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Physics | Class 11 | Chapter 10 - Thermal Properties of Matter

Thermal Properties of Matter

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Socho tum chai garam kar rahe ho, haath se cup pakda toh garam laga, thandi bottle pakdi toh thandi. Ye garam-thanda ka ehsaas hi heat aur temperature hai. Par ek lohe ki rod ko dhoop mein rakho toh woh failti hai, thande mein sikudti hai. Paani ko 0 pe rakho toh baraf, 100 pe bhap.

Ye chapter isi ka khel hai: garam karne pe cheezein kyun failti hain? Kitni heat lagegi paani garam karne mein? Ek bartan se dusre mein garmi kaise jaati hai conduction se, hawa se convection se, bina hawa ke radiation se? Aur suraj ki roshni ka rang kyun temperature batata hai?

Jaise kitchen mein steel ka karchhi garam ho jaata hai handle tak, ya fan ke neeche garam hawa upar jaati hai, ya dhoop mein kaala kapda zyada garam lagta hai — sab isi chapter ke formulas se samjhenge.

Heat = energy transfer due to temp difference, Temp = measure of hotness, Scales C F K relation C/100=(F-32)/180=(K-273.15)/100, Ideal gas PV=μRT absolute zero -273.15C. Expansion ΔL=αLΔT β=2α γ=3α water anomalous 0-4C. Specific heat Q=mcΔT calorimetry heat lost=gained Q=mL latent. Transfer conduction Q/t=KAΔT/L, convection hot rises cold sinks, radiation Stefan E=σAeT⁴ Wien λmT=2.9e-3 mK Newton cooling dT/dt=-k(T-Ts).

1. Temperature Scales & Ideal Gas - Thanda Garam Ka Scