Refrigerators and heat pumps — COP: Question bank

1.7.20 · D5 · Physics › Thermodynamics › Refrigerators and heat pumps — COP

Neeche har jagah use hone wale symbols ka reminder (sab positive magnitudes per cycle):

$Q_{C}$ = cold reservoir (temperature $T_{C}$ ) se bahar nikali gayi heat,
$Q_{H}$ = hot reservoir (temperature $T_{H}$ ) mein daali gayi heat,
$W$ = tumhara diya hua work, jahan $Q_{H} = Q_{C} + W$ ,
COP = "benefit ÷ cost" — fridge ke liye cooling $Q_{C} / W$ , heat pump ke liye heating $Q_{H} / W$ .
In formulas mein temperatures kelvin mein hoti hain: $T (K) = T (° C) + 273$ .

True or false — justify karo

Khuli door wala fridge kamre ko time ke saath thanda karta hai.

False. Wo

Q_{H} = Q_{C} + W

heat kamre mein dump karta hai jabki saamne wali hawa se sirf

Q_{C}

remove karta hai; net effect kamre ko exactly

W

jitna garm karna hai — wohi work jo motor spend karti hai.

COP

ek se kam ho sakta hai.

True fridge ke liye — agar

Q_{C} < W

ho (bada temperature hill), toh

COP_{r e f} = Q_{C} / W < 1

. Lekin heat pump kabhi 1 se neeche nahin ja sakta kyunki

COP_{h p} = COP_{r e f} + 1 \geq 1

Fixed

T_{H}

aur

T_{C}

ke liye, ek real refrigerator Carnot se zyada COP rakh sakta hai.

False. Carnot value

T_{C} / (T_{H} - T_{C})

woh ceiling hai jo second law set karta hai; koi bhi real device entropy generate karta hai, isliye uska COP strictly lower hota hai.

Ek hi hardware se bana heat pump aur refrigerator hamesha exactly 1 ke difference wala COP rakhte hain.

True. Dono ek hi

W

share karte hain; unke numerators ka difference

Q_{H} - Q_{C} = W

hai, isliye

COP_{h p} - COP_{r e f} = W / W = 1

exactly hota hai — machine kitni bhi ideal ho, is par depend nahin karta.

Jab

T_{C} \to T_{H}

(bahut chhota gap), maximum COP infinity tak jata hai.

True. Denominator

T_{H} - T_{C} \to 0

hota hai, isliye

COP^{C a r n o t} \to \infty

— almost koi hill nahin toh almost koi work nahi lagti. Ye kabhi actually infinity nahin pahunchta kyunki bilkul zero gap mein heat usefully move nahin hoti.

Work input

W

badhana hamesha

Q_{C}

badhata hai.

Ek general law ke taur par False — ye device par depend karta hai. Zyada

W

zyada heat move kar sakta hai, lekin ek fixed real machine ke liye COP

= Q_{C} / W

roughly temperatures ki property hai, isliye

Q_{C}

sirf

W

ke saath tab scale karta hai jab COP fixed rahe; zyada spend karke Carnot ceiling cross nahin kar sakte.

Heat pump, 1:1 electric resistance heater se behtar hai.

True. Heater

W

ko exactly

W

heat mein badalta hai (COP = 1). Heat pump

Q_{H} = Q_{C} + W > W

deliver karta hai, kyunki wo bahar se free heat

Q_{C}

bhi kheenchta hai.

Agar

Q_{C} = 0

ho (cold se kuch bhi extract nahin), toh machine ek perfect heater hai.

Aadha sach —

Q_{C} = 0

ke saath

Q_{H} = W

milta hai, yaani sara work heat ban jaata hai aur

COP_{h p} = 1

hota hai. Ye sirf ek resistance heater hai; "heat pump advantage" khatam ho gaya kyunki kuch bhi uphill pump nahin ho raha.

Error dhundho

"COP 1 se kam hona chahiye kyunki engine ki efficiency 1 se kam hoti hai."

Error ye hai ki COP ko efficiency ki tarah treat kiya ja raha hai. Efficiency useful work out ÷ heat in hoti hai (converted energy ka fraction); COP heat moved ÷ work paid hai — ek leverage ratio jo usually 1 se zyada hota hai.

