Refrigerators and heat pumps — COP: Worked examples

1.7.20 · D3 · Physics › Thermodynamics › Refrigerators and heat pumps — COP

Shuru karne se pehle, woh char tools jo hum baar baar use karte hain (har ek parent mein pehle se earn kiya hua hai):

Scenario matrix

Har exam question COP par in cells mein se exactly ek mein aata hai (A–H). Neeche ka map unhe us move ke hisaab se group karta hai jo unhe solve karta hai — wahi do-move routing jo neeche wala mnemonic use karta hai. Worked examples saarey aath cells ko cover karte hain.

#	Cell (case class)	Kya tricky hai	Example
A	$T_{H}, T_{C}$ diya → Carnot COP nikalo	plain plug-in, kelvin	Ex 1
B	Celsius diya → convert karna padega	°C→K trap	Ex 2
C	Non-ideal device: COP & $W$ diya → $Q_{C}, Q_{H}$ nikalo	first law, Carnot nahi	Ex 3
D	Usi machine ka heat-pump viewpoint	$+ 1$ relation, "beats a heater"	Ex 4
E	Limiting case: $T_{C} \to T_{H}$ (gap → 0)	COP → ∞, kyun	Ex 5
F	Degenerate case: $T_{C} \to 0$ K ya bahut bada gap	COP → 0, third-law hint	Ex 6
G	Door-open fridge (real-world twist)	room ka net heating	Ex 7
H	Real vs ideal ratio (exam twist)	Carnot ka fraction achieve kiya	Ex 8

Figure — Refrigerators and heat pumps — COP

Yeh figure Carnot fridge COP ko vertical axis par plot karti hai (labelled $COP_{r e f}^{C}$ , log scale par taaki bade aur chote dono values fit ho sakein) horizontal axis ke against — temperature gap $T_{H} - T_{C}$ kelvin mein, ek fixed hot side $T_{H} = 300 K$ ke liye. Ise padhte waqt interrogate karo:

Far left (blue curve upar jaati hai): gap $\approx 1 K$ hai, isliye denominator tiny hai aur COP 200 se bhi upar explode hoti hai — yeh exactly wahi green dot Ex 5 hai ( $T_{C} \to T_{H}$ , COP $\to \infty$ ).
Middle: white Ex 8 dot gap $30 K$ par, COP $= 9$ ; yellow Ex 1 dot gap $50 K$ par, COP $= 5$ . Notice karo ki curve yahan pehle se steeply girr rahi hai.
Dashed white line COP $= 1$ par: iske neeche, ek fridge us work se kam heat move karti hai jo woh pay karti hai. Sirf far-right red Ex 6 dot (gap $297 K$ ) is line ke neeche girta hai, COP $\approx 0.01$ par.
Slope padhna: gap har baar double hone par COP roughly half ho jaata hai — "hill" badhti hai aur leverage collapse hoti hai. Yeh single trend cells A, E, F aur H sab ko ek saath explain karta hai.

Cell A — plain Carnot plug-in

Worked example Ex 1 · Freezer ek warm room ke against

Ek freezer apna andar $T_{C} = 250 K$ par rakhta hai jabki kitchen $T_{H} = 300 K$ hai. Maximum possible $COP_{r e f}$ kya hai?

Forecast: gap $50 K$ hai — na tiny, na huge. Guess: COP thoda 4 se upar? Apna guess likho.

Do temperatures identify karo: $T_{C} = 250$ , $T_{H} = 300$ (pehle se kelvin mein hain). Yeh step kyun? Carnot COP entropy $Q / T$ se bana hai, jo sirf absolute kelvin scale par sense deta hai — formula touch karne se pehle units check karo.
Climb compute karo $T_{H} - T_{C} = 300 - 250 = 50 K$ . Yeh step kyun? Denominator hamesha woh hill hai jis par heat ko pump karna hai; ise pehle isolate karna arithmetic slips se bachata hai.
$COP_{r e f}^{C} = \frac{T _{C}}{T _{H} - T _{C}} = \frac{250}{50} = 5$ apply karo. Yeh step kyun? Yeh ceiling hai — in temperatures ke saath koi bhi real freezer COP $= 5$ se behtar nahi kar sakta.

