Thermal Properties of Matter: Class 11 Physics Chapter 10 Hinglish mein

Physics Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 22 Oct 2026 · Updated 22 Oct 2026

120 min
1 simulation
29 practice questions
35 MCQs

Is chapter mein aap kya seekhenge

Temperature aur heat mein farak samajhna, aur temperatures ko Celsius, Fahrenheit aur Kelvin scales ke beech convert karna
PV = μRT ki madad se samjhao ki constant-volume gas thermometer kaise absolute zero aur Kelvin scale tak le jaata hai
Linear, area aur volume expansion aur thermal stress calculate karo, aur paani ka anomalous expansion samjhao
Specific heat capacity, molar specific heat aur latent heat se calorimetry ke problems solve karo, changes of state ke saath bhi
Heating curves aur phase diagrams padho, aur samjhao ki pressure melting aur boiling points ko kaise khiskata hai
Ek rod aur jude hue rods mein conduction se steady heat flow calculate karo, aur roz ki zindagi mein convection aur radiation samjhao
Garam bodies aur stars par Wien's displacement law aur Stefan–Boltzmann law lagao
Newton's law of cooling batao, uska exponential form derive karo aur usse cooling times ka andaaza lagao

NCERT Physics Class 11, Chapter 10, pp. 202–225 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

10.1 Introduction

Ek mechanic steel ke ring gear ko flywheel par fit karta hai: woh ring ko itna garam karta hai ki woh flywheel par sarak jaaye, phir thanda hone deta hai taaki woh kaske pakad le. Chennai ke Marina beach par dopahar ki hawa samundar se aati hai, lekin der raat ko woh samundar ki taraf jaati hai. Pressure cooker Kochi se kahin zyada Leh mein kaam aata hai, aur steam ka ek jhonka ubalte paani ki chheent se zyada jala sakta hai.

Har paheli is par tiki hai ki jab temperature difference ki wajah se energy andar ya bahar flow karti hai to matter kaise respond karta hai. Yeh chapter dikhata hai ki temperature kaise naapa jaata hai, matter kaise expand hota hai, garam karne aur changes of state mein kitni energy lagti hai, heat kaise travel karti hai, aur chai ka cup kitni jaldi thanda hota hai.

10.2 Temperature aur heat

Nahane ke paani mein ungli dalo aur turant pata chal jaata hai ki woh garam hai ya thanda, lekin tum 38 °C aur 41 °C mein farak nahi bata sakte, aur pighle iron ya liquid nitrogen ko haath se test karna to mushkil hi hai. Garam aur thande ka hamara ehsaas aasaani se dhokha bhi kha jaata hai, jaisa neeche wali activity dikhati hai. Physics iski jagah temperature (ताप) use karti hai: ek number, jo kisi instrument se ek tay scale par padha jaata hai, aur batata hai ki body kitni garam hai.

Try karke dekho

Kya tum apni ungliyon par bharosa kar sakte ho?

Tumhe chahiye

teen bowls
ice cubes
nal ka paani
gungunaa paani (garam nahi)

Pehle guess karo

Beech wale bowl ka nal ka paani har haath ko kaisa lagega?

Dono haathon ko ek jaisaLeft haath ko garam, right ko thandaLeft haath ko thanda, right ko garamDono haathon ko thanda

Ek bowl mein paani aur kuch ice cubes bharo, doosre mein nal ka paani, aur teesre mein itna gungunaa paani jise chhoona aaraamdeh ho (lagbhag 40 °C).
Apna left haath thande bowl mein aur right haath garam bowl mein ek minute ke liye rakho.
Ab dono haath ek saath beech wale bowl mein dalo.

May ki ek dopahar ko abhi-abhi ubla doodh ka pan aur lunch bag ka ek frozen gel pack kitchen ki ek hi shelf par rakho. Doodh thanda hota hai, pack pighal kar garam hota hai, aur ek-do ghante baad dono room temperature par aa jaate hain (Fig. 10.1). Energy doodh se room mein, aur room se pack mein flow hui hai. Jo energy kisi system aur uske surroundings ke beech unke temperatures alag hone ki wajah se pass hoti hai, use heat (ऊष्मा) kehte hain. Jab temperatures barabar ho jaate hain, net flow ruk jaata hai, aur hum kehte hain ki bodies thermal equilibrium mein hain.

2/3Heat doodh se nikalti hai aur thande pack mein jaati hai.

Fig. 10.1Heat hamesha garam body se thandi body ki taraf flow karti hai, aur temperatures barabar hone par ruk jaati hai.

95 degrees Celsius par doodh ka pan aur minus 5 degrees par frozen gel pack 32 degrees wale room mein ek shelf par rakhe hain. Step 1 shuruaati temperatures dikhata hai. Step 2 arrows jodta hai: heat doodh se room mein aur room se thande pack mein flow karti hai. Step 3 dono ko 32 degrees par dikhata hai, bina arrows ke, thermal equilibrium mein.

Heat transit mein energy hai, isliye uski SI unit joule (J) hai. Koi body heat contain nahi karti, bilkul waise hi jaise bank account mein deposits 'rakhe' nahi hote: body mein internal energy hoti hai, aur heat use badalne ka ek tareeka hai. Jo energy kisi aur wajah se boundary cross karti hai, jaise paani ko mathne wale stirrer ka kiya hua work, woh heat nahi hai (Chapter 11 ise aage badhata hai). Temperature ki SI unit kelvin (K) hai; roz ki zindagi mein hum degree Celsius (°C) bhi use karte hain. Baaki chapter yeh dekhta hai ki heat matter ke saath kya karti hai: use garam karti hai, phulaati hai, aur pighlaati ya ubaalti hai.

10.3 Temperature ka measurement

Digital fever thermometer ek chhote bead ka electrical resistance padhta hai jo temperature ke saath badalta hai; laboratory thermometer rangeen alcohol ke thread ki length padhta hai; infrared 'gun' surface se nikalne wala radiation padhti hai. Har ek kisi thermometric property par tika hai, yaani aisi measurable quantity jo temperature ke saath lagataar aur dohraane layak tareeke se badalti hai. Liquid-in-glass thermometer ke liye liquid, aksar mercury ya alcohol, lagbhag linearly expand hona chahiye, taaki thread par barabar steps ka matlab temperature ke barabar steps ho.

Thread ki length ko number mein badalne ke liye scale chahiye, aur scale ko do fixed points chahiye: aisi physical ghatnaayein jo har baar ek hi fixed temperature par hon aur jinhe aasaani se dohraaya ja sake. Traditional jodi hai ice point, jahan pure ice aur paani saath rehte hain, aur steam point, jahan pure paani ubalta hai, dono standard atmospheric pressure par. Celsius scale inhe 0 °C aur 100 °C kehta hai aur beech ke gap ko 100 degrees mein baantta hai. Fahrenheit scale inhe 32 °F aur 212 °F kehta hai, isliye wahi gap 180 degrees ka hota hai (Fig. 10.2).

2/4Paani 100 °C ya 212 °F par ubalta hai: 100 Celsius degrees 180 Fahrenheit degrees ke barabar hain.

Fig. 10.2Ek temperature, teen numbers. Celsius degree ka size kelvin ke barabar hai aur Fahrenheit degree ka 1.8 guna.NCERT Fig. 10.4 · p. 204

Celsius, Fahrenheit aur Kelvin scales ke teen parallel rulers, aise aligned ki barabar temperatures ek ke upar ek hon, ice point se pehle ek break ke saath. Step 1: ice point, 0 °C, 32 °F, 273.15 K. Step 2: steam point, 100 °C, 212 °F, 373.15 K, brackets dikhate hain ki dono points ke beech 100 aur 180 degrees hain. Step 3: body temperature, 37 °C, 98.6 °F, 310.15 K. Step 4: absolute zero, −273.15 °C, −459.67 °F, 0 K.

Dono scales linear hain, isliye koi temperature har scale par ice point se steam point tak ke raaste ka same fraction hota hai:

Celsius–Fahrenheit conversion

\frac{t_{F} - 32}{180} = \frac{t_{C}}{100}

(10.1)

Rearrange karne par, $t_{F} = \frac{9}{5} t_{C} + 32.$ $t_{F}$ aur $t_{C}$ ka graph ek seedhi line hai jiska slope $9 / 5$ hai aur jo $t_{F}$ axis ko 32 par kaatti hai. Yaad rakho ki changes sirf slope se convert hote hain: 10 Celsius degrees ki badhat 18 Fahrenheit degrees ki badhat hai.

Example

Bukhaar padhna

Ek digital thermometer 102.2 °F dikhata hai. (a) Yeh degrees Celsius mein kitna hai? (b) Kis temperature par Celsius aur Fahrenheit thermometer same number dikhate hain?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Chai ke cup ka temperature 20 Celsius degrees girta hai. Yeh kitne Fahrenheit degrees girta hai?

20683611

Answer dekho

Answer (C) 36

Differences sirf slope se convert hote hain:

20 \times 9 / 5 = 36.

32 jodna (68 paane ke liye) change ki jagah temperature convert karne wali galti hai.

10.4 Ideal-gas equation aur absolute temperature

Ek mercury thermometer aur ek alcohol thermometer ko 0 °C aur 100 °C par calibrate karo, phir dono ko us paani mein dalo jise mercury wala 50 °C batata hai. Alcohol wala thoda alag padhta hai, kyunki dono liquids bilkul proportional tareeke se expand nahi hote, aur kisi ka bhi sahi hone ka daava behtar nahi hai. Gases is sawaal ko suljhaati hain. Kam density par hydrogen, helium aur nitrogen sab temperature par ek hi tarah respond karti hain, isliye unse bane thermometers aapas mein agree karte hain, aur gas natural standard ban jaati hai.

Gas ki fixed maatra ke liye teen quantities uski state batati hain: pressure $P$ , volume $V$ aur temperature. 1662 mein Robert Boyle ne bataya ki steady temperature par product $P V$ constant rehta hai: Boyle's law. 1780s mein Jacques Charles ne paaya ki steady pressure par volume temperature ke saath-saath badhta hai, yaani $V / T$ constant hai, basharte $T$ sahi zero se gina jaaye (woh hum neeche dhoondhenge): Charles' law. Dono milkar kisi diye gaye sample ke liye $P V / T$ ko constant banate hain. Gas ki maatra double karne se $P V / T$ double hota hai, isliye yeh moles ki sankhya $μ$ ke proportional hai:

Ideal-gas equation

P V = μ R T

(10.2)

jahan

$μ$: gas ke moles ki sankhya
$R$: universal gas constant, $8.31 J m o l^{- 1} K^{- 1}$
$T$: absolute temperature, kelvin mein

Real gases kam density par Eq. (10.2) ko kaafi kareeb se follow karti hain; jo gas ise bilkul theek follow kare use ideal gas kehte hain. Volume fixed rakho to equation $P \propto T$ deti hai: band gas ka pressure thermometric property ban jaata hai. Yahi constant-volume gas thermometer ka principle hai.

