Thermal expansion — linear, area, volumetric: Worked examples

1.7.7 · D3 · Physics › Thermodynamics › Thermal expansion — linear, area, volumetric

Yeh page parent topic ke liye workout gym hai. Parent ne teeno formulas atoms se banaye the. Yahan hum ensure karte hain ki koi bhi situation tumhe surprise na kar sake — har sign, har dimension, har degenerate case, har real-world twist ka apna fully worked example hai.

Shuru karne se pehle, ek reminder har symbol ka jo hum use karenge, taaki kuch bhi unexplained na rahe:

Scenario matrix

Is topic ke har problem ko in cells mein se ek (ya blend) mein daala ja sakta hai. Neeche ke examples mein cell label diya gaya hai, aur milke ye poora grid cover karte hain.

Cell	Kya alag hai	Kaun sa trap test karta hai
C1 · Linear, heating	1-D rod, $Δ T > 0$	simple formula
C2 · Sign flip (cooling)	$Δ T < 0$ → object shrinks	yaad rakhna ki length girti hai
C3 · Area / the hole	2-D, ek hole metal ki tarah expand karta hai	sochna ki hole shrink karta hai
C4 · Volume + which coefficient	3-D, $α$ diya hai na ki $γ$	$\times 3$ karna bhool jaana
C5 · Degenerate: $Δ T = 0$	koi temperature change nahi	answer hai "kuch nahi hilda"
C6 · Differential expansion	do materials, stretches ka difference	bimetallic strip / gap fit
C7 · Constrained (no room)	rod expand nahi ho sakti → stress aata hai	strain $\to$ force, links Stress and strain
C8 · Limiting / word problem	real object, engineering decision	sahi dimension choose karna
C9 · Anomaly ( $γ < 0$ )	water 0–4 °C heating pe shrinks	$γ$ ka sign

Worked examples

C1 — Linear expansion, seedha heating

C2 — Sign flip: cooling se shrinkage hoti hai

C3 — Area, aur famous hole

C4 — Volume, coefficient trap

Worked example Thermometer bulb mein mercury

Ek bulb mein $1 0^{\circ} C$ par $0.200 cm^{3}$ mercury hai. $6 0^{\circ} C$ tak warm kiya. Mercury ke volume mein change? $γ_{Hg} = 1.8 \times 1 0^{- 4} K^{- 1}$ (seedha $γ$ diya gaya hai).

Forecast: Kaun sa coefficient use karoge — aur kya yeh $\times 1$ , $\times 2$ , ya $\times 3$ situation hai?

Volume → humein $γ$ chahiye. Yahan $γ$ directly diya gaya hai, toh multiply karne ki zaroorat nahi. Yeh step kyun? Trap yeh habit hai: bahut se problems mein $α$ diya hota hai aur tumhe $\times 3$ karna padta hai. Dhyan se padho — yahan coefficient already volumetric hai.
$Δ T = 50 K$ .
$Δ V = V_{0} γ Δ T = 0.200 \times 1.8 \times 1 0^{- 4} \times 50 = 1.8 \times 1 0^{- 3} cm^{3}$ .

Verify: Naya volume $0.2018 cm^{3}$ ; fractional $0.9%$ . Mercury steel se ~30× zyada volume mein expand karta hai — isliye yeh ek achha thermometer fluid hai. Aur bade expansion ke liye Gas laws dekho. ✓

C5 — Degenerate case: $Δ T = 0$

C6 — Differential expansion (bimetallic idea / fitting a gap)

Worked example Aluminium shaft par steel ring

$2 0^{\circ} C$ par ek aluminium shaft ka diameter $50.00 mm$ hai aur ek steel ring ka inner diameter $49.98 mm$ hai (slide karne ke liye bahut tight). Dono ko equally heat karo. Kis $Δ T$ par diameters just match karenge? $α_{Al} = 2.3 \times 1 0^{- 5}$ , $α_{steel} = 1.2 \times 1 0^{- 5} K^{- 1}$ .

Forecast: Dono ko heat karo — kya gap close hoga ya open? (Kaun zyada teezi se bada hoga?)

Heating ke baad diameters: shaft $50.00 (1 + α_{Al} Δ T)$ , ring $49.98 (1 + α_{steel} Δ T)$ . Yeh step kyun? Har metal apne $α$ se expand karta hai. Ring chhota shuru hoti hai lekin slower expand karti hai — gap actually galat direction mein chauda hota hai. Isliye hume cool karna hoga, ya numbers ko fit-check ke liye padhna hoga.
Unhe equal set karo: $50.00 (1 + 2.3 \times 1 0^{- 5} Δ T) = 49.98 (1 + 1.2 \times 1 0^{- 5} Δ T)$
Expand karo aur $Δ T$ collect karo: $50.00 - 49.98 = Δ T (49.98 \times 1.2 \times 1 0^{- 5} - 50.00 \times 2.3 \times 1 0^{- 5})$ $0.02 = Δ T (5.9976 \times 1 0^{- 4} - 1.15 \times 1 0^{- 3}) = Δ T (- 5.502 \times 1 0^{- 4})$ $Δ T = \frac{0.02}{- 5.502 \times 1 0 ^{- 4}} \approx - 36.4 K$