Verify: fridge formula cold temperature $T_{C}$ upar rakhti hai (benefit = cooling), giving $250/50 = 5$ ; heat-pump twin uski jagah hot temperature $T_{H}$ upar rakhta hai, $COP_{h p} = \frac{T _{H}}{T _{H} - T _{C}} = \frac{300}{50} = 6 = 5 + 1$ ✓ ( $+ 1$ relation holds hai).

Cell B — Celsius trap

Worked example Ex 2 · Air-conditioner °C mein quoted

Ek AC ek room ko $T_{C} = 20 ° C$ tak cool karta hai jabki $T_{H} = 40 ° C$ par bahar heat dump karta hai. Maximum $COP_{r e f}$ nikalo. Ek student likhta hai $\frac{20}{40 - 20} = 1.0$ . Kya yeh sahi hai?

Forecast: naive Celsius answer $1.0$ hai. Lekin $\sim 300 K$ absolute scale par $20 K$ gap chhota hai, isliye true COP 1 se bahut zyada hona chahiye. Predict karo: around 14–15?

Dono ko kelvin mein convert karo exact offset $273.15$ use karke: $T_{C} = 20 + 273.15 = 293.15 K$ , $T_{H} = 40 + 273.15 = 313.15 K$ . Yeh step kyun? $Q \propto T$ entropy $Q / T$ se aaya; Celsius numbers ke ratios meaningless hain kyunki °C ka ek arbitrary zero hai. Entropy dekho. (Rough work ke liye $+ 273$ rounding theek hai, lekin strict thermodynamics context mein $273.15$ use karo.)
Climb: $T_{H} - T_{C} = 313.15 - 293.15 = 20 K$ . (Note: gap °C ya K dono mein same hai — lekin numerator nahi hai!) Yeh step kyun? Temperature differences dono scales mein identical hain; temperature values nahi hain — yahi poora trap hai.
$COP_{r e f}^{C} = \frac{293.15}{20} = 14.66$ . Yeh step kyun? Yeh honest answer hai; student ka $1.0$ ~15 ka factor off tha.

Verify: correct ratio gap $20$ ko absolute numerator $293.15$ ke saath use karta hai, giving $14.66$ , galat Celsius answer se 14× se zyada. Ek bade absolute floor par tiny hill ⇒ big COP, Ex 1 ke trend ke consistent.

Cell C — non-ideal device, heats nikalo

Worked example Ex 3 · Ek real fridge, COP aur work diya hua

Ek real refrigerator ka $COP_{r e f} = 3.5$ hai aur uska compressor $W = 120 J$ work karta hai per cycle. $Q_{C}$ (andar se kheechi gayi heat) aur $Q_{H}$ (room mein dump ki gayi heat) nikalo.

Forecast: benefit ÷ cost $= 3.5$ , isliye $Q_{C}$ work ka $3.5 \times$ hona chahiye $\approx 420 J$ , aur $Q_{H}$ thoda zyada. Guess $Q_{H} \approx 540 J$ ?

Definition $COP_{r e f} = \frac{Q _{C}}{W}$ se, $Q_{C} = COP_{r e f} \times W = 3.5 \times 120 = 420 J$ solve karo. Yeh step kyun? Humein actual COP diya gaya hai, isliye hum general definition use karte hain (Carnot nahi) — yeh kisi bhi device ke liye hold karta hai, ideal ya non-ideal.
First law $Q_{H} = Q_{C} + W = 420 + 120 = 540 J$ use karo. Yeh step kyun? Ek poore cycle mein working substance apni starting state mein wapas aati hai, isliye uski internal energy unchanged hai — "heat in equals heat out" — yahi reason hai ki balance $Q_{H} = Q_{C} + W$ hold karta hai chahe fridge kitni bhi achi ho. First Law of Thermodynamics dekho.