Aise thermometer ke liye pressure aur Celsius temperature ka graph banao to ek seedhi line milti hai (Fig. 10.3). Alag gas ya alag maatra alag slope deti hai. Real gases ka pressure zero ke aas-paas pahunchne se bahut pehle hi woh liquid ban jaati hain, lekin agar har line ko neeche ki taraf badhao to sab ek hi temperature, −273.15 °C, par zero pressure tak pahunchti hain. Wahan ideal gas koi pressure nahi daalegi, aur isse kam temperature ka koi matlab nahi banta. Is limit ko absolute zero kehte hain.

3/3Har line −273.15 °C par zero pressure tak pahunchti hai: absolute zero.

Fig. 10.3Alag-alag kam density wale gas samples ki lines, badhane par, sab −273.15 °C par zero pressure tak pahunchti hain.NCERT Fig. 10.3 · p. 204

Pressure aur Celsius temperature. Step 1: ek gas sample ke 0 aur 100 °C ke beech naape gaye points ek seedhi line par hain. Step 2: do aur samples alag slope wali seedhi lines dete hain. Step 3: teeno lines, dashed lines ki tarah badhaane par, ek hi point, −273.15 °C, par zero pressure tak pahunchti hain, jise absolute zero mark kiya gaya hai.

Lord Kelvin (William Thomson) ne ek aisa scale suggest kiya jiska zero absolute zero hai aur jiski degree ka size Celsius degree ke barabar hai. Is par temperatures kelvin mein likhe jaate hain, bina degree sign ke:

Kelvin aur Celsius

T = t_{C} + 273.15

(10.3)

Temperature difference ki value kelvin aur Celsius degrees mein same hoti hai, isliye $J k g^{- 1} K^{- 1}$ jaisi unit Celsius changes ke saath kaam karti hai. Lekin $P V = μ R T$ mein, aur har us law mein jo $T$ se multiply ya divide karta hai, tumhe kelvin hi use karna hoga.

Example

Gas thermometer padhna

Constant-volume helium thermometer ka bulb paani ke triple-point cell mein, $273.16 K$ par, $50.0 k P a$ pressure dikhata hai. Ubalte paani ke upar ki steam mein yeh $68.3 k P a$ dikhata hai. Steam ka temperature kelvin aur degrees Celsius mein nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Kya aap jaante hain?NCERT Points to ponder 1 · p. 223

273.16 kyun, aur 273.15 kyun?

Ice aur steam points pressure ke saath aur ghuli hui hawa ya salts ke saath khisak jaate hain, isliye modern thermometry paani ke triple point ko pasand karti hai: woh akela temperature aur pressure jahan ice, liquid paani aur water vapour saath rehte hain. 1954 se 2019 tak kelvin ko is point ki value theek 273.16 K maan kar define kiya gaya tha. Celsius temperature $t_{C} = T - 273.15$ se define hota hai, jisse triple point 0.01 °C par aata hai. Tab 1 atm par ice lagbhag theek 0 °C par pighalti hai, aur paani lagbhag 99.97 °C par ubalta hai, ab bilkul 0 aur 100 nahi.

2019 se kelvin ko iske badle Boltzmann constant ki value se fix kiya gaya hai, $k_{B} = 1.380649 \times 10^{- 23} J / K .$ Paani ka triple point ab ek naapi jaane wali quantity hai: ab bhi 273.16 K, lagbhag ek kelvin ke das-hazaarve hisse tak.

Quick check

2 mein se 1

Constant volume par rakhi gas ka pressure 27 °C par

1.20 \times 10^{5} P a

hai. Kis Celsius temperature par uska pressure

1.60 \times 10^{5} P a

hoga?

°C

Answer dekho

Answer 127 °C

Kelvin use karo:

T_{2} = 300 K \times 1.60 / 1.20 = 400 K,

yaani

127^{\circ} C .

27 °C ko seedha scale karne se galat 36 °C milta.

10.5 Thermal expansion

Poles ke beech lage bijli ke cables December ki subah se zyada May ki dopahar mein saaf taur par jhukte hain. Concrete ki sadkein alag-alag slabs mein banaayi jaati hain, jinke beech patle gaps ek naram sealant se bhare hote hain. Fridge se nikle mote glass tumbler mein seedha ubalta paani daalo to woh crack ho sakta hai, kyunki uski andar ki surface bahar ki surface ke garam hone se pehle hi phool jaati hai. Zyadatar substances garam karne par expand aur thanda karne par contract hote hain; temperature ke saath size ke is badlaav ko thermal expansion kehte hain.

Length ke badlaav ko hum linear expansion, surface ke badlaav ko area expansion aur volume ke badlaav ko volume expansion kehte hain (Fig. 10.4). $l$ length wali rod ko chhote $Δ T$ se garam karne par experiments dikhate hain ki length ka fractional change $Δ T$ ke proportional hota hai:

Fig. 10.4Linear, area aur volume expansion (bahut badha-chadha kar). Dashed outlines thande sizes dikhati hain.NCERT Fig. 10.5 · p. 205

Teen panels jinmein badhat badha-chadha kar dikhaayi gayi hai: ek rod jo apni dashed thandi outline se delta l lambi hai, ek square tile jo dono directions mein delta A badhi hai, aur ek cube jo teeno directions mein delta V badha hai.

Linear expansion

\frac{Δ l}{l} = α_{l} Δ T

(10.4)

Constant $α_{l}$ , yaani coefficient of linear expansion ya linear expansivity, material par depend karta hai aur iski unit $K^{- 1}$ hai. Table 10.1 mein typical values di gayi hain. Utne hi garam karne par copper Pyrex glass se lagbhag paanch guna zyada expand hota hai, isiliye Pyrex dishes fridge se seedha oven mein jaane par bhi bach jaati hain. Invar, ek iron–nickel alloy, lagbhag bilkul nahi badalta, isliye ise measuring tapes aur precision instruments mein use kiya jaata hai.

Table 10.1Room temperature ke aas-paas coefficients of linear expansion ki typical values

Material	$α_{l}$ $(10^{- 5} K^{- 1})$
Aluminium	2.3
Brass	1.9
Copper	1.7
Steel	1.2
Concrete	1.2
Khidki ka glass	0.9
Pyrex glass	0.33
Invar	0.12

Volume ke liye hum coefficient of volume expansion ko isi tarah define karte hain:

Volume expansion

\frac{Δ V}{V} = α_{V} Δ T

(10.5)

$α_{l}$ ki tarah, $α_{V}$ bhi material ki property hai, lekin yeh bilkul constant nahi hai. Copper jaise solid ke liye yeh bahut kam temperatures par bahut chhota hota hai, solid ke garam hone ke saath badhta hai aur sirf high temperatures par jaakar sthir hota hai. Roz ki ranges mein hum dono coefficients ko constants maante hain. Table 10.2 dikhata hai ki liquids solids se kahin zyada expand hote hain. Ethanol mercury se lagbhag chhah guna zyada expand hota hai, isiliye alcohol thermometer ka thread har degree par itna zyada khiskata hai.

Table 10.220 °C ke aas-paas coefficients of volume expansion ki typical values

Substance	$α_{V}$ $(K^{- 1})$
Steel	$3.6 \times 10^{- 5}$
Khidki ka glass	$2.7 \times 10^{- 5}$
Mercury	$1.8 \times 10^{- 4}$
Paani	$2.1 \times 10^{- 4}$
Glycerine	$5.0 \times 10^{- 4}$
Petrol	$9.5 \times 10^{- 4}$
Ethanol	$1.1 \times 10^{- 3}$
Hawa (constant pressure)	$3.4 \times 10^{- 3}$

Paani: sabse alag

Paani mashhoor exception hai. Iski density lagbhag 4 °C par sabse zyada hoti hai (theek-theek 3.98 °C par, 999.97 kg m⁻³), isliye 0 °C aur 4 °C ke beech paani garam hone par sikudta hai aur thanda hone par phoolta hai. Jamne par volume mein lagbhag 9 % ka aur jump aata hai, isiliye ice tairti hai aur freezer mein rakhi bhari glass bottle crack ho jaati hai. Dono anokhi baatein hydrogen bonds se aati hain, jo ice ke molecules ko ek khule framework, jismein kaafi khaali jagah hoti hai, mein lock kar dete hain; is framework ke tukde jamne se thoda upar ke paani mein bhi dikhne lagte hain.

Socho, sardi shuru hote waqt Ladakh ki ek jheel ka kya hota hai (Fig. 10.5). Density tay karti hai ki kaun sa paani upar rahega. Jab tak surface 4 °C se garam hai, thanda hone se woh bhaari hota hai, isliye neeche doobta hai aur jheel tab tak ulat-palat hoti hai jab tak poora paani 4 °C ke aas-paas na aa jaaye. Aur thanda hua paani ab neeche ke 4 °C wale paani se halka hai, isliye sabse thandi layer upar tairti hai, 0 °C tak pahunchti hai aur jam kar ek dhakkan ban jaati hai. Ice heat ko kam conduct karti hai, isliye yeh dhakkan aage ki cooling ko dheema kar deta hai, aur jheel ka tal poori sardi 4 °C ke aas-paas rehta hai, aur machhliyan zinda rehti hain. Agar koi liquid thanda hone par lagataar dense hota jaata, to woh tal se upar ki taraf poora jam jaata.

4/4Upar ice banti hai aur neeche ke 4 °C wale paani ko insulate karti hai.

Fig. 10.5Paani 4 °C ke aas-paas sabse dense hota hai, isliye jheel upar se jamti hai aur uske tal ka paani liquid rehta hai.NCERT Fig. 10.7 · p. 206

Ek jheel ka cross-section, saath mein ek chhota density curve jo 4 °C par peak karta hai. Step 1: 10 °C wala surface ka paani thanda hokar doobta hai jabki garam paani upar aata hai. Step 2: poori jheel 4 °C par hai. Step 3: 4 °C se neeche wali thandi layer upar rehti hai. Step 4: ice jheel ko dhak leti hai, theek uske neeche ka paani 0 °C par hai aur tal ka paani, ek machhli ke saath, 4 °C par hai.

Gases

Gases solids ya liquids se kahin zyada expand hoti hain, aur unki expansivity temperature par bahut depend karti hai. Ideal gas ke liye yeh seedha gas equation se nikalti hai.

DerivationIdeal gas ki volume expansivity

Pressure fixed rakho aur temperature ko $Δ T$ se badalne do. Gas equation $P V = μ R T$ se milta hai $P Δ V = μ R Δ T .$ Ise $P V = μ R T$ se divide karo:

\frac{Δ V}{V} = \frac{Δ T}{T}

Eq. (10.5) se compare karne par:

Constant pressure par ideal gas

α_{V} = \frac{1}{T}

(10.6)

0 °C (273 K) par isse milta hai $3.7 \times 10^{- 3} K^{- 1},$ aur room temperature (lagbhag 300 K) par $3.3 \times 10^{- 3} K^{- 1} .$ Yeh paani ya mercury ki value ka lagbhag pandrah guna aur steel ki value ka lagbhag sau guna hai. Liquid ke ulat, gas ki expansivity garam hone par saaf taur par kam hoti hai.

