Temperature — thermal equilibrium, thermometers, scales: Exercises

1.7.1 · D4 · Physics › Thermodynamics › Temperature — thermal equilibrium, thermometers, scales

Level 1 — Recognition

Exercise 1.1

Ek metal ka doorknob aur ek wooden door ek hi thande kamre mein raat bhar bina chuye rakhe hain. Doorknob tumhare haath ko zyaada thanda lagta hai. Actually mein kaun zyaada temperature par hai?

Recall Solution

Koi bhi nahi — dono same temperature par hain. Dono objects itni der se usi kamre mein hain ki unhone hawa ke saath (aur ek doosre ke saath bhi, Zeroth Law of Thermodynamics ke zariye) thermal equilibrium achieve kar li hai. "Zyaada thanda lagta hai" ka sirf matlab hai ki metal tumhare garam haath se heat zyaada tezi se khinchta hai kyunki woh heat ka achha conductor hai. Feeling rate of heat flow ke baare mein hai, temperature ke baare mein nahi.

Exercise 1.2

Paani freeze hota hai aur paani boil hota hai (1 atm par). Har scale par readings fill karo.

	Ice point	Steam point
Celsius	?	?
Fahrenheit	?	?
Kelvin	?	?

Recall Solution

	Ice point	Steam point
Celsius	$0 ° C$	$100 ° C$
Fahrenheit	$32 ° F$	$212 ° F$
Kelvin	$273.15 K$	$373.15 K$

Kyun: Celsius aur Kelvin dono ice→steam gap ko 100 equal steps mein baantte hain, isliye ek degree dono mein same size ka hai; Kelvin bas origin ko $273.15$ neeche shift kar deta hai. Fahrenheit 180 steps use karta hai, isliye uska degree chhota hota hai.

Level 2 — Application

Exercise 2.1

$t_{C} = 20 ° C$ (ek comfortable room) ko Fahrenheit aur Kelvin mein convert karo.

Recall Solution

Fahrenheit: $t_{F} = \frac{9}{5} (20) + 32 = 36 + 32 = 68 ° F$ . Kelvin: $T = 20 + 273.15 = 293.15 K$ . Fahrenheit ke liye do steps kyun: $\frac{9}{5}$ se multiply karo taaki interval size stretch ho (F degrees chhote hote hain), phir $32$ add karo origin shift karne ke liye (F ice point ko $32$ par rakhta hai, $0$ par nahi).

Exercise 2.2

Ek weather report kehti hai ki $t_{F} = 5 ° F$ hai. Celsius mein convert karo.

Recall Solution

$t_{C} = \frac{5}{9} (t_{F} - 32) = \frac{5}{9} (5 - 32) = \frac{5}{9} (- 27) = - 15 ° C$ . Pehle 32 kyun subtract karte hain: origin shift ko pehle undo karna zaroori hai rescaling se — $- 32$ Fahrenheit ka offset hatata hai, phir $\frac{5}{9}$ interval ko Celsius size par wapas le aata hai.

Exercise 2.3

$T = 300 K$ ko Celsius aur Fahrenheit mein convert karo.

Recall Solution

Celsius: $t_{C} = 300 - 273.15 = 26.85 ° C$ . Fahrenheit: $t_{F} = \frac{9}{5} (26.85) + 32 = 48.33 + 32 = 80.33 ° F$ . Celsius se kyun guzarna padta hai: Fahrenheit formula Celsius ke terms mein likha hai, isliye Kelvin → Celsius → Fahrenheit ek safe two-hop route hai.

Level 3 — Analysis

Exercise 3.1

Ek garam din par ek room $22 ° C$ se $34 ° C$ tak warm hota hai. Is change ko Kelvin aur Fahrenheit degrees mein express karo.

Recall Solution

Celsius change hai $Δ t_{C} = 34 - 22 = 12 ° C$ . Kelvin mein: $Δ T = Δ t_{C} = 12 K$ . Fahrenheit mein: $Δ t_{F} = \frac{9}{5} (12) = 21.6 ° F$ . $+ 273.15$ ya $+ 32$ kyun nahi: ek interval do readings ka difference hota hai. Dono readings same offset carry karti hain, isliye subtract karne par woh cancel ho jaata hai. Sirf slope ( $1$ Kelvin ke liye, $\frac{9}{5}$ Fahrenheit ke liye) bachta hai.

Exercise 3.2

Kin temperature par Celsius aur Fahrenheit scales same number read karte hain?

Recall Solution

$t_{F} = t_{C} = x$ set karo: $x = \frac{9}{5} x + 32 \Rightarrow x - \frac{9}{5} x = 32 \Rightarrow - \frac{4}{5} x = 32 \Rightarrow x = - 40.$ To $- 40 ° C = - 40 ° F$ . Yeh kaisa dikhta hai (figure dekho): $t_{C}$ – $t_{F}$ plot par do straight lines exactly ek baar $(- 40, - 40)$ par cross karti hain.

Figure — Temperature — thermal equilibrium, thermometers, scales

Exercise 3.3

Ek faulty Celsius thermometer melting ice mein $2 ° C$ aur boiling water mein (1 atm par) $102 ° C$ read karta hai. Kisi liquid mein yeh $52 ° C$ read karta hai. Sahi Celsius temperature kya hai?

Recall Solution

Faulty scale phir bhi linear hai, isliye ice se steam tak fraction of the way dono scales par same hona chahiye: $\frac{t _{true} - 0}{100 - 0} = \frac{t _{faulty} - 2}{102 - 2} .$ Plug in $t_{faulty} = 52$ : $t_{true} = 100 \cdot \frac{52 - 2}{100} = 100 \cdot \frac{50}{100} = 50 ° C .$ Ratio method kyun: koi bhi do linear scales apne fixed points ke beech fractions par agree karti hain, isliye un fractions ko equate karna ek ko doosre mein convert kar deta hai bina $a$ aur $b$ ko alag-alag jaane.