W = Q_{C} - Q_{H}

, isliye work = jo heat li minus jo dump ki."

Sign error. Device cold se liye se zyada hot dump karta hai, isliye

Q_{H} > Q_{C}

aur surplus paid work hai:

W = Q_{H} - Q_{C} > 0

Q_{C} - Q_{H}

likhne se negative "work" milta hai, jo ek driven pump ke liye impossible hai.

"4 °C par aur 27 °C room mein fridge ka Carnot COP

4/ (27 - 4) \approx 0.17

hai."

Galat scale.

Q / T

entropy bookkeeping ko absolute temperature chahiye. Convert karo:

277/ (300 - 277) \approx 12

, na ki

0.17

. Celsius ka zero arbitrary hai, isliye °C ke ratios yahan meaningless hain.

"Ek reversible fridge ka

Q_{C} / Q_{H} = T_{C} / T_{H}

hota hai, isliye energy conservation violate hoti hai kyunki

Q_{C} < Q_{H}

Koi violation nahin —

Q_{C} < Q_{H}

zaroori hai: difference

Q_{H} - Q_{C} = W

wohi work hai jo tumne pay ki, exactly jaisa first law demand karta hai. Reversibility ratio fix karta hai; first law missing

W

supply karta hai.

"Refrigerator aur heat pump same machine hain, isliye unka COP same hoga."

Hardware same hai, lekin benefit alag hai. Fridge

Q_{C}

(cooling) ki parwah karta hai, pump

Q_{H}

(heating) ki — alag numerators milte hain — isliye

COP_{h p} = COP_{r e f} + 1

, kabhi equal nahin.

"Kyunki real fridge ke liye

Δ S_{u ni v er se} = 0

hai, hum directly

Q_{C} / T_{C} = Q_{H} / T_{H}

use kar sakte hain."

Sirf reversible fridge ka

Δ S_{u ni v er se} = 0

hota hai. Real fridge entropy generate karta hai (

Δ S_{u ni v er se} > 0

), isliye

Q_{H} / Q_{C} > T_{H} / T_{C}

— hot par zyada heat dump hoti hai per unit cold heat li gayi — matlab real device work waste karta hai aur COP lower hota hai.

Why questions

Second law zero-work wala fridge kyun forbid karta hai?

Zero-work heat transfer cold se hot mein entropy

Q_{C} / T_{C}

cold body se le jaata aur sirf

Q_{C} / T_{H}

(chhota) hot par dump karta, universe ki entropy ghataata — forbidden. Tumhe

W

pay karna padta hai taaki extra dumped heat entropy balance carry kare.

Dono Carnot COPs ka denominator same

T_{H} - T_{C}

kyun hota hai?

Dono COPs (heat of interest)

/ W

ke barabar hain, aur reversible cycle ke liye

W = Q_{H} - Q_{C}

jo temperature relation ko

(T_{H} - T_{C})

ke factor mein badal deta hai. "Climb"

T_{H} - T_{C}

common cost hai; sirf top (kaun si heat tumhe chahiye) badalta hai.

Bada temperature gap fridge ko worse (lower COP) kyun banata hai?

Bada gap matlab zyada steep hill jiske upar heat push karni hai. Reversibility force karta hai ki per unit li gayi heat zyada dump ho, isliye zyada work "climb par kharch" hoti hai — denominator

T_{H} - T_{C}

badhta hai jabki numerator ghatta hai, COP girta hai.

Hum

Q_{H}

Q_{C}

W

ko positive magnitudes kyun treat kar sakte hain bina sign confusion ke?

Kyunki humne already physically har arrow ki direction fix kar li hai (cold in, hot out, work in) energy diagram mein. Ek baar direction haath se set ho jaaye, toh algebra

Q_{H} = Q_{C} + W

sirf sizes use karta hai, competing sign convention se bacha jaata hai.

COP_{h p}

ka hard floor 1 kyun hai lekin

COP_{r e f}

ka nahin?

