Temperature — thermal equilibrium, thermometers, scales: Worked examples

1.7.1 · D3 · Physics › Thermodynamics › Temperature — thermal equilibrium, thermometers, scales

Yeh page parent topic ka ek companion drill page hai. Parent ne ideas banaye; yahan hum har tarah ke questions dhundte hain jo temperature conversions aur thermometers throw kar sakte hain — har sign, har degenerate input, har limiting case — aur har ek ko zero se work out karte hain.

The scenario matrix

Is chapter ka har temperature problem inhi case-classes mein se ek hai. Neeche ke chhe examples inhe saare hit karte hain.

Cell	Case class	Tricky kyun hai	Example
A	Positive value, C→F	plain slope + shift	Ex 1
B	Negative value / below zero	signs survive the $\times \frac{9}{5}$	Ex 2
C	Ek difference (interval), reading nahi	offsets ( $+ 32$ , $+ 273.15$ ) cancel hone chahiye	Ex 3
D	Degenerate / crossing point	do lines milti hain — ek equation	Ex 4
E	Gas thermometer, absolute scale	ratio, shift nahi; limiting $P \to 0$	Ex 5
F	Real-world word problem + exam twist	reading vs. difference sahi se choose karo	Ex 6

Case A — ek positive reading, Celsius → Fahrenheit

Worked example Ex 1 · Cell A

Ek pleasant room $t_{C} = 20 ° C$ par hai. Yeh Fahrenheit scale par kya hoga?

Forecast: 32 se zyada (ice point ke upar), aur har Celsius degree $\frac{9}{5}$ Fahrenheit degree ke barabar hai, toh guess karo "high 60s mein kahin."

Step 1 — Reading or Rise? Yeh ek reading hai (ek where), isliye hum full formula use karte hain. Yeh step kyun? Room scale par ek fixed spot par baitha hai; hum koi change measure nahi kar rahe. Offsets andar rehte hain.

Step 2 — Interval size scale karo. $\frac{9}{5} \times 20 = 36$ Yeh step kyun? $\frac{9}{5}$ ice point ke upar gap ka size Celsius degrees se chote Fahrenheit degrees mein convert karta hai.

Step 3 — Origin shift karo. $t_{F} = 36 + 32 = 68 ° F$ Yeh step kyun? Fahrenheit ka ice point $32$ hai, $0$ nahi, isliye hum $32$ add karte hain "ice ke upar degrees" se actual Fahrenheit reading par jaane ke liye.

Verify: Reverse karo: $t_{C} = \frac{5}{9} (68 - 32) = \frac{5}{9} (36) = 20 ° C$ . ✓ Round-trip exactly wapas aa jaata hai.

Case B — ek negative reading, signs survive hone chahiye

Case C — ek temperature difference (yahan log trip karte hain)

Worked example Ex 3 · Cell C

Ek experiment ke dauran ek metal block $Δ t_{C} = 40 ° C$ cool hota hai. Is change ko Kelvin aur Fahrenheit degrees mein express karo.

Forecast: Kelvin mein ek change Celsius change ke exactly barabar hota hai — toh $40$ K. Fahrenheit mein yeh bada hona chahiye (F degrees chote hain), toh guess karo $\approx 72$ .

Step 1 — Reading or Rise? Yeh ek Rise hai (word "by" = ek change). Yeh step kyun? Hume bataya gaya hai temperature kitna move kiya, kahan landa nahi. Offsets cancel hone chahiye.

Step 2 — Kelvin change. $Δ T = Δ t_{C} = 40 K$ Yeh step kyun? Kelvin aur Celsius ka degree size identical hai; $+ 273.15$ ek fixed origin shift hai jo difference lete waqt subtract ho jaata hai: $(T_{2} - T_{1}) = (t_{2} + 273.15) - (t_{1} + 273.15)$ .

Step 3 — Fahrenheit change. $Δ t_{F} = \frac{9}{5} Δ t_{C} = \frac{9}{5} (40) = 72 ° F$ Yeh step kyun? Sirf slope $\frac{9}{5}$ ek difference mein survive karta hai; $+ 32$ usi tarah cancel hota hai. Kisi change mein kabhi $32$ mat add karo.