Volume aur linear expansion ko jodna

Derivation

α_{V} = 3 α_{l}

kyun

$l$ side wala ek cube lo jo isotropic solid ka bana hai, yaani jo har direction mein barabar expand hota hai. Ise $Δ T$ se garam karne par har edge lambi hoti hai, $Δ l = α_{l} l Δ T,$ isliye

\begin{aligned} Δ V & = (l + Δ l)^{3} - l^{3} \\ ≃ 3 l^{2} Δ l \end{aligned}

(10.7)

Poora expansion hai $3 l^{2} Δ l + 3 l (Δ l)^{2} + (Δ l)^{3} .$ Bahut zyada garam karne par bhi $Δ l / l$ sirf lagbhag $10^{- 3}$ hota hai, isliye aakhri do terms pehle wale se kam se kam hazaar guna chhote hain, aur hum unhe chhod dete hain. $V = l^{3}$ se divide karne par:

\frac{Δ V}{V} = 3 \frac{Δ l}{l} = 3 α_{l} Δ T

(10.8)

α_{V} = 3 α_{l}

(10.9)

Patli sheet ke liye yahi reasoning area expansivity $2 α_{l}$ deti hai, jaisa Example 10.1 dikhata hai.

Example 10.1NCERT Example 10.1 · p. 207

Steel tile ki area expansivity

Ek patli rectangular steel tile ki sides $a$ aur $b$ hain (Fig. 10.6). Dikhao ki iska coefficient of area expansion, $(Δ A / A) / Δ T,$ $2 α_{l}$ hai.

Fig. 10.6Extra area do patli strips aur ek chhota corner square hai.NCERT Fig. 10.8 · p. 207

a aur b sides wala ek rectangle uske expansion ko badha-chadha kar banaya gaya hai. delta b chaudi ek patli strip upar ki taraf chalti hai, jiska area a guna delta b hai; delta a chaudi ek patli strip right side ke neeche chalti hai, jiska area b guna delta a hai; aur delta a guna delta b area wala ek chhota square upar right corner mein hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.2NCERT Example 10.2 · p. 207

Ring gear ko shrink-fit karna

Ek steel ring gear ka inner diameter $30^{\circ} C$ par $299.70 m m$ hai. Ise $300.00 m m$ diameter wale flywheel rim par sarakna hai. Ring ko kis temperature tak garam karna hoga? Steel ke liye $α_{l} = 1.2 \times 10^{- 5} K^{- 1}$ lo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Thermal stress

Ab expansion ko rok do. Ek steel rod ko do na hilne wali walls ke beech fix karo aur garam karo (Fig. 10.7). Walls ko use theek utni length peechhe dhakelni padegi jitni woh badhti, isliye rod mein compressive strain aata hai: $Δ l / l = α_{l} Δ T .$ Hooke's law (Chapter 8) se rod mein thermal stress hai

Thermal stress

\frac{F}{A} = Y α_{l} Δ T

jahan

$Y$: material ka Young's modulus

Stress sirf material aur temperature change par depend karta hai, rod ki length ya motaai par nahi. Force, $F = A Y α_{l} Δ T,$ cross-section ke saath badhta hai aur bahut bada ho sakta hai. Isiliye pehle railway lines rails ke beech gaps chhod kar bichhaayi jaati thi, isiliye aajkal ki welded rails ek dhyaan se chune gaye temperature par bichhaayi jaati hain aur bhaari concrete sleepers se baandhi jaati hain, aur isiliye bridges rollers aur expansion joints par tike hote hain.

3/3Walls use peechhe dhakelti hain: compressive stress $Y α_{l} Δ T .$

Fig. 10.7Clamp ki hui rod badh nahi sakti, isliye walls use dabaati hain: thermal stress

Y α_{l} Δ T .

Ek steel rod do rigid walls ke beech theek fit hoti hai. Step 1: thandi, koi force nahi. Step 2: garam, ek dashed ghost dikhata hai ki yeh right wall ke paar delta l badhti. Step 3: walls badhat ko rokti hain aur dono ends par andar ki taraf forces F lagaati hain, jisse compressive stress Y alpha delta T aata hai.

Example

Bridge girder mein stress

$120 c m^{2}$ cross-section wala ek steel girder sardi ki subah dono ends par rigidly fix kiya jaata hai. Dopahar tak woh $25 K$ garam ho jaata hai. Thermal stress aur har support par force nikalo. Yeh values lo: $Y = 2.0 \times 10^{11} P a$ aur $α_{l} = 1.2 \times 10^{- 5} K^{- 1} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Ek steel plate mein ek circular hole hai. Plate ko garam karne par hole ka diameter

kam hota hai, kyunki steel hole ke andar expand hota haibadhta hai, jaise hole steel se bhara hosame rehta haisirf tab badhta hai jab plate patli ho

Answer dekho

Answer (B) badhta hai, jaise hole steel se bhara ho

Thermal expansion plate ki har length ko ek hi factor se badhata hai, hole ka kinaara bhi shaamil hai. Steel "andar ki taraf" nahi behta.

10.6 Specific heat capacity

Khaali steel ka tawa gas par ek minute mein itna garam kyun ho jaata hai ki chhoo na sako, jabki bagal wale burner par paani ka pan sirf gungunaa hone mein kai minute leta hai? Kisi cheez ko garam karne ke liye zaroori heat sirf flame par depend nahi karti. Steady rate se energy dene wala heater tumhe sirf time naap kar heat ki maatraon ko compare karne deta hai.

ActivityNCERT 10.6 · p. 208

Zaroori heat kaun tay karta hai?

Tumhe chahiye

ek immersion heater ya chhota hot plate
ek steel ka bowl
ek thermometer
ek stopwatch ya phone timer
paani aur coconut oil
ek kitchen scale

Pehle guess karo

Usi heater se 200 g paani ko 10 °C garam hone mein 2 minute lagte hain. 400 g ko 10 °C garam hone mein lagbhag kitna time lagega?

1 minute2 minute4 minute8 minute

Bowl mein 200 g paani dalo, uska temperature note karo, heater on karo aur time naapo ki paani ko 10 °C garam hone mein kitna time lagta hai. Garam hote waqt dheere-dheere hilaate raho.
Same temperature ke taaze paani se phir shuru karo aur 20 °C rise ka time naapo.
400 g paani aur 10 °C rise ke saath dohraao.
200 g coconut oil aur 10 °C rise ke saath dohraao, pehle bowl ko saaf karke sukha lo.

Table 10.3Usi heater ke saath heating times

Run	Liquid aur mass	Rise (°C)
1	paani, 200 g	10
2	paani, 200 g	20
3	paani, 400 g	10
4	coconut oil, 200 g	10

Tumhari entries apne aap save hoti hain.

Fig. 10.8Same heater, alag samples: zyada mass ya bada specific heat matlab dheema rise.

Ek steady heater ke liye temperature rise ko heating time ke against plot kiya gaya hai. Teen seedhi lines origin se shuru hoti hain. Sabse steep 200 g coconut oil ki hai, jo 200 g paani se lagbhag do guna tezi se badhti hai. Beech wali line 200 g paani ki hai. Sabse kam dhalaan wali 400 g paani ki hai, jo 200 g paani se aadhi tezi se badhti hai.

To teen cheezein matter karti hain: kitna material hai ( $m$ ), use kitna garam kiya gaya ( $Δ T$ ) aur woh kis cheez ka bana hai. Kisi body ki heat capacity $S$ woh heat hai jo use har kelvin garam hone ke liye chahiye:

Heat capacity

S = \frac{Δ Q}{Δ T}

(10.10)

Heat capacity kisi object ki hoti hai: bhari bucket ki cup se zyada hoti hai. Material ki property paane ke liye mass se divide karo. Specific heat capacity $s$ woh heat hai jo prati kilogram prati kelvin chahiye:

Specific heat capacity

s = \frac{S}{m} = \frac{1}{m} \frac{Δ Q}{Δ T}

(10.11)

Yaani $Δ Q = m s Δ T,$ jo tab tak sahi hai jab tak substance ek hi state mein rahe. $s$ ki SI unit hai $J k g^{- 1} K^{- 1} .$ Yeh substance par aur thoda uske temperature par depend karti hai.

Definition

Specific heat capacity

Kisi substance ke unit mass ka temperature, bina change of state ke, ek kelvin badhaane ke liye zaroori heat; $s = Δ Q / (m Δ T) .$

ExamplePaani: lagbhag $4186 J k g^{- 1} K^{- 1} .$

Jab substance ki maatra moles $μ$ mein gini jaati hai, to prati mole prati kelvin zaroori heat molar specific heat capacity $C$ hai, $J m o l^{- 1} K^{- 1}$ mein:

Molar specific heat capacity

C = \frac{1}{μ} \frac{Δ Q}{Δ T}

(10.12)

Gas ke liye answer is par depend karta hai ki tum use kaise garam karte ho. Rigid container mein garam karne par gas saari heat apni internal energy badhaane mein lagaati hai; isse $C_{V}$ milta hai, constant volume par molar specific heat. Constant pressure par garam karne par woh expand bhi hoti hai aur apne surroundings ko peechhe dhakelti hai, isliye use har kelvin ke liye zyada heat chahiye; isse $C_{P}$ milta hai. Table 10.5 mein har gas ke liye $C_{P} - C_{V}$ lagbhag $8.3 J m o l^{- 1} K^{- 1},$ yaani gas constant $R$ ke barabar aata hai. Chapter 11 samjhata hai kyun.

Table 10.4Room temperature aur 1 atm ke aas-paas specific heat capacities

Substance	$s$ $(J k g^{- 1} K^{- 1})$
Paani	4186
Ethanol	2440
Ice (0 °C ke aas-paas)	2100
Aluminium	900
Glass	840
Sookhi ret	830
Iron	450
Copper	385
Mercury	140
Lead	128

Table 10.5Room temperature ke aas-paas kuch gases ki molar specific heats (J mol⁻¹ K⁻¹)

Gas	$C_{P}$	$C_{V}$	$C_{P} - C_{V}$
Helium	20.8	12.5	8.3
Argon	20.8	12.5	8.3
Nitrogen	29.1	20.8	8.3
Oxygen	29.4	21.1	8.3
Carbon dioxide	37.1	28.8	8.3

Samundar kinaare Mumbai ke ek din ko Thar desert ke Jaisalmer ke ek din se compare karo. Mumbai mein raat dopahar se mushkil se thandi hoti hai; Jaisalmer mein temperature aksar 15 °C se zyada badal jaata hai. Farak samundar ka hai. Paani ka specific heat ret ya chattaan ka lagbhag paanch guna hai, isliye Arabian Sea din ki dhoop ko temperature mein thodi si badhat ke saath sokh leta hai aur raat bhar dheere-dheere heat waapas deta hai, jabki sookhi ret jaldi garam aur jaldi thandi hoti hai. Yahi property paani ko car engines ka coolant aur rubber ki hot-water bottle ki filling banaati hai.

Example

Nahane ka paani garam karna

Ek $2.0 k W$ electric geyser $15 L$ paani ko $18^{\circ} C$ se $45^{\circ} C$ tak garam karta hai. Heat losses ko ignore karo, to kitna time lagega?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Copper aur aluminium ke barabar masses ko same maatra mein heat milti hai. Kaun zyada garam hota hai?

Aluminium, kyunki woh kam dense haiDono barabar garam hote hain, kyunki masses barabar hainAluminium, kyunki woh behtar conduct karta haiCopper, kyunki uska specific heat chhota hai

Answer dekho

Answer (D) Copper, kyunki uska specific heat chhota hai

Δ T = Δ Q / (m s) .

Copper

(385 J k g^{- 1} K^{- 1})

s

aluminium (

900

) se chhota hai, isliye woh do guna se bhi zyada garam hota hai. Barabar masses ka matlab barabar garmaahat nahi hai.

10.7 Calorimetry

Kisi system ko itna achhe se lapeto ki koi heat andar ya bahar na ja sake, to hum use isolated kehte hain. Energy usse bahar nahi ja sakti, isliye jab uske parts ke beech heat flow karti hai, to garam parts ka khoya har joule thande parts mein mil jaata hai. Calorimetry isi hisaab-kitaab par tiki hai, jise method of mixtures bhi kehte hain:

Principle of calorimetry

garam parts ki khoi heat = thande parts ki mili heat

Calorimetry ka matlab hai heat ka measurement, aur calorimeter iske liye use hone wala bartan hai (Fig. 10.9). Har part ka ek kaam hai:

Copper ya aluminium ka patla cup. Yeh khud kam heat leta hai aur jaldi apne andar ki cheezon ke temperature par aa jaata hai.
Usi metal ka stirrer, taaki paani, cup aur koi bhi sample ek common temperature par pahunch jaayein.
Ek outer jacket, jiske gap mein glass wool ya foam bhara hota hai, taaki room se andar aane ya bahar jaane wali heat ka leak dheema ho.
Ek dhakkan jismein thermometer ke liye ek chhed hota hai.

Cup apne andar ki cheezon ke saath garam hota hai, isliye use bhi ginna padta hai. $m_{c}$ mass aur $s_{c}$ specific heat wala cup prati kelvin utni heat sokhta hai jitni $m_{c} s_{c} / s_{w}$ mass ka paani sokhta; yeh mass cup ka water equivalent hai. Asal mein kuch heat phir bhi room mein leak hoti hai, isliye dhyaan se kiye gaye experiments jaldi kiye jaate hain aur room temperature se thoda neeche se shuru hote hain.

2/3Heat block se paani aur cup mein flow karti hai.

Fig. 10.9Agar koi heat bahar na nikle, to garam block ki khoi heat paani aur cup ki mili heat ke barabar hoti hai.

Insulated jacket ke andar paani ka ek metal cup, stirrer aur thermometer ke saath. Step 1: 98 °C par ek metal block 24 °C wale paani ke upar latka hai. Step 2: block paani mein hai aur arrows dikhate hain ki heat usse paani aur cup mein flow karti hai. Step 3: sab kuch steady 28 °C padhta hai.

Example 10.3NCERT Example 10.3 · p. 209

Kisi metal ka specific heat nikalna

Ek $0.120 k g$ metal cylinder ko water bath mein $98.0^{\circ} C$ tak garam karke jaldi se $0.080 k g$ mass wale aluminium calorimeter mein utaara jaata hai, jismein $0.200 k g$ paani hai, jiska temperature hai $24.0^{\circ} C .$ Final steady temperature hai $28.0^{\circ} C .$ Metal ka specific heat nikalo. Lo $s_{w} = 4186$ aur $s_{A l} = 900 J k g^{- 1} K^{- 1} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

2 mein se 1

Negligible heat capacity wale insulated flask mein 80 °C par

200 g

paani ko 20 °C par

300 g

paani ke saath milaaya jaata hai. Final temperature, °C mein, kya hai?

°C

Answer dekho

Answer 44 °C

Khoi heat = mili heat:

200 (80 - T) = 300 (T - 20),

isliye

500 T = 22 000

aur

T = 44^{\circ} C .

Yeh 20 °C ke zyada kareeb hai kyunki thanda paani zyada hai.

10.8 Change of state

Nimbu-paani ke glass mein ice cubes paani ban jaate hain; kettle ka paani steam ban jaata hai; sardi ki subah tumhari saans ka water vapour waapas chhoti-chhoti boondon mein badal jaata hai. Har ek change of state hai, yaani solid, liquid aur gas ke beech badlaav, aur har ek tab hota hai jab substance apne surroundings se heat leta hai ya unhe heat deta hai. Changes of state mein ek baat common hai jise tum khud check kar sakte ho.

ActivityNCERT Fig. 10.9 · p. 210

Ice ko minute-dar-minute garam karna

Tumhe chahiye

kuchli hui ice
stand par ek steel ya heat-proof glass ka beaker
−10 °C se 110 °C tak padhne wala thermometer
ek hot plate ya dheemi flame
ek stirrer aur ek ghadi

Pehle guess karo

Jab tak beaker mein ice aur paani ka mixture hai, aur heater on hai, thermometer

lagataar badhega0 °C ke aas-paas rahegagirega, kyunki pighalne mein heat lagti haipehle se tezi se badhega

Beaker ko seedha freezer se nikali kuchli hui ice se bharo aur uska temperature note karo. Yeh 0 °C se neeche ho sakta hai.
Ise dheere aur steady garam karo, lagataar hilaate raho, aur har minute thermometer padho.
Tab tak chalu rakho jab tak paani kuch minute ubal na le. Note karo ki aakhri ice kab gaayab hoti hai aur ubalna kab shuru hota hai.
Temperature ko time ke against plot karo.

Table 10.6Temperature readings

Reading	Time (min)	Temperature (°C)	Ice bachi?
1
2
3
4
5
6

Tumhari entries apne aap save hoti hain.

Abhi nahi kar sakte? Simulation try karo · Fig. 10.10

Pehle naam: solid se liquid melting (ya fusion) hai, aur liquid se solid freezing. Tumhara graph dikhata hai ki inmein khaas kya hai. Flat step ka matlab hai ki jab tak ice aur paani dono maujood hain, woh ek fixed temperature, melting point, par thermal equilibrium mein saath rehte hain. Har pure substance ka apna melting point hota hai. Kyunki yeh pressure ke saath thoda khisakta hai, data tables normal melting point dete hain, jo is pressure par naapa jaata hai: $1 a t m = 1.013 \times 10^{5} P a .$

Equilibrium mein do states ka temperature aur pressure same hota hai; unke beech jo alag hai woh density hai, yaani prati mole volume. 0 °C par paani mein tairti ice roz ki misaal hai: same temperature, same pressure, lekin ice lagbhag 9 % kam dense hai.

SimulationIce se steam, particle-dar-particle

Band container mein paani ke molecules, particles ki tarah bane hue, ek thermometer aur ek temperature slider ke saath jo sample ko garam ya thanda karta hai. Ice ke particles lattice mein vibrate karte hain, liquid ke particles ek doosre ke paas se phisalte hain aur steam ke particles azaadi se udte hain. Jab sample pighalta ya ubalta hai, thermometer 0 °C ya 100 °C par ruka rehta hai aur ek label batata hai ki kitne percent ki state badal chuki hai.

Fig. 10.10Sample ko garam ya thanda karne ke liye temperature slider khiskao, aur dekho ki har change of state par thermometer kaise rukta hai.

Try karo

−20 °C wali ice se shuru karo aur slider ko 30 °C par set karo. Thermometer kahan rukta hai, aur intezaar ke dauraan label kya batata hai?
Ab 120 °C set karo. 100 °C par ruke time ko 0 °C par ruke time se compare karo. Kis change of state ko zyada heat chahiye?
120 °C se waapas 50 °C tak thanda karo. Neeche aate waqt 100 °C par kya hota hai?

Pressure melting points ko badalta hai, aur ice ke liye yeh asar 'ulta' chalta hai. Ice paani se kam dense hai, isliye use dabaane se liquid ko fayda hota hai aur melting point kam hota hai, har extra atmosphere par lagbhag 0.007 °C. Ek badhiya experiment iska result dikhata hai.

ActivityNCERT Fig. 10.10 · p. 210

Ek wire jo ice ko kaatti hai lekin use saabut chhod deti hai

Tumhe chahiye

lagbhag 20 cm lamba ice ka block (loaf tin mein paani jamao)
do eentein ya stools
lagbhag 60 cm patli copper ya steel wire
do bhari hui 2 L paani ki bottles

Pehle guess karo

Lagbhag ek ghante baad, sabse zyada chance hai ki wire

ab bhi block ke upar tiki hogiblock ke neeche hogi, aur block do tukdon mein kat chuka hogablock ke neeche hogi, aur block ab bhi ek hi tukde mein hogablock ke beech mein jam chuki hogi

Ice block ko do supports ke aar-paar rakho, uska beech ka hissa bina support ke rahe.
Wire ko block ke beech ke upar latkao, har end par ek bhari bottle baandh kar.
Ise kisi thandi, chhaayadaar jagah par chhod do aur har das minute mein dekho.

Wire dheere-dheere ice mein dhansti hai aur neeche se nikal jaati hai, phir bhi block ek hi tukde mein rehta hai. Patli wire ke neeche pressure zyada hai, isliye theek neeche ki ice pighal jaati hai, jabki woh 0 °C se thodi neeche hai. Pighla paani wire ke upar ki taraf ghoom kar pahunchta hai, jahan pressure phir normal hai, aur phir jam jaata hai, aur latent heat chhodta hai jise wire neeche conduct karke agli layer pighlaati hai. Pressure mein pighalna aur pressure hatne par phir jamna regelation (पुनर्हिमायन) hai. Copper wire steel wali se tez kaam karti hai kyunki woh us heat ko behtar conduct karti hai; dhaaga to mushkil se kaam karta hai.

2/3Wire ke neeche ice pighalti hai aur uske upar phir jam jaati hai.

Fig. 10.11Pressure wire ke neeche ki ice pighlaata hai; paani uske upar phir jam jaata hai.NCERT Fig. 10.10 · p. 211

Ice ka ek block do supports par rakha hai aur uske upar ek wire hai, jiske dono ends par wazan bandha hai. Step 1: wire upar tiki hai. Step 2: ek close-up circle wire ke neeche high pressure, uske chaaron taraf behti pighle paani ki patli film, aur uske upar phir jamna dikhata hai. Step 3: wire block ke neeche nikal chuki hai, block saabut hai, aur jahan se wire guzri wahan ek halki line hai.

Ice skating ko aksar isi asar se samjhaaya jaata hai, ki blade pressure se pighle paani par phisalta hai. Patle blade par skater ka weight melting point ko kam to karta hai, lekin bahut thoda, jabki skating −20 °C par bhi achhe se hoti hai. Measurements dikhate hain ki phislan wali layer ka zyadatar hissa friction se aayi heat se aur ice ki surface par ek patli film se aata hai jo 0 °C se neeche bhi liquid jaisi hoti hai.

Garam karna jaari rakho to 100 °C ke aas-paas doosra flat step aata hai. Ab liquid vapour mein badal raha hai, jise vaporisation kehte hain; boiling mein vapour ke bubbles liquid ke andar hi bante hain. Liquid aur vapour ek fixed temperature, boiling point, par equilibrium mein saath rehte hain, jab tak liquid khatam na ho jaaye.

