signature hai — yeh exactly second-law boundary par baithti hai. Koi bhi real unit $Δ S > 0$ dikhayega aur 80 J se zyaada ki zarurat hogi.

Level 5 — Mastery

L5.1

Ek heat pump ek aise ghar ko warm karta hai jo bahar heat $8000 J$ per second (8 kW) ki rate se leak karta hai. Steady temperature hold karne ke liye, pump ko $Q_{H} = 8 kW$ deliver karna hoga. Yeh $T_{C} = 268 K$ aur $T_{H} = 298 K$ ke beech Carnot ka 50% par run karta hai. Electrical power $W_{in}$ nikalo jo yeh draw karta hai. Ek 1:1 resistive heater se compare karo.

Recall Solution

Step 1 — Carnot COP. $COP_{h p}^{C a r n o t} = \frac{298}{298 - 268} = \frac{298}{30} \approx 9.9 \overline{3}$ (bar: $9.9333 \dots = \frac{149}{15}$ ). Step 2 — real COP. $COP_{r e a l} = 0.50 \times 9.9 \overline{3} \approx 4.9 \overline{6}$ (yaani $4.9666 \dots = \frac{149}{30}$ ). Step 3 — power drawn. Har second, $\dot{W}_{in} = \frac{Q _{H}}{COP _{r e a l}} = \frac{8000}{4.9 6} \approx 1610.7 W$ . Step 4 — compare. Ek resistive heater ka COP $= 1$ hota hai, isliye woh poora $8000 W$ draw karega. Heat pump $\approx 1611 W$ use karta hai — same warmth ke liye electricity lagbhag 5× kam. Fark woh free heat $Q_{C}$ hai jo thandi bahar se khींchi jaati hai.

L5.2

Do engineers argue karte hain. Engineer X insist karta hai ki ek fridge $COP_{r e f} = \infty$ ke saath banaya ja sakta hai "agar hum ise perfectly banayein." Engineer Y kehta hai impossible. Carnot formula use karke batao ki kaun si ek physical condition mein $COP_{r e f} \to \infty$ , aur kyun second law usse reach karna forbid karta hai.

Recall Solution

$COP_{r e f}^{C a r n o t} = \frac{T _{C}}{T _{H} - T _{C}} \to \infty ⟺ T_{H} - T_{C} \to 0, i.e. T_{H} = T_{C} .$ Limit kyun blow up hoti hai. Jaise dono reservoirs same temperature ke paas aate hain, "hill" gayab ho jaati hai — heat move karne mein essentially koi work nahi lagta. Yeh unreachable kyun hai. Agar $T_{H} = T_{C}$ ho toh koi temperature difference nahi aur pump karne ke liye distinct "cold" aur "hot" bodies nahi; heat ko pumping ki zarurat hi nahi hogi. Kisi bhi useful fridge ke liye $T_{H} > T_{C}$ chahiye, isliye denominator strictly positive hai aur COP finite hai. Engineer Y sahi hai: $\infty$ ek limit hai, kabhi achievable value nahi.

L5.3

Ek Carnot refrigerator ( $T_{C} = 260 K$ , $T_{H} = 300 K$ ) har cycle mein $Q_{C} = 780 J$ remove karta hai. Iska exhaust heat $Q_{H}$ ek Carnot heat engine ko input ke roop mein feed kiya jaata hai jo usi $T_{H} = 300 K$ aur ek colder sink $T_{L} = 200 K$ ke beech operate karta hai. Engine jo work produce karta hai woh nikalo, aur isliye engine ke output ko credit karne ke baad fridge chalane ki net work cost nikalo.

Recall Solution

Pehle symbols. Fridge $W_{in}$ consume karta hai; engine $W_{o u t}$ produce karta hai. Yeh direction mein opposite hain, isliye hum inhe alag symbols ke roop mein rakhte hain aur sirf end mein combine karte hain.

Fridge side. $COP_{r e f} = \frac{260}{300 - 260} = \frac{260}{40} = 6.5$ . Fridge chalane ka work: $W_{in} = \frac{Q _{C}}{COP _{r e f}} = \frac{780}{6.5} = 120 J$ . Exhaust: $Q_{H} = Q_{C} + W_{in} = 780 + 120 = 900 J$ .

Engine side. Ek Carnot engine jo $300 K$ par $Q_{H} = 900 J$ leta hai aur $200 K$ par reject karta hai uski efficiency hai $η = 1 - \frac{T _{L}}{T _{H}} = 1 - \frac{200}{300} = \frac{1}{3} .$ Work produced: $W_{o u t} = η Q_{H} = \frac{1}{3} \times 900 = 300 J$ .

Net cost. Engine $W_{o u t} = 300 J$ work wapas deta hai jabki fridge ko sirf $W_{in} = 120 J$ chahiye tha: $W_{n e t} = W_{in} - W_{o u t} = 120 - 300 = - 180 J .$ Negative cost ka matlab hai combination surplus work produce karta hai — lekin dhyan do ki engine apni reject heat $Q_{H} - W_{o u t} = 900 - 300 = 600 J$ $200 K$ sink mein dump karta hai, jo ek teesra reservoir hai. Koi free lunch nahi: surplus us heat ke liye pay kiya jaata hai jo bahut thande $200 K$ sink tak neeche flow hoti hai. (Heat Engines and Efficiency aur Carnot Cycle use karta hai.)

Connections

Carnot Cycle — upar use ki gayi har COP ceiling supply karta hai.
Second Law of Thermodynamics — isliye L2.3 aur L5.2 fail hote hain; isliye COP finite hai.
Entropy — $Q / T$ bookkeeping L4.3 ke peeche.
Heat Engines and Efficiency — L5.3 mein reversed device.
First Law of Thermodynamics — $Q_{H} = Q_{C} + W_{in}$ throughout use kiya gaya.