Verify: units sab joules ✓. Room ko $540 J$ milta hai jabki andar sirf $420 J$ kho jaata hai — missing $120 J$ exactly woh electricity hai jo tumne pay ki, ab heat mein convert ho gayi. $Q_{H} > Q_{C}$ ✓ (surplus $= W$ ).

Cell D — same machine, heat-pump viewpoint

Worked example Ex 4 · Heat pump vs plain electric heater

Ex 3 ki exact machine ab ek ghar ko garam karne ke liye heat pump ki tarah bech di gayi hai. Har cycle mein woh abhi bhi $W = 120 J$ karta hai aur $Q_{H} = 540 J$ deliver karta hai. $COP_{h p}$ kya hai, aur yeh $\$$1-per-joule electric heater (COP$ =1$) se kaise compare karta hai?

Forecast: heating COP fridge COP plus one hona chahiye, isliye $3.5 + 1 = 4.5$ . Matlab same electricity ke liye plain heater se ~4.5× heat.

$COP_{h p} = \frac{Q _{H}}{W} = \frac{540}{120} = 4.5$ use karo. Yeh step kyun? Heat pump ke liye benefit delivered heat $Q_{H}$ hai, cooling nahi — different numerator, same cost $W$ .
Relation $COP_{h p} = COP_{r e f} + 1 = 3.5 + 1 = 4.5$ ✓ se cross-check karo. Yeh step kyun? Confirm karta hai ki humne numerators mix nahi kiye; dono COPs denominator $W$ share karte hain aur unke numerators $Q_{H} - Q_{C} = W$ se differ karte hain.

Verify: ek plain electric heater $120 J$ electricity ko $120 J$ heat mein turn karta hai (COP $= 1$ ). Yeh pump $540 J$ deliver karta hai — 4.5× zyada — kyunki yeh bahar se free heat bhi kheenchta hai. Yahi extra hai isliye heat pumps paise bachate hain. Heat Engines and Efficiency dekho.

Cell E — limiting case: gap zero tak shrink ho jaata hai

Worked example Ex 5 ·

$T_{C} \to T_{H}$ hone par kya hota hai? Ek heat pump ek cellar ko garam karta hai jo pehle se bahar jaisa hi garam hai: $T_{H} = 291 K$ , $T_{C} = 290 K$ . Phir imagine karo $T_{C} \to 291$ . $COP_{h p}^{C}$ kya karta hai?

Forecast: hill almost flat hai, isliye heat pumping almost free hona chahiye → COP enormous, infinity ki taraf jaata hua. Guess: 291?

Climb $= T_{H} - T_{C} = 291 - 290 = 1 K$ . Yeh step kyun? $1 K$ denominator wahi hai jo ratio ko explode karta hai; ise isolate karna mechanism dikhata hai.
$COP_{h p}^{C} = \frac{291}{1} = 291$ . Yeh step kyun? Almost koi work nahi chahiye — tum heat ko bilkul bhi lift nahi kar rahe.
Limit $T_{C} \to T_{H}$ lo: denominator $\to 0^{+}$ , numerator $T_{H}$ rehta hai, isliye $COP_{h p}^{C} \to + \infty$ . Yeh step kyun? Yeh figure mein curve ka left end hai — koi gap nahi matlab koi hill nahi matlab (ideally) infinite leverage. Second Law of Thermodynamics abhi bhi ek literal zero-work transfer forbid karta hai, isliye infinity sirf mathematical ceiling hai.

Verify: figure par, jaise gap → 0 blue curve upar shoot karti hai ✓. Fridge twin: $COP_{r e f}^{C} = 290/1 = 290 = 291 - 1$ ✓.