Verify (lambe tarike se karo offset cancel prove karne ke liye): Maan lo block $t_{C} : 90 \to 50$ gaya. F mein: $90 \to \frac{9}{5} (90) + 32 = 194$ ; $50 \to \frac{9}{5} (50) + 32 = 122$ . Difference $194 - 122 = 72$ . ✓ $+ 32$ subtraction par gayab ho gaya.

Case D — degenerate "same reading" case

Worked example Ex 4 · Cell D

Kya koi aisi temperature hai jahan Celsius aur Fahrenheit thermometers same number dikhayein? Use nikalo.

Forecast: Do lines ke alag slopes hain ( $1$ vs $\frac{9}{5}$ ) isliye woh exactly ek baar cross karti hain — negatives mein kahin, kyunki $0 ° C$ par F already $32$ read karta hai (F aage hai), isliye woh pehle equal rahi hongi, zero se neeche.

Step 1 — "Same reading" ko ek equation mein likho. Common value ko $x$ hone do: $x = \frac{9}{5} x + 32$ Yeh step kyun? "Dono scales par same number" matlab $t_{F} = t_{C} = x$ substitute karo conversion mein — ek unknown, ek equation, do straight lines ka ek crossing point.

Step 2 — $x$ terms collect karo. $x - \frac{9}{5} x = 32 \Rightarrow - \frac{4}{5} x = 32$ Yeh step kyun? Dono $x$ 's ko ek side laana single unknown ko isolate karta hai.

Step 3 — Solve karo. $x = 32 \times (- \frac{5}{4}) = - 40$ Yeh step kyun? $- \frac{4}{5}$ se divide karo (yani $- \frac{5}{4}$ se multiply karo) $x$ ko release karne ke liye.

Verify: $\frac{9}{5} (- 40) + 32 = - 72 + 32 = - 40$ . ✓ Toh $- 40 ° C = - 40 ° F$ — woh ek jagah jahan scales agree karte hain.

Figure — Temperature — thermal equilibrium, thermometers, scales

Figure do conversion lines ko exactly ek point par cross karte hua dikhata hai — $- 40$ par red dot. Uske upar Fahrenheit zyada read karta hai; uske neeche (thanda) Celsius zyada read karta hai.

Case E — gas thermometer, absolute scale, aur limit $P \to 0$

Worked example Ex 5 · Cell E

Ek constant-volume gas thermometer paani ke triple point ( $T_{tr} = 273.16 K$ ) par pressure $P_{tr} = 3.0 \times 1 0^{4} Pa$ read karta hai. Ek hot bath mein yeh $P = 4.5 \times 1 0^{4} Pa$ read karta hai. (a) Bath temperature nikalo. (b) Yeh thermometer absolute zero par kya pressure read karta?

Forecast: Pressure factor $1.5$ se badh gaya, isliye temperature bhi utni hi badhni chahiye: $\approx 410$ K. Aur absolute zero ko $P = 0$ correspond karna chahiye — "absolute" ka matlab yahi hai.

Step 1 — Single-fixed-point absolute law use karo. Constant volume par, $T = T_{tr} \frac{P}{P _{tr}}$ Yeh step kyun? Dilute gas ke liye fixed volume par, pressure directly absolute temperature ke proportional hai. Ek fixed point poori line fix karta hai kyunki line origin se guzarni chahiye. Isliye gas thermometer absolute scale define karta hai — koi shift nahi, sirf ek ratio.

Step 2 — (a) plug in karo. $T = 273.16 \times \frac{4.5 \times 1 0 ^{4}}{3.0 \times 1 0 ^{4}} = 273.16 \times 1.5 = 409.74 K$ Yeh step kyun? Pressure ratio $1.5$ hai, aur temperature identically scale hoti hai kyunki relation ek pure proportion hai.

Step 3 — (b) ka jawab, limit. $T \to 0$ set karo: $0 = T_{tr} \frac{P}{P _{tr}} \Rightarrow P = 0 Pa$ Yeh step kyun? Absolute zero defined hai us temperature ke roop mein jahan extrapolated gas pressure zero hit karta hai. Ek gas negative pressure se push nahi kar sakti, isliye yeh hard floor hai — tum isse thanda nahi ja sakte.

Verify (units + sanity): $K \times \frac{Pa}{Pa} = K$ ✓ (pressure ratio dimensionless hai). Aur $409.74 K > 273.16 K$ , correct hai kyunki bath triple point se zyada hot hai. ✓

Figure pressure ko absolute temperature ke against plot karta hai: ek straight line origin se guzarti hai. Use $P = 0$ tak extend karne par exactly $T = 0$ par utarti hai — woh intercept hai absolute zero.

Case F — real-world word problem exam twist ke saath

Worked example Ex 6 · Cell F

Ek weather report kehta hai aaj ka high $95 ° F$ hai aur aaj raat ka low $59 ° F$ hai. (a) High ko Celsius mein convert karo. (b) Ek physics exam poochta hai: "Temperature raat bhar kitne Celsius degrees girti hai?" Ek student har reading convert karta hai aur subtract karta hai. Shortcut dikhao aur confirm karo ki woh match karta hai.

Forecast: $95 ° F$ garm hai — Celsius mein mid-30s ke aas paas. Drop $95 - 59 = 36$ F-degrees hai; kyunki C-degrees bade hain, Celsius drop 36 se chota hona chahiye — guess karo $\approx 20$ .

Step 1 — High convert karo (ek Reading). $t_{C} = \frac{5}{9} (95 - 32) = \frac{5}{9} (63) = 35 ° C$ Yeh step kyun? Ek reading full inverse formula use karta hai: $32$ origin subtract karo, phir $\frac{5}{9}$ se Celsius-sized degrees mein scale down karo.

Step 2 — Overnight drop ek Rise (change) hai, isliye sirf slope use karo. $Δ t_{F} = 95 - 59 = 36 ° F, Δ t_{C} = \frac{5}{9} Δ t_{F} = \frac{5}{9} (36) = 20 ° C$ Yeh step kyun? Yeh exam twist hai: ek difference, isliye $+ 32$ / $- 32$ cancel hota hai. Interval ko sirf $\frac{5}{9}$ se convert karo.

Step 3 — Lambe tarike se confirm karo (dono convert karo, subtract karo). $59 ° F \to \frac{5}{9} (59 - 32) = \frac{5}{9} (27) = 15 ° C; 35 - 15 = 20 ° C$ Yeh step kyun? Har reading convert karke phir subtract karna shortcut se same answer dena chahiye — "offsets cancel" ka poora point yahi hai.

Verify: Shortcut $= 20 ° C$ , lambe tarike se $= 20 ° C$ . ✓ Aur high wapas convert hota hai: $\frac{9}{5} (35) + 32 = 63 + 32 = 95 ° F$ . ✓

Quick self-test

Recall Ek

change ke liye kaunsa formula, aur kyun? Celsius→Fahrenheit change ke liye, sirf $Δ t_{F} = \frac{9}{5} Δ t_{C}$ use karo ::: kyunki $+ 32$ ek origin shift hai jo do readings subtract karne par cancel ho jaata hai.

Recall Celsius aur Fahrenheit same number kahan read karte hain?

$- 40$ par ::: $x = \frac{9}{5} x + 32$ solve karne par $x = - 40$ milta hai, isliye $- 40 ° C = - 40 ° F$ .

Recall Gas thermometer ratio use karta hai, shift kyun nahi?

Kyunki constant volume par $P \propto T$ origin se guzarti line ke saath ::: isliye ek single fixed point poora absolute (Kelvin) scale set karta hai.

Recall Body temp

$37 ° C$ Fahrenheit mein? $98.6 ° F$ ::: from $\frac{9}{5} (37) + 32 = 66.6 + 32$ .

Dekho bhi: Zeroth Law of Thermodynamics, Heat and Internal Energy, Thermal Expansion (mercury/gas thermometers isi par rely karte hain), aur Ideal Gas Law gas thermometer ke $P \propto T$ foundation ke liye.