Verify: $Δ T$ negative hai — tumhe ~36 °C cool karna hoga (lagbhag $- 1 6^{\circ} C$ tak), kyunki zyada teezi se expand hone wala aluminium shaft heating par steel ring se aage nikal jaata hai. Sign ne sahi bataya ki "heat galat direction hai." Yahi differential cell ka poora point hai. ✓

C7 — Constrained rod → thermal stress

Worked example Clamp ki hui steel bar

Ek steel bar $2 0^{\circ} C$ par do rigid walls ke beech clamp ki gayi hai taaki woh lamba na ho sake. Ise $7 0^{\circ} C$ tak heat kiya jaata hai. Compressive stress dhundho. $α_{steel} = 1.2 \times 1 0^{- 5} K^{- 1}$ , Young's modulus $Y = 2.0 \times 1 0^{11} Pa$ .

Forecast: Bade hone ki jagah nahi, toh "wanted expansion" ki energy kahan jaati hai?

Agar free hota, toh $\frac{Δ L}{L _{0}} = α Δ T$ se strain aata. Yeh step kyun? Walls is strain ko zero par force karte hain, isliye wall ko magnitude $α Δ T$ ka ek equal aur opposite compressive strain impose karna padta hai.
Stress $= Y \times strain = Y α Δ T$ (Stress and strain se). Yeh step kyun? Elastic solids mein stress aur strain Young's modulus $Y$ se linked hain.
$σ = 2.0 \times 1 0^{11} \times 1.2 \times 1 0^{- 5} \times 50 = 1.2 \times 1 0^{8} Pa = 120 MPa$ .

Verify: $120 MPa$ ek real, bada stress hai (steel ki yield ~250 MPa ke kareeb) — isliye bridges aur rails ko expansion joints ki zaroorat hoti hai. Dhyan do ki $L_{0}$ cancel ho jaata hai: thermal stress length par depend nahi karta. ✓

C8 — Word/limiting problem: sahi dimension choose karna

Worked example Kaunsa coefficient ek filled tank use karta hai?

Glass ka ek tank, volume $2000 cm^{3}$ , $1 5^{\circ} C$ par glycerine se poora bhara hai. $4 5^{\circ} C$ tak heat kiya. Kitna glycerine overflow hoga? $γ_{glycerine} = 5.3 \times 1 0^{- 4}$ , $α_{glass} = 9 \times 1 0^{- 6} K^{- 1}$ .

Forecast: Kya liquid ka poora expansion overflow karta hai, ya tank ki khud ki growth "room bana" leti hai?

Liquid $Δ V_{liq} = V_{0} γ_{liq} Δ T$ se bada hona chahta hai.
Lekin container ki cavity bhi badi hoti hai, apne volume coefficient $γ_{glass} = 3 α_{glass} = 2.7 \times 1 0^{- 5} K^{- 1}$ se. Yeh step kyun? Cavity ek "hole" hai → yeh 3-D mein solid glass ki tarah expand karti hai. Overflow difference hai (apparent expansion).
$Δ V_{spill} = V_{0} (γ_{liq} - γ_{glass}) Δ T = 2000 (5.3 \times 1 0^{- 4} - 2.7 \times 1 0^{- 5}) (30)$ . $= 2000 \times 5.03 \times 1 0^{- 4} \times 30 = 30.18 cm^{3} .$

Verify: Positive → liquid overflow karta hai (uska $γ$ glass se kaafi zyada hai). Agar (galti se) tank ki growth ignore karte toh $31.8 cm^{3}$ milta — container ka correction ~5% ghatata hai. Difference ka structure C6 jaisa hai. ✓

C9 — Anomaly: negative $γ$

Recap

Recall Kaun si cell kaun si thi?

C1 plain linear ::: rail, $Δ L = 4.32 mm$ C2 cooling / sign flip ::: Al wire $2.875 mm$ shrinks C3 hole (area) ::: washer hole $3.0204 cm^{2}$ tak bada hota hai C4 $γ$ seedha diya gaya ::: mercury $Δ V = 1.8 \times 1 0^{- 3} cm^{3}$ C5 $Δ T = 0$ ::: kuch nahi badalta C6 differential ::: fit ke liye ~36 K cool karna hoga C7 constrained ::: thermal stress $120 MPa$ , length se independent C8 apparent expansion ::: glycerine overflow $30.18 cm^{3}$ C9 anomaly ::: water 0→4 °C par $0.272 cm^{3}$ sikodata hai

Connections

Thermal expansion — linear, area, volumetric — parent jo yahan use hua har formula banaya.
Stress and strain — constrained-rod cell (C7) yahan hai.
Anomalous expansion of water — negative- $γ$ cell (C9).
Gas laws — jahan bahut bada expansion appear hota hai aur $Δ T$ absolute hona chahiye.
Pendulum and simple harmonic motion — ek lambi hoti rod se timing errors.