Level 4 — Synthesis

Exercise 4.1

Ek constant-volume gas thermometer water ke triple point par ( $T_{tr} = 273.16 K$ ) pressure $P_{tr} = 4.0 \times 1 0^{4} Pa$ read karta hai. Ek hot oil bath mein yeh $P = 5.5 \times 1 0^{4} Pa$ read karta hai. Bath temperature Kelvin aur Celsius mein nikalo.

Recall Solution

Constant volume par absolute (single-fixed-point) definition temperature ko pressure ke proportional banati hai: $T = T_{tr} \frac{P}{P _{tr}} = 273.16 \times \frac{5.5 \times 1 0 ^{4}}{4.0 \times 1 0 ^{4}} = 273.16 \times 1.375 = 375.60 K .$ Celsius mein: $t_{C} = 375.60 - 273.15 = 102.45 ° C$ . Sirf pressure scale kyun karte hain: Ideal Gas Law fixed volume par $P \propto T$ (absolute) banata hai, isliye pressure ratio hi temperature ratio hai. Yahi wajah hai ki gas thermometer true Kelvin scale define karta hai.

Exercise 4.2

Fixed volume dilute gas ka pressure $P = P_{0} (1 + α t_{C})$ follow karta hai jahan $α = \frac{1}{273.15} ° C^{- 1}$ hai. Dikhao ki $P = 0$ tak extrapolate karne par absolute zero milta hai, aur uski value batao.

Recall Solution

$P = 0$ set karo: $0 = P_{0} (1 + α t_{C}) \Rightarrow 1 + α t_{C} = 0 \Rightarrow t_{C} = - \frac{1}{α} = - 273.15 ° C .$ Yeh kaisa dikhta hai (figure dekho): $P$ vs $t_{C}$ ki straight line, axis se peeche extend karne par, $- 273.15 ° C$ par zero pressure hit karti hai. Kyunki ek real gas negative pressure nahi push kar sakti, koi bhi temperature iske neeche nahi ho sakta — yeh absolute zero $= 0 K$ hai.

Level 5 — Mastery

Exercise 5.1

Do gas thermometers, X (constant volume) aur ek liquid thermometer Y, dono $0 ° C$ aur $100 ° C$ par agree karne ke liye calibrate kiye gaye hain. True temperature $50 ° C$ par, thermometer Y $48 ° C$ read karta hai kyunki uska liquid perfectly linearly expand nahi karta. Jabki gas thermometer X $50 ° C$ read karta hai. (a) Kaun sa thermometer "true" absolute scale define karta hai, aur kyun? (b) Is point par Y, X se kitna disagree karta hai?

Recall Solution

(a) Gas thermometer X true scale define karta hai. Saare dilute gases same universal $P$ – $T$ line follow karte hain (yeh kinetic-theory result hai Ideal Gas Law ke peeche), isliye woh sab ek doosre se agree karte hain. Y jaisa liquid sirf apne do fixed points par linear hone ke liye define kiya jaata hai; unke beech uska real expansion thoda curve karta hai, isliye woh drift karta hai. (b) Disagreement $= t_{X} - t_{Y} = 50 - 48 = 2 ° C$ . Y midpoint par $2 ° C$ kam read karta hai. Gases kyun jeette hain: low pressure par molecules bahut kam interact karte hain, isliye behaviour substance-independent ho jaata hai — yahi ek absolute standard ki definition hai.

Exercise 5.2

Ek steel bridge span installation par $t_{C} = 15 ° C$ par hai. Garmiyon mein yeh $45 ° C$ tak pahunchta hai; sardiyon mein, $- 10 ° C$ tak. (a) Installation se summer aur winter changes Kelvin mein do. (b) Yeh changes thermal-expansion formula $Δ L = L_{0} α_{L} Δ T$ mein feed hote hain. Explain karo kyun hum yahan $Δ T$ Kelvin ya $Δ t_{C}$ mein interchangeably use kar sakte hain.

Recall Solution

(a) Summer change: $Δ t_{C} = 45 - 15 = 30 ° C \Rightarrow Δ T = 30 K$ . Winter change: $Δ t_{C} = - 10 - 15 = - 25 ° C \Rightarrow Δ T = - 25 K$ (span contract karta hai). (b) Kyunki ek Kelvin degree aur ek Celsius degree same size ke hain (dono ice→steam ko 100 parts mein split karte hain). Sirf origin $273.15$ se differ karta hai, aur kisi bhi difference mein origin shift cancel ho jaata hai. Isliye $Δ T = Δ t_{C}$ exactly — Thermal Expansion formula dono ke saath identical numbers deta hai. Yeh kyun matter karta hai: agar tumne carelessly absolute temperature $T = 318 K$ daala instead of change $Δ T = 30 K$ , tum expansion ko tenfold overestimate kar dete.

Recall Har use kiye gaye tool ka one-line summary

Convert readings ::: $t_{F} = \frac{9}{5} t_{C} + 32$ , $T = t_{C} + 273.15$ Convert changes ::: $Δ t_{F} = \frac{9}{5} Δ t_{C}$ , $Δ T = Δ t_{C}$ (offsets cancel ho jaate hain) Linear thermometers compare karo ::: fractions equate karo $\frac{t - ice}{steam - ice}$ Gas thermometer ::: $T = T_{tr} P / P_{tr}$ constant volume par Absolute zero ::: $P = P_{0} (1 + α t_{C})$ ko $P = 0$ tak extrapolate karo $\Rightarrow - 273.15 ° C$