COP_{h p} = Q_{H} / W = (Q_{C} + W) / W = COP_{r e f} + 1 \geq 1

; chahe kuch pump na kare (

Q_{C} = 0

), work khud delivered heat ban jaati hai. Fridge sirf

Q_{C}

count karta hai, jo

W

se neeche ja sakta hai, isliye uska COP 1 se kam ho sakta hai.

Heat pump ka advantage mild weather mein sabse bada aur extreme cold mein sabse chhota kyun hota hai?

COP

T_{H} / (T_{H} - T_{C})

hai; warmer bahar (bada

T_{C}

, chhota gap) denominator tiny banata hai aur COP huge. Extreme cold mein gap badh jaata hai, COP 1 ki taraf girta hai, aur pump ek simple heater se thoda hi behtar rehta hai.

Edge cases

T_{C} \to 0

K (cold reservoir absolute zero ke paas) par

COP_{r e f}

ka kya hoga?

COP_{r e f} = T_{C} / (T_{H} - T_{C}) \to 0

. Absolute zero ke paas kisi cheez se heat ke aakhri bits nikalna enormous work maangta hai — third law ki ek echo: finite work se 0 K nahin pahuncha ja sakta.

Agar

W = 0

ho lekin

T_{C} < T_{H}

, toh model kya predict karta hai?

W = 0

ke saath,

COP_{r e f} = Q_{C} /0

undefined hai (division by zero). Physically, zero work input wala driven fridge uphill pump hi nahin kar sakta — yeh "infinite COP" exactly wahi forbidden second-law case hai.

Kya Carnot heat pump ka

COP_{h p} = 1

exactly ho sakta hai?

Sirf degenerate limit

T_{C} = 0

K mein, jo

T_{H} / (T_{H} - 0) = 1

deta hai. Kisi bhi real cold reservoir ke liye jahan

T_{C} > 0

ho, Carnot heat pump 1 se zyada hota hai — wo hamesha kuch free outside heat kheechta hai.

T_{C} = T_{H}

(koi gap nahin) ke instant par fridge kya kar raha hai?

Carnot COP infinity tak blow up karta hai, matlab infinitesimal work heat move karti hai. Lekin koi temperature difference na ho toh heat flow ka koi reason hi nahin tha — ye woh idealized boundary hai jahan "cost" vanish hoti hai kyunki task khud vanish ho jaata hai.

Agar koi real device COP exactly Carnot value report kare, toh kya true hona chahiye?

Wo reversible hona chahiye — zero entropy generate karta ho, heat sirf infinitesimal temperature differences ke across exchange karta ho, aur infinitely slowly run karta ho. Finite speed aur finite gaps wale real machines hamesha short fall karte hain.

Heat pump ke liye, kya

Q_{C}

(cold se kheenchi gayi heat)

Q_{H}

(hot par deliver ki gayi heat) se zyada ho sakti hai?

Nahin. Kyunki

Q_{H} = Q_{C} + W

jahan

W \geq 0

hai, hamesha

Q_{H} \geq Q_{C}

hota hai. Deliver ki gayi heat kam se kam extracted heat ke barabar hoti hai, equality sirf tab hoti hai jab

W = 0

ho.

Recall Har trap ka one-line summary

Baar baar aane wale jhooth: (1) COP ek efficiency hai (nahin — ye leverage hai, aksar >1); (2) Celsius use karo (nahin — kelvin, kyunki $Q / T$ ko absolute zero chahiye); (3) fridge aur heat pump COPs equal hain (nahin — exactly 1 se differ karte hain); (4) work zero ya negative ho sakti hai (nahin — $W = Q_{H} - Q_{C} > 0$ ); (5) bada gap better hai (nahin — climb $T_{H} - T_{C}$ cost hai).

Connections

Refrigerators and heat pumps — COP — parent note jise ye traps stress-test karte hain.
Heat Engines and Efficiency — COP (leverage) ko efficiency (fraction) se contrast karo.
Carnot Cycle — reversible ceiling ka source jise har trap reference karta hai.
Second Law of Thermodynamics — woh law jo "no free uphill heat" ke peeche hai.
Entropy — $Q / T$ bookkeeping jis par sign traps rely karte hain.
First Law of Thermodynamics — $Q_{H} = Q_{C} + W$ jo sign errors ko khatam karta hai.