DemonstrationNCERT Fig. 10.11 · p. 212

Jab chaaho tab ubalna

Tumhe chahiye

ek round-bottomed borosilicate flask, aadha paani se bhara
ek burner, stand aur clamp
do chhed wala stopper, jismein thermometer aur rubber tubing wali chhoti outlet tube ho
ek saada stopper
barf jaise thande paani ka jug

Pehle guess karo

Step 3 mein, 85 °C paani wale band, ulte flask ke chaaron taraf barf jaisa thanda towel lapeta jaata hai. Andar ka paani

shaant rahega, kyunki woh thanda ho raha haiphir se ubalne lagegaupar se jam jaayegaturant flask ko crack kar dega

Flask ko garam karo. Pehle ghuli hui hawa ke chhote bubbles dikhte hain. Phir neeche bade steam bubbles bante hain, lekin thande paani mein upar uthte hue woh sikud kar gaayab ho jaate hain. Aakhir mein steam bubbles surface tak pahunchte hain aur paani ubalta hai, thermometer 100 °C ke aas-paas hota hai.
Outlet tube ko sirf kuch seconds ke liye dabaa kar band karo, bubbling dekho, aur turant chhod do.
Heat band karo aur paani ko lagbhag 85 °C tak thanda hone do. Stopper ki jagah saada stopper lagao, flask ko clamp mein ulta karo, aur uske chaaron taraf barf jaise thande paani mein bhiga towel lapeto.

Outlet dabaane se steam band ho jaati hai aur pressure badhta hai, aur ubalna ruk jaata hai: zyada pressure par paani ko ubalne ke liye zyada temperature chahiye. Ulte flask ko thanda karne se andar ka vapour condense hota hai, pressure girta hai, aur 85 °C wala paani phir ubalta hai. Boiling point pressure badhne par badhta hai aur pressure girne par girta hai. Yeh bhi dekho ki flask ke andar ki steam dikhaayi nahi deti; outlet par safed badal chhoti boondein hain jo steam ke hawa mein thanda hone par banti hain.

Kisi se bhi poochho jisne Leh mein rajma pakaaya ho. Lagbhag 3500 m upar hawa ka pressure sea level ki value ka lagbhag do-tihaai hota hai aur paani lagbhag 88 °C par ubalta hai, isliye dal ghanton pakti rehti hai. Pressure cooker iska ulta karta hai: uska wazan wala valve andar ka pressure atmospheric ka lagbhag do guna rakhta hai, paani lagbhag 120 °C par ubalta hai, aur khaana bahut kam time mein pak jaata hai. 1 atm par boiling point normal boiling point hai.

Kuch solids liquid stage ko poori tarah chhod dete hain. Khule puja box mein rakha camphor (कपूर) dheere-dheere gaayab ho jaata hai, aur solid carbon dioxide, dry ice, −78.5 °C par seedha gas ban jaata hai; uske aas-paas ka safed kohra hawa ka water vapour hai jo thand mein condense ho raha hai. Solid se seedha vapour mein badlaav sublimation hai; iodine aur naphthalene balls bhi sublime hoti hain.

Kya aap jaante hain?NCERT Box: Triple point · p. 211

Phase diagrams aur triple point

Pressure ko temperature ke against plot karo aur mark karo ki kahan kaun si state stable hai, to tumhe phase diagram milta hai (Fig. 10.12). Teen curves ise solid, liquid aur vapour regions mein baantte hain. Fusion curve par solid aur liquid saath rehte hain, vaporisation curve par liquid aur vapour saath rehte hain, aur sublimation curve par solid aur vapour saath rehte hain. Teeno curves triple point par milte hain, jo akela aisa temperature aur pressure hai jahan teeno states ek saath equilibrium mein hoti hain. Paani ke liye yeh 273.16 K (0.01 °C) aur 611.66 Pa hai, lagbhag 0.006 atm. Vaporisation curve critical point par khatam hota hai, paani ke liye 374 °C aur 218 atm, jiske aage liquid aur vapour mein farak nahi kiya ja sakta.

Paani ka fusion curve left ki taraf jhukta hai: zyada pressure, kam melting point, yaani regelation wala asar. Carbon dioxide ka triple point −56.6 °C aur 5.11 atm par hai. Kyunki yeh pressure 1 atm se zyada hai, khuli hawa mein liquid carbon dioxide nahi reh sakta, aur dry ice pighalne ki jagah sublime hoti hai.

3/4Kam pressure, kam boiling point: Leh 88 °C, pressure cooker 120 °C.

Fig. 10.12Paani ka phase diagram (scale ke hisaab se nahi). Har curve aisi conditions ka set hai jahan do states saath rehti hain.NCERT Box: Triple point · p. 211

Paani ka schematic phase diagram, pressure aur temperature. Step 1: fusion, vaporisation aur sublimation curves solid, liquid aur vapour regions ko alag karte hain; vaporisation curve critical point par khatam hota hai. Step 2: triple point jahan woh milte hain, 273.16 K aur 611.66 Pa. Step 3: vaporisation curve par points, 0.65 atm aur 88 °C, 1 atm aur 100 °C, aur 2 atm aur 120 °C. Step 4: left ki taraf jhuke fusion curve ke paas ek upar wala arrow: zyada pressure, kam melting point.

10.8.1 Latent heat

Fig. 10.13 ke flat steps ko phir se dekho: heater ab bhi energy daal raha hai, phir bhi thermometer nahi hilta. Woh energy molecules ko rearrange karne mein jaati hai, solid ke framework ko dheela karne ya liquid ke molecules ko alag kheenchne mein, aur use latent heat kehte hain, Latin ke 'hidden' (chhupa hua) se. Change of state ke liye zaroori heat badalne wale mass ke proportional hoti hai:

Latent heat

Q = m L

(10.13)

jahan

$L$: us change ke liye substance ki latent heat, $J k g^{- 1}$ mein

Definition

Latent heat

Jab koi substance constant temperature aur pressure par state badalta hai, to prati unit mass absorb ya release hone wali heat; $L = Q / m$ .

Example0 °C par 1 kg ice pighlaane mein lagte hain $3.34 \times 10^{5} J .$

Latent heat $L$ pressure par depend karti hai aur 1 atm par di jaati hai. Solid se liquid ke liye yeh latent heat of fusion $L_{f}$ hai; liquid se vapour ke liye latent heat of vaporisation $L_{v}$ . Change ulta hone par utni hi heat waapas nikalti hai: jamta paani prati kilogram $L_{f}$ chhodta hai aur condense hoti steam $L_{v}$ chhodti hai. Table 10.7 kuch substances ki values deta hai.

5/5100 °C ke upar steam garam hoti hai; uska slope phir steep hai.

Fig. 10.13Paani ka heating curve (scale ke hisaab se nahi). Flat parts changes of state hain; slopes

1 / (m s)

hain.NCERT Fig. 10.12 · p. 213

Paani ko di gayi heat ke against temperature, scale ke hisaab se nahi. Step 1: ek steep line jab ice −20 °C se 0 °C tak garam hoti hai. Step 2: 0 °C par ek flat step jab woh pighalti hai. Step 3: kam steep line jab paani 100 °C tak garam hota hai. Step 4: 100 °C par ek lamba flat step jab woh ubalta hai. Step 5: ek steep line jab steam 100 °C se upar garam hoti hai.

Temperature aur di gayi heat ke graph par flat steps changes of state hain aur dhalaan wale parts ka slope hai $Δ T / Δ Q = 1 / (m s) .$ Ice aur steam ki lines paani ki line se zyada steep hain kyunki ice aur steam ka specific heat liquid paani ka lagbhag aadha hai.

Table 10.71 atm par melting aur boiling points aur latent heats (latent heats 10⁵ J/kg mein)

Substance	Melting point (°C)	$L_{f}$	Boiling point (°C)	$L_{v}$
Paani	0	3.34	100	22.6
Ethanol	−114	1.1	78	8.5
Mercury	−39	0.11	357	2.9
Nitrogen	−210	0.26	−196	2.0
Lead	327	0.23	1749	8.6
Copper	1085	2.1	2562	47

Paani ke liye $L_{f} = 3.34 \times 10^{5} J / k g$ aur $L_{v} = 2.26 \times 10^{6} J / k g .$ Ek kilogram ubalte paani ko steam banaane mein ek kilogram ice pighlaane se lagbhag saat guna heat lagti hai, aur us paani ko 0 °C se 100 °C tak garam karne se paanch guna se zyada. Ise ulta socho: skin par condense hoti ek kilogram steam thanda hona shuru hone se pehle hi wahan $2.26 \times 10^{6} J$ daal deti hai, jo pehle se 100 °C wala paani nahi kar sakta. Isliye idli steamer ya pressure-cooker ki seeti se nikla steam ka jhonka ubalte paani ki chheent se zyada buri tarah jalaata hai.

Example 10.4NCERT Example 10.4 · p. 213

Latent heat of fusion naapna

Negligible heat capacity wale insulated flask mein $0^{\circ} C$ par $80 g$ ice ko $400 g$ paani mein daala jaata hai, jiska temperature hai $40.0^{\circ} C .$ Jab saari ice pighal jaati hai, mixture ka temperature hota hai $20.0^{\circ} C .$ Ice ki latent heat of fusion nikalo. Lo $s_{w} = 4186 J k g^{- 1} K^{- 1} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.5NCERT Example 10.5 · p. 214

Freezer ki ice se steam tak

1 atm par $- 10^{\circ} C$ ki $2.0 k g$ ice ko $100^{\circ} C$ ki steam banaane mein kitni heat lagti hai? Use karo $s_{ice} = 2100,$ $s_{w} = 4186 J k g^{- 1} K^{- 1},$ $L_{f} = 3.34 \times 10^{5}$ aur $L_{v} = 2.26 \times 10^{6} J / k g .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Jab 100 °C par

50 g

steam condense hokar 100 °C ka paani banti hai, to kitni heat, kilojoules mein, release hoti hai?

(L_{v} = 2.26 \times 10^{6} J / k g)

Answer dekho

Answer 113 kJ

Q = m L_{v} = 0.050 \times 2.26 \times 10^{6} = 1.13 \times 10^{5} J = 113 k J .

10.9 Heat transfer

Heat garam se thande ki taraf chalti energy hai, do bodies ke beech ya ek hi body ke do parts ke beech. Yeh teen alag tareekon se travel karti hai, aur gas stove par doodh ka pan teeno ek saath dikhata hai (Fig. 10.14): usmein rakhi steel ki kalchhi se conduction, ghoomte doodh mein convection, aur radiation jo flame ke paas rakhe haath par mehsoos hota hai.

3/3Flame tumhare haath ko seedha garam karti hai, kisi medium ki zaroorat nahi.

Fig. 10.14Heat travel karne ke teen tareeke: conduction, convection aur radiation.NCERT Fig. 10.13 · p. 214

Gas flame par doodh ka pan, usmein steel ki kalchhi aur ek taraf ek haath. Step 1: kalchhi ke saath ek arrow conduction dikhata hai. Step 2: doodh mein loops dikhate hain ki garam doodh beech mein upar uthta hai aur thanda doodh sides par neeche jaata hai: convection. Step 3: flame se lehradaar arrows seedha haath tak pahunchte hain: radiation.