Cell F — degenerate case: cold reservoir near absolute zero

Worked example Ex 6 · Bahut low temperatures reach karne ki koshish

Ek cryo-fridge $T_{H} = 300 K$ par ek room mein dump karta hai aur ek sample ko $T_{C} = 3 K$ tak cool karna hai. $COP_{r e f}^{C}$ nikalo. Phir pucho: $T_{C} \to 0$ hone par kya hota hai?

Forecast: huge gap ( $297 K$ ) upar tiny cold temperature ke saath → COP tiny hona chahiye, 1 se kaafi neeche. Guess ~0.01?

Climb $= 300 - 3 = 297 K$ . Yeh step kyun? Near-maximal hill curve ke far-right par rehta hai — jahan COP sabse chhoti hoti hai.
$COP_{r e f}^{C} = \frac{3}{297} = 0.01010$ . Yeh step kyun? Har 1 J heat remove karne ke liye ~99 J work pay karna padta hai — extremely expensive cooling.
Limit $T_{C} \to 0 K$ : numerator $\to 0$ , denominator $\to T_{H}$ , isliye $COP_{r e f}^{C} \to 0$ . Yeh step kyun? Absolute zero ke near heat ke last scraps remove karna unbounded work per joule cost karta hai — yeh third law ka fingerprint hai. Yeh figure ki curve ka right end hai.

Verify: $3/297 \approx 0.0101$ , sach mein $≪ 1$ — un rare cases mein se ek jahan ek refrigerator COP 1 se neeche hoti hai ✓ (parent ke "COP usually $> 1$ " se contrast karo). Heat-pump twin $= 0.0101 + 1 = 1.0101$ , abhi bhi 1 se thoda upar ✓.

Cell G — door-open fridge (real-world twist)

Worked example Ex 7 · Kya ek fridge apni door khuli rakh kar kitchen cool kar sakta hai?

Tum ek fridge ko ek sealed kitchen mein door khulke chalta chhod dete ho. Har cycle mein woh (ab khule) andar se $Q_{C} = 300 J$ remove karta hai aur uska motor $W = 90 J$ use karta hai. Kya kitchen cooler, warmer, ya same rehti hai?

Forecast: intuition kehta hai "cold air bahar aata hai → cooler." Physics kehta hai fridge ke peeche dump hone wala $Q_{H}$ dekho. Step 2 padhne se pehle guess karo.

Room mein dump ki gayi heat hai $Q_{H} = Q_{C} + W = 300 + 90 = 390 J$ per cycle. Yeh step kyun? First law phir se — har joule jo fridge remove karti hai plus har joule electricity coils ke peeche heat ban jaati hai.
Lekin door khuli hone ke saath, woh $Q_{C} = 300 J$ jo usne "remove" ki woh usi room se li gayi jisme woh dump karti hai. Sealed kitchen mein har cycle mein net heat added $= Q_{H} - Q_{C} = W = 90 J$ . Yeh step kyun? Cooling aur dumping ab ek connected space mein hoti hai, isliye woh cancel ho jaati hain — sirf paid work chhod kar, jo cancel nahi ho sakti.
Net effect: $+ 90 J$ heat per cycle → kitchen warm ho jaati hai. Yeh step kyun? Ek open-door fridge $90 J$ -per-cycle heater hai. Sirf ek band door (andar tha thanda, room se alag) actually khaana thanda rakhti hai.

Verify: net heating $= W = 90 J > 0$ ✓. Yeh Second Law of Thermodynamics se match karta hai: tum ek device use karke ek closed room cool nahi kar sakte — jo work tum feed karte ho uske liye heat ke alawa koi jagah nahi hai.

Cell H — real fraction of Carnot (exam twist)

Worked example Ex 8 · Mera real fridge ideal se kitna acha hai?

Ek fridge $T_{C} = 270 K$ aur $T_{H} = 300 K$ ke beech operate karta hai. Uska measured $COP_{r e f} = 6.0$ hai. Yeh Carnot maximum ka kitna fraction achieve karta hai? Agar yeh $Q_{C} = 600 J$ per cycle move karta hai, toh woh actually kitna work pay karta hai, aur ek ideal device kitna pay karta?

Forecast: $30 K$ gap ke saath Carnot COP $270/30 = 9$ hai. Toh real wala 6 par ideal ka roughly two-thirds achieve karta hai, aur same cooling ke liye ideal se zyada work pay karta hai.

Carnot ceiling: $COP_{r e f}^{C} = \frac{T _{C}}{T _{H} - T _{C}} = \frac{270}{300 - 270} = \frac{270}{30} = 9.0$ . Yeh step kyun? Yeh in do temperatures ke liye best-possible bar set karta hai; inhi do reservoirs ke beech koi bhi real device is ceiling se neeche COP rakhta hai — isliye measured $6.0$ better be $9.0$ se kam (hai bhi).
Ideal ka fraction $= \frac{COP _{r e f}}{COP _{r e f}^{C}} = \frac{6.0}{9.0} = 0.6667$ , yaani Carnot ka $66.7%$ . Yeh step kyun? Yeh "second-law efficiency" ek-number summary hai ki real machine perfection ke kitni close pahunchi — exam ka favourite quantity.
$Q_{C} = 600 J$ ke liye real work: definition $COP_{r e f} = \frac{Q _{C}}{W}$ ko rearrange karke $W_{r e a l} = \frac{Q _{C}}{COP _{r e f}} = \frac{600}{6.0} = 100 J$ milta hai. Yeh step kyun? Humein actual COP aur actual cooling di gayi hai, isliye definition (Carnot nahi) true electricity paid deti hai.
Same $600 J$ ke liye ideal work: $W_{i d e a l} = \frac{Q _{C}}{COP _{r e f}^{C}} = \frac{600}{9.0} = 66.7 J$ . Yeh step kyun? Ek perfect Carnot device same heat ko kam work mein move karta hai; $W_{r e a l}$ aur $W_{i d e a l}$ ke beech gap woh energy hai jo irreversibility mein waste ho gayi. Carnot Cycle dekho.

Verify: $6 < 9$ isliye real $<$ Carnot ✓ (kabhi violate nahi hua). Real work $100 J >$ ideal $66.7 J$ ✓ — tum imperfection ke liye har cycle mein $33.3 J$ penalty pay karte ho, exactly woh $\frac{1}{0.6667} - 1 = 50%$ extra work jo second-law efficiency predict karta hai.

Recall Matrix par quick self-test

Door khuli fridge ke baare mein konsa cell poochta hai? ::: Cell G — room ka net heating work $W$ ke barabar hota hai. Jab temperature gap zero tak shrink hota hai, Carnot COP kya karta hai? ::: Cell E — yeh $+ \infty$ tak diverge karta hai. Kab ek refrigerator ka COP legitimately 1 se neeche gir sakta hai? ::: Cell F — bahut bade gap ke saath bahut thanda $T_{C}$ (e.g. $T_{C} = 3 K$ , $T_{H} = 300 K$ gives $\approx 0.0101$ ). Jab bhi temperatures °C mein diye gaye hon toh pehla step kya hai? ::: Kelvin mein convert karo (Cell B): $T (K) = T (° C) + 273.15$ .

Connections

Parent: Refrigerators & Heat Pumps — COP — yeh page uska worked-example drill hai.
Carnot Cycle — Cells A, B, E, F, H mein use ki gayi ceiling supply karta hai.
First Law of Thermodynamics — Cells C, D, G mein $Q_{H} = Q_{C} + W$ use kiya gaya.
Second Law of Thermodynamics — kyun COP $\to \infty$ sirf ek limit hai aur kyun ek open-door fridge room cool nahi kar sakta.
Entropy — Cell B mein Celsius trap ke peeche $Q / T$ bookkeeping.
Heat Engines and Efficiency — leverage-ratio COP ko fractional efficiency se contrast karta hai.