10.9.1 Conduction

Ubalti dal ke bartan mein steel ki kalchhi chhod do to jaldi hi uska handle pakadne layak nahi rehta, jabki usmein upar kuch beh kar nahi gaya. Energy steel mein ek particle se doosre particle tak pahunchaayi gayi hai: garam end par tezi se hilte particles apne dheeme padosiyon se takraate hain, aur metals mein free electrons zyadatar energy le jaate hain, isiliye metals itne achhe conductors hain. Yeh conduction hai, kisi material mein garam se thande regions ki taraf heat ka transfer, bina material ki bulk motion ke. Liquids solids se kam achha conduct karte hain, aur gases bahut kam.

Conduction ko numbers mein dekhne ke liye ek standard arrangement lo (Fig. 10.15): uniform cross-section $A$ aur length $L$ wala ek bar, jiski sides lagged hain taaki heat sirf ends se nikal sake, jo $T_{C}$ aur $T_{D}$ ( $T_{C} > T_{D}$ ) par do bade reservoirs se sate hain. Kaafi der chhodo to bar ek steady state mein aa jaata hai jismein time ke saath kuch nahi badalta: temperature ek end se doosre end tak barabar girta hai, aur heat thande end se theek utni tezi se nikalti hai jitni tezi se garam end mein aati hai. Aise bars par experiments ek simple rule dete hain. Flow ka rate, yaani heat current $H$ , cross-section ya temperature difference double karne par double ho jaata hai, aur length double karne par aadha:

Heat conduction

H = K A \frac{T_{C} - T_{D}}{L}

(10.14)

jahan

$K$: material ki thermal conductivity
$H$: heat current, watts mein

3/3Bar ke har section se same heat current $H$ guzarta hai.

Fig. 10.15Steady state mein temperature bar ke saath linearly girta hai, aur har section se same heat current guzarta hai.NCERT Fig. 10.14 · p. 215

Ek lagged bar T C par garam reservoir ko T D par thande reservoir se jodta hai, uske neeche temperature aur position ka graph hai. Step 1: shuru hone ke thodi der baad sirf garam end garam hai; curve tezi se girta hai phir flat ho jaata hai. Step 2: steady state mein graph T C se T D tak ek seedhi line hai. Step 3: arrows bar ke saath same heat current H dikhate hain.

Constant $K$ material ki thermal conductivity (ऊष्मा चालकता) hai. Iski SI unit hai $J s^{- 1} m^{- 1} K^{- 1},$ yaani $W m^{- 1} K^{- 1} .$ Yeh temperature ke saath thoda badalti hai, lekin aam ranges mein hum ise constant maante hain. Table 10.8 iski bahut badi range dikhata hai: copper shaant hawa se lagbhag 15 000 guna behtar heat conduct karta hai.

Table 10.8Room temperature ke aas-paas thermal conductivities

Material	$K$ $(W m^{- 1} K^{- 1})$
Silver	429
Copper	401
Aluminium	237
Brass	109
Steel	50
Lead	35
Ice	2.2
Glass	1.0
Concrete	0.8
Paani	0.60
Lakdi	0.12
Thermocole	0.035
Hawa (shaant)	0.026

Designers table ke dono siron ka use karte hain. Garam chai wala steel ka glass pakadna mushkil hota hai jabki usi chai wala mitti ka kulhad theek rehta hai, kyunki kulhad ki deewar tumhari ungliyon tak heat kahin dheere pahunchaati hai. Ice-cream wale apna maal thermocol boxes mein rakhte hain, aur razai tumhe garam rakhti hai, mainly kyunki dono hawa ko band karte hain, jo sabse kharab conductors mein se ek hai. Doosri taraf, laptop apne processor ki heat ko copper heat pipes se aluminium fins tak le jaata hai, taaki chip kabhi overheat na ho.

DerivationSeries mein rods: thermal resistance

Eq. (10.14) ko aise likho: $T_{C} - T_{D} = H L / (K A) .$ Quantity $R = L / (K A)$ bar ka thermal resistance hai: temperature difference voltage ka role nibhaata hai aur $H$ current ka. Do bars ko end to end jodo aur lag karo. Steady state mein dono se same $H$ guzarna chahiye, warna junction garam ya thanda hota jaata, isliye dono temperature drops seedhe jud jaate hain:

Series mein thermal resistances

T_{1} - T_{2} = H (R_{1} + R_{2}), R = \frac{L}{K A}

Series mein resistances jud jaate hain, theek waise hi jaise tum Class 12 mein electric circuits ke liye paoge. Agar bars same do reservoirs ke beech side by side hon, to iski jagah heat currents jud jaate hain.

Example 10.6NCERT Example 10.6 · p. 215

Junction kahan sabse garam hai?

$20.0 c m$ lambi copper rod aur $10.0 c m$ lambi steel rod, dono ka cross-section $4.0 c m^{2}$ , end to end jodi jaati hain aur lag ki jaati hain. Copper ka free end $100^{\circ} C$ ki steam mein aur steel ka free end $0^{\circ} C$ ki pighalti ice mein rakha jaata hai. Junction ka steady temperature aur heat current nikalo. Lo $K_{C u} = 400$ aur $K_{steel} = 50 W m^{- 1} K^{- 1} .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 10.7NCERT Example 10.7 · p. 216

Barabar length ki do rods

Barabar length $L$ aur cross-section $A$ wali do rods, jinki conductivities $K_{1}$ aur $K_{2}$ hain, end to end jodi jaati hain aur lag ki jaati hain. Rod 1 ka free end $T_{1}$ par hai aur rod 2 ka $T_{2}$ par, $T_{1} > T_{2}$ . (i) Junction temperature $T_{0}$ nikalo. (ii) $2 L$ length wali ek single rod ki conductivity $K^{'}$ nikalo jo same heat current le jaaye. (iii) Ek aluminium-alloy rod $(K_{1} = 200 W m^{- 1} K^{- 1})$ aur ek steel rod $(K_{2} = 50 W m^{- 1} K^{- 1}),$ dono $0.25 m$ lambi, $A = 4.0 c m^{2},$ $T_{1} = 400 K$ aur $T_{2} = 300 K$ ke liye dono values aur heat current calculate karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Fig. 10.16Do materials chuno, unki lengths aur end temperatures set karo, aur junction temperature aur heat current padho.

Quick check

3 mein se 1

1.0 m^{2}

area aur

4.0 m m

motaai wale khidki ke kaanch ke dono faces ke temperatures mein

5.0 K

ka farak hai. Agar

K = 1.0 W m^{- 1} K^{- 1},

to heat current, watts mein, kya hai?

Answer dekho

Answer 1250 W

H = K A Δ T / L = 1.0 \times 1.0 \times 5.0 / 0.0040 = 1250 W .

Patla glass heat ko tezi se jaane deta hai, isiliye double glazing hawa ki ek layer band karti hai.

10.9.2 Convection

Stove par paani ke pan mein chutki bhar haldi daalo aur kanon ko dekho: woh beech se upar uthte hain aur sides se neeche aate hain, gol-gol. Paani khud chal raha hai aur apne saath heat le ja raha hai. Yeh convection (संवहन) hai, aur yeh sirf liquids aur gases mein ho sakta hai. Neeche ka paani garam hota hai, expand hota hai aur kam dense ho jaata hai, isliye buoyancy use upar dhakelti hai; zyada dense, thanda paani uski jagah lene neeche aata hai aur phir woh garam hota hai. Aisa convection current sirf conduction se kahin tezi se pan mein heat phailaata hai. Kyunki yeh buoyancy par tika hai, is natural convection ko gravity chahiye: International Space Station par candle lambi flame ki jagah ek dheemi, gol chamak ki tarah jalti hai.

Forced convection mein fan ya pump fluid ko chalaata hai. Ceiling fan, convection oven aur laptop ke andar ka fan, sab garam surfaces ke paas se hawa dhakel kar heat ko tezi se door le jaate hain. Tumhari body bhi ek forced-convection system chalaati hai: dil blood pump karta hai, jo kaam kar rahi muscles aur organs se heat ikattha karke skin tak le jaata hai, aur poori body ko lagbhag same temperature par rakhta hai.

Dopahar chaar baje Marina beach par khade ho jao to hawa Bay of Bengal se lagataar andar ki taraf chalti hai; aadhi raat ke baad wahan khade ho to ek halki hawa samundar ki taraf jaati hai (Fig. 10.17). Iski wajah uneven heating se chalne wala convection loop hai. Dopahar ki dhoop mein ret samundar ki surface se kahin zyada garam ho jaati hai, kyunki samundar ka bada specific heat aur lagataar hilna uske temperature ko lagbhag steady rakhte hain. Garam ret apne upar ki hawa ko garam karti hai; woh hawa buoyant hokar upar chadhti hai, oonchaai par samundar ki taraf phailti hai, thande paani ke upar neeche aati hai aur surface ke saath sea breeze ban kar lautti hai. Andhera hone ke baad ret jaldi heat kho deti hai jabki samundar garam rehta hai, isliye loop ulta chalta hai aur kinaare se land breeze chalti hai.

Fig. 10.17Din mein zameen zyada garam hoti hai aur hawa samundar se andar aati hai; raat mein loop ulat jaata hai.NCERT Fig. 10.17 · p. 217

Coast ke do panels. Din: dhoop hai, garam hawa zameen ke upar uthti hai, oonchaai par samundar ki taraf jaati hai, thande samundar ke upar neeche aati hai aur surface par sea breeze ban kar andar ki taraf chalti hai. Raat: chaand nikla hai, hawa garam samundar ke upar uthti hai aur surface ki hawa zameen se samundar ki taraf land breeze ban kar chalti hai.

Is loop ko poore hemisphere jitna bada karo to trade winds milti hain, woh steady north-easterly hawaayein jo sailing ships ko tropics ke aar-paar le jaati thi. Equator par dhoop sabse tez hoti hai, isliye wahan ki hawa upar uthti hai aur oonchaai par poles ki taraf jaati hai; rotation na hota to woh poles ke paas neeche aakar seedhe waapas behti. Lekin Earth ghoomti hai. Equator par ek point lagbhag 1670 km/h se east ki taraf chalta hai aur pole ke paas ka point lagbhag hilta hi nahi, aur is ghoomti surface ke upar chalne wali hawa ek taraf mud jaati hai. Isliye oonchaai wala flow lagbhag 30° latitude par neeche aata hai, aur zameen ke paas ka waapsi flow mud kar north-east se aane wali hawa ban jaata hai.

10.9.3 Radiation

Chhaanv se nikal kar dhoop mein aao aur tumhari skin turant garam ho jaati hai. 150 million kilometre ki lagbhag khaali space mein tumhe Sun se koi material nahi jodta, isliye na conduction iske liye zimmedaar ho sakta hai na convection. Energy radiation ki tarah aati hai: electromagnetic waves, yaani oscillate karte electric aur magnetic fields jo vacuum mein bhi light ki speed se travel karte hain, yaani $3 \times 10^{8} m / s .$ Sunlight ko hum tak pahunchne mein lagbhag 8.3 minute lagte hain. Is tarah aane wali energy ko radiant energy kehte hain; electromagnetic waves tum Class 12 mein padhoge.

Tum bhi radiate karte ho. Har object, solid, liquid ya gas, sirf garam hone ki wajah se electromagnetic waves emit karta hai, aur yeh thermal radiation temperature ke saath tezi se badhta hai: tumhari body invisible infrared deti hai, toaster ke elements orange chamakte hain, aur Sun safed chamakta hai. Kisi surface par girne wala radiation kuch waapas uchhal jaata hai aur kuch sokh liya jaata hai. Matt black surfaces lagbhag saara sokh leti hain aur, same temperature par, sabse zyada bhejti bhi hain; chamkeeli, halke rang ki surfaces zyadatar reflect karti hain aur kam emit karti hain.

Isiliye Rajasthan ke gaon ke ghar aksar chune se safed pute hote hain aur safed 'cool-roof' paint upar ki manzil ke kamron ko thanda rakhta hai, jabki chhat par lage solar water heaters ki absorber plates aur box-type solar cooker ke bartan matt black paint kiye jaate hain.

Vacuum flask, ya thermos, har mode ko ek-ek karke hara kar chai ko ghanton garam ya ice ko thanda rakhta hai (Fig. 10.18):

Conduction aur convection: bottle double-walled glass ki hai jiski walls ke beech vacuum hai, isliye dono mein se koi bhi gap ke aar-paar heat nahi le ja sakta.
Radiation: gap ki taraf wali glass surfaces silvered hain. Garam andar ki wall se aane wala radiation waapas andar reflect hota hai, aur bahar se aane wala radiation door reflect ho jaata hai.
Bache hue leaks: heat ab bhi stopper aur neck se conduct ho sakti hai, isliye stopper cork ya plastic ka hota hai aur bottle insulating pads par tiki hoti hai.

3/3Silvered walls radiation ko waapas chai ki taraf uchhaal deti hain.

Fig. 10.18Vacuum flask conduction aur convection ko vacuum se, aur radiation ko silvered walls se rokta hai.

Vacuum flask ka cross-section: double glass walls jinke beech vacuum hai, silvered faces, ek stopper aur base pads, andar chai. Step 1: vacuum ko paar karne ki koshish karta arrow kata hua hai, yaani koi conduction nahi. Step 2: ek upar wala arrow stopper se rukta hai aur khaali gap mein ghoomne ke liye hawa nahi hai, yaani koi convection nahi. Step 3: lehradaar radiation arrows silvered walls se takra kar waapas chai mein jaate hain.

10.9.4 Blackbody radiation

Heater coil ki chamak ko prism se guzaaro to ek rang nahi balki rangon ki ek patti milti hai: thermal radiation wavelengths ki ek continuous range mein phaila hota hai, jinmein se zyadatar invisible hain. Energy kaise bantti hai, yeh temperature par depend karta hai. Fig. 10.19 teen temperatures par ek perfect emitter ki prati unit area prati unit wavelength radiate ki gayi energy plot karta hai. Har curve ek peak tak chadhta hai aur phir dheere-dheere girta hai; temperature badhne par poora curve upar uthta hai aur uska peak chhoti wavelengths ki taraf khisakta hai.

3/36000 K par yeh lagbhag 485 nm par peak karta hai; peaks chhoti wavelengths ki taraf badhte hain.

Fig. 10.19Garam bodies har wavelength par zyada radiate karti hain, aur unka peak chhoti wavelengths ki taraf jaata hai.NCERT Fig. 10.18 · p. 218

0 se 3000 nm tak wavelength ke against prati unit wavelength energy, visible band shaded aur line styles ka legend ke saath. Step 1: neecha dotted 4000 K curve jo lagbhag 725 nm par peak karta hai. Step 2: dashed 5000 K curve, lagbhag teen guna ooncha, jo lagbhag 580 nm par peak karta hai. Step 3: kahin ooncha solid 6000 K curve jo lagbhag 485 nm par peak karta hai, aur ek dotted line teeno peaks ko jodti hai jab woh chhoti wavelengths ki taraf jaate hain.

Peak par wavelength, $λ_{m}$ , absolute temperature ke inversely proportional hai. Yeh Wien's displacement law hai:

Wien's displacement law

λ_{m} T = b

(10.15)

jahan

$b$: Wien's constant, $2.9 \times 10^{- 3} m K$

Wien's law kisi bhi tez garam ki gayi cheez ke rangon ka sequence samjhata hai, jaise electric stove ki coil jo pehle halki laal, phir orange chamakti hai, aur filament jo peela-safed chamakta hai: temperature badhne par peak infrared se visible ki taraf aata hai. Isse astronomers kisi star ke rang se uski surface ka temperature bhi padh lete hain. Sirius, jiska spectrum ultraviolet mein lagbhag 290 nm par peak karta hai, ki surface lagbhag 10 000 K par hai; laal Betelgeuse, jo lagbhag 830 nm par peak karta hai, lagbhag 3500 K par hai; Sun, jo lagbhag 500 nm par peak karta hai, lagbhag 5800 K par hai. Tum bhi radiate karte ho: 34 °C wali skin infrared mein lagbhag 9.4 μm par peak karti hai, aur thermal camera yahi dekhta hai.

In curves ka shape universal hai: yeh sirf temperature se tay hota hai, aur perfect emitter kitna bada hai, uska shape kya hai aur woh kis cheez ka bana hai, isse koi farak nahi padta. Aise perfect emitter ko blackbody kehte hain. Garam furnace ki deewar mein ek chhota chhed bahut achha blackbody hai. In curves ko samjhaana unnisvi sadi ki physics ke bas ki baat nahi thi. Max Planck 1900 mein tabhi safal hue jab unhone maana ki energy chhote packets, yaani quanta, mein emit hoti hai, aur isi se quantum physics shuru hui, jisse tum Class 12 mein miloge.

Fig. 10.20Temperature ko drag karo, ya koi star ya lamp chuno, aur dekho ki peak kaise khisakta hai aur total power kaise tezi se badhti hai.

Ek garam body kul kitni power radiate karti hai? Surface area $A$ aur absolute temperature $T$ wale perfect radiator ke liye prati second emit hone wali energy hai

Stefan–Boltzmann law

H = A σ T^{4}

(10.16)

jahan

$σ$: Stefan–Boltzmann constant, $5.67 \times 10^{- 8} W m^{- 2} K^{- 4}$

Josef Stefan ne 1879 mein experimental data se yeh law nikala, aur Ludwig Boltzmann ne 1884 mein ise theory se derive kiya. Absolute temperature double karne se radiate hone wali power 16 guna ho jaati hai. Real surfaces same temperature par blackbody se kam radiate karti hain, isliye hum likhte hain

H = A e σ T^{4}

(10.17)

jahan emissivity $e$ , 0 aur 1 ke beech, surface par depend karti hai: lamp black aur human skin ke liye 0.95 se upar, polished silver ke liye 0.1 se neeche. Kyunki achha absorber achha emitter hota hai, wahi $e$ aane wale radiation ka absorb hone wala fraction bhi batata hai. $120 c m^{2}$ surface area wali ek orange-garam heater rod, $1100 K$ par aur $e \approx 0.9$ ke saath, radiate karti hai lagbhag $0.9 \times 0.012 \times 5.67 \times 10^{- 8} \times 1100^{4} \approx 900 W,$ jo 1 kW heater ke output ka zyadatar hissa hai.

Body apne surroundings se radiation absorb bhi karti hai. Agar woh $T$ par hai aur uske surroundings $T_{s}$ par, to radiation se energy khone ka net rate hai

Net radiation loss

H = e σ A (T^{4} - T_{s}^{4})

(10.18)

Example

Earth tak kitni power pahunchti hai?

Sun ko $6.96 \times 10^{8} m$ radius aur $5770 K$ surface temperature wala blackbody maano. (a) Woh kul kitni power radiate karta hai, nikalo. (b) $1.50 \times 10^{11} m$ door Earth par, Sun ki taraf face karti surface par prati square metre pahunchne wali power nikalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Ek star ka spectrum

400 n m

par peak karta hai. Kelvin mein uski surface ka temperature estimate karo.

(b = 2.9 \times 10^{- 3} m K)

Answer dekho

Answer 7250 K

T = b / λ_{m} = 2.9 \times 10^{- 3} / 4.00 \times 10^{- 7} \approx 7250 K,

jo Sun se garam hai, jiska peak lagbhag 500 nm par hai.

10.10 Newton's law of cooling

Counter par chhoda gaya filter coffee ka steel tumbler pehle jaldi heat khota hai, phir dheere aur dheere, jab tak woh room temperature par na aa jaaye. Cooling ka rate is par kaise depend karta hai ki woh kitna garam hai? Thermometer aur ghadi se tum yeh pata laga sakte ho.

ActivityNCERT 10.10 · p. 219

Thande hote cup ka time naapna

Tumhe chahiye

ek steel tumbler, jiske cardboard dhakkan mein do chhed hon
ek thermometer (0–100 °C) aur ek patla stirrer
kettle se garam paani
ek stopwatch ya phone timer

Pehle guess karo

Experiment ke kis hisse mein paani sabse tezi se thanda hoga?

Pehle kuch minuteExperiment ke beech meinAakhri kuch minuteYeh poore samay ek steady rate se thanda hoga

Thermometer par room temperature note karo.
Lagbhag 250 mL garam paani, lagbhag 70 °C par, tumbler mein dalo, use dhak do, aur thermometer aur stirrer ko chhedon mein ghusao.
Dheere se hilao aur kam se kam 20 minute tak har 2 minute par temperature padho.
Har reading ke liye excess temperature (paani minus room) nikalo aur use time ke against plot karo.

Table 10.9Cooling readings

Time (min)	Paani ka temperature (°C)	Room se excess (°C)
0
2
4
8
12
20

Tumhari entries apne aap save hoti hain.

Isaac Newton ne yeh rule 1701 mein high temperatures naapne par ek anonymous paper mein bataya. Newton's law of cooling ke mutabik, koi body us rate se heat khoti hai jo uske temperature $T_{2}$ aur uske surroundings ke temperature $T_{1}$ ke difference ke proportional hai:

Newton's law of cooling

- \frac{d Q}{d t} = k (T_{2} - T_{1})

(10.19)

Constant $k > 0$ is se tay hota hai ki kitni surface khuli hai aur woh kaisi hai: chamkeeli ya dull, shaant hawa mein ya hawa ke jhonke mein. Yeh law ek approximation hai jo tab kaam karta hai jab body apne surroundings se bahut zyada garam na ho, zyada se zyada kuch das kelvin, jaise table par chai ka cup ya walls se heat khota room. Tab heat nikalne ka har raasta temperature difference ke saath lagbhag barabar badhta hai. Radiation dikhata hai kyun: jab $T_{2}$ , $T_{1}$ ke kareeb ho to $T_{2}^{4} - T_{1}^{4} = (T_{2} - T_{1}) (T_{2} + T_{1}) (T_{2}^{2} + T_{1}^{2}) \approx 4 T_{1}^{3} (T_{2} - T_{1})$ , isliye Eq. (10.18) bhi linear ho jaati hai.

DerivationLaw se exponential tak

Maano body ka mass $m$ aur specific heat $s$ hai. Jab $d t$ time mein uska temperature $d T_{2}$ badalta hai, to uska heat content is rate se badalta hai

\frac{d Q}{d t} = m s \frac{d T_{2}}{d t}

(10.20)

Ise Eq. (10.19) mein dalo aur variables alag karo:

\frac{d T_{2}}{T_{2} - T_{1}} = - \frac{k}{m s} d t = - K d t

(10.21)

jahan

$K$: cooling constant $k / m s$

Surroundings ko fixed $T_{1}$ par rakhte hue integrate karo:

\ln (T_{2} - T_{1}) = - K t + c

(10.22)

ya, exponentials lene par,

T_{2} = T_{1} + C^{'} e^{- K t}

(10.23)

jahan

$C^{'}$: $e^{c}$ , $t = 0$ par excess temperature

Excess temperature exponentially decay karta hai: yeh har barabar time interval mein same fraction se girta hai. Yeh har $(\ln 2) / K$ mein aadha ho jaata hai, chahe shuruaati value kuch bhi ho.

Law test karne ke liye surroundings ek jaane-pehchaane, steady temperature par rehne chahiye. Ek practical arrangement mein garam paani wala copper calorimeter ek double-walled vessel ke andar rakha jaata hai jiski walls ke beech paani hota hai. Ek thermometer calorimeter ka paani, $T_{2}$ , padhta hai, aur doosra walls ka paani, $T_{1}$ . $\ln (T_{2} - T_{1})$ ka time ke against graph $- K$ slope wali seedhi line hai (Fig. 10.21), jaisa Eq. (10.22) predict karti hai.

3/3Excess ke log ka time ke against graph $- K$ slope wali seedhi line hai.

Fig. 10.21Excess temperature exponentially decay karta hai, isliye uska logarithm seedhi line mein girta hai.NCERT Fig. 10.20 · p. 220

Step 1: thande hote paani ke liye temperature aur time, ek curve jo pehle tezi se girta hai aur room temperature ki taraf flat hota jaata hai, jo dashed dikhaaya gaya hai. Step 2: shuru mein ek steep tangent aur baad mein ek halka tangent dikhate hain ki cooling rate excess temperature ke saath girta hai. Step 3: excess temperature ke natural log ka time ke against graph, ek seedhi line jo minus K slope ke saath neeche jaati hai.

Example 10.8NCERT Example 10.8 · p. 221

Chai ka cup thanda hone mein kitna time?

$30^{\circ} C$ wale room mein chai $4.0$ minute mein $80^{\circ} C$ se $70^{\circ} C$ tak thandi hoti hai. Use $60^{\circ} C$ se $50^{\circ} C$ tak girne mein kitna time lagega, nikalo. (a) Average temperatures use karke solve karo, jaisa chhote intervals ke liye aam hai, aur (b) exponential law se.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Newton's law of cooling follow karne wali body ke liye

\ln (T_{2} - T_{1})

ka time ke against graph hai

positive slope wali seedhi lineek curve jo flat ho jaata hainegative slope wali seedhi lineek horizontal line

Answer dekho

Answer (C) negative slope wali seedhi line

Eq. 10.22 hai

\ln (T_{2} - T_{1}) = - K t + c

- K

slope wali seedhi line. Flat hota curve khud

T_{2}

ka graph hai.

Recap

Summary aur key points

Temperature naapta hai ki body kitni garam hai; heat woh energy hai jo temperature difference ki wajah se transfer hoti hai, aur thermal equilibrium par flow ruk jaata hai (Section 10.2).
Temperature scale ko ek thermometric property aur fixed points chahiye; $t_{F} = \frac{9}{5} t_{C} + 32$ (Section 10.3).
Kam density wali gases $P V = μ R T$ follow karti hain. Constant-volume gas thermometers absolute zero, −273.15 °C, tak extrapolate hote hain, jo Kelvin scale ka zero hai: $T = t_{C} + 273.15$ (Section 10.4).
$Δ l / l = α_{l} Δ T$ aur $Δ V / V = α_{V} Δ T$ ; isotropic solid ke liye $α_{V} = 3 α_{l}$ aur area expansivity $2 α_{l}$ hai; constant pressure par ideal gas ke liye $α_{V} = 1 / T$ (Section 10.5).
Paani 4 °C ke aas-paas sabse dense hota hai, isliye jheelein upar se neeche ki taraf jamti hain (Paani: sabse alag).
Expand hone se roki gayi rod mein thermal stress $Y α_{l} Δ T$ hota hai (Thermal stress).
$Δ Q = m s Δ T$ ; molar specific heat $C = Δ Q / (μ Δ T),$ aur gases ke liye $C_{P} - C_{V} \approx R$ (Section 10.6).
Isolated system mein garam parts ki khoi heat thande parts ki mili heat ke barabar hoti hai (Section 10.7).
Change of state ke dauraan temperature constant rehta hai aur $Q = m L$ ; pressure melting aur boiling points ko khiskata hai, aur triple point par teeno states saath rehti hain (Section 10.8, Section 10.8.1).
Conduction: steady state mein $H = K A (T_{C} - T_{D}) / L$ ; series mein thermal resistances $L / K A$ jud jaate hain (Section 10.9.1).
Convection fluid ke bulk flow se heat le jaata hai, natural ya forced; radiation electromagnetic waves ki tarah travel karta hai aur use kisi medium ki zaroorat nahi (Section 10.9.2, Section 10.9.3).
Wien's law $λ_{m} T = b$ ; Stefan–Boltzmann law $H = A e σ T^{4}$ , net loss $e σ A (T^{4} - T_{s}^{4})$ ke saath (Section 10.9.4).
Newton's law of cooling, $- d Q / d t = k (T_{2} - T_{1}),$ chhote temperature differences ke liye $T_{2} = T_{1} + C^{'} e^{- K t}$ deta hai (Section 10.10).

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercises 10.1	Practice Q1
NCERT Exercises 10.2	Practice Q2
NCERT Exercises 10.3	Practice Q3
NCERT Exercises 10.4	Practice Q4
NCERT Exercises 10.5	Practice Q5
NCERT Exercises 10.6	Practice Q6
NCERT Exercises 10.7	Practice Q7
NCERT Exercises 10.8	Practice Q8
NCERT Exercises 10.9	Practice Q9
NCERT Exercises 10.10	Practice Q10
NCERT Exercises 10.11	Practice Q11
NCERT Exercises 10.12	Practice Q12
NCERT Exercises 10.13	Practice Q13
NCERT Exercises 10.14	Practice Q14
NCERT Exercises 10.15	Practice Q15
NCERT Exercises 10.16	Practice Q16
NCERT Exercises 10.17	Practice Q17
NCERT Exercises 10.18	Practice Q18
NCERT Exercises 10.19 (a)–(c)	Practice Q19
NCERT Exercises 10.19 (d)–(e)	Practice Q20
NCERT Exercises 10.20	Practice Q21

Quick check: MCQs

10.2 Temperature aur heat3 questions
10.3 Temperature ka measurement2 questions
10.4 Ideal-gas equation aur absolute temperature2 questions
Thermal stress3 questions
10.6 Specific heat capacity3 questions
10.7 Calorimetry2 questions
10.8 Change of state3 questions
10.8.1 Latent heat3 questions
10.9.1 Conduction3 questions
10.9.3 Radiation4 questions
10.9.4 Blackbody radiation4 questions
10.10 Newton's law of cooling3 questions

Glossary · 15

Key terms (English + Hindi)

absolute zeroपरम शून्यparam shoonya: Sabse kam sambhav temperature, 0 K ya −273.15 °C, jahan ideal gas koi pressure nahi daalegi.Chapter mein pehli baar yahan aaya →

coefficient of linear expansionरैखिक प्रसार गुणांकrekhik prasaar gunaank: Prati kelvin length mein fractional badhat, $α_{l} = Δ l / (l Δ T) .$ Chapter mein pehli baar yahan aaya →
coefficient of volume expansionआयतन प्रसार गुणांकaayatan prasaar gunaank: Prati kelvin volume mein fractional badhat, $α_{V} = Δ V / (V Δ T)$ ; isotropic solid ke liye $α_{V} = 3 α_{l} .$ Chapter mein pehli baar yahan aaya →
convectionसंवहनsanvahan: Fluid ke bulk flow se heat transfer, jo buoyancy (natural) ya pump ya fan (forced) se chalta hai.Chapter mein pehli baar yahan aaya →

emissivityउत्सर्जकताutsarjakta: Kisi surface ke emit kiye radiation ka same temperature wale blackbody ke radiation se ratio, 0 aur 1 ke beech.Chapter mein pehli baar yahan aaya →

heatऊष्माushma: Temperature difference ki wajah se transfer hone wali energy, joules mein naapi jaati hai.Chapter mein pehli baar yahan aaya →

latent heatगुप्त ऊष्माgupt ushma: Constant temperature par change of state mein prati unit mass li ya di gayi heat, $L = Q / m$ .Chapter mein pehli baar yahan aaya →

molar specific heat capacityमोलर विशिष्ट ऊष्मा धारिताmolar vishisht ushma dhaarita: Prati mole prati kelvin zaroori heat; gas ke liye $C_{P}$ , $C_{V}$ se zyada hota hai.Chapter mein pehli baar yahan aaya →

regelationपुनर्हिमायनpunarhimaayan: Pressure mein ice ka pighalna aur pressure hatne par phir jam jaana.Chapter mein pehli baar yahan aaya →

specific heat capacityविशिष्ट ऊष्मा धारिताvishisht ushma dhaarita: 1 kg substance ko 1 K garam karne ke liye zaroori heat, J kg⁻¹ K⁻¹ mein.Chapter mein pehli baar yahan aaya →
sublimationऊर्ध्वपातनoordhvapaatan: Solid se seedha vapour mein badlaav, jaise dry ice ya camphor mein.Chapter mein pehli baar yahan aaya →

temperatureतापtaap: Ek tay scale par number jo batata hai ki body kitni garam hai; heat zyada temperature se kam temperature ki taraf flow karti hai.Chapter mein pehli baar yahan aaya →
thermal conductivityऊष्मा चालकताushma chaalakta: $H = K A Δ T / L$ mein constant $K$ ; material kitni achhi tarah heat conduct karta hai, W m⁻¹ K⁻¹ mein.Chapter mein pehli baar yahan aaya →
thermal stressतापीय प्रतिबलtaapiya pratibal: Aisi body mein stress jise garam ya thanda kiya jaaye lekin size badalne na diya jaaye; clamp ki hui rod ke liye $Y α_{l} Δ T$ .Chapter mein pehli baar yahan aaya →
triple pointत्रिक बिंदुtrik bindu: Woh akela temperature aur pressure jahan solid, liquid aur vapour saath rehte hain; paani ke liye 273.16 K aur 611.66 Pa.Chapter mein pehli baar yahan aaya →

Thermal Properties of Matter: Class 11 Physics Chapter 10 Hinglish mein

Is chapter mein aap kya seekhenge

10.1 Introduction

10.2 Temperature aur heat

10.3 Temperature ka measurement

10.4 Ideal-gas equation aur absolute temperature

10.5 Thermal expansion

Paani: sabse alag

Gases

Volume aur linear expansion ko jodna

Thermal stress

10.6 Specific heat capacity

10.7 Calorimetry

10.8 Change of state

10.8.1 Latent heat

10.9 Heat transfer

10.9.1 Conduction

10.9.2 Convection

10.9.3 Radiation

10.9.4 Blackbody radiation

10.10 Newton's law of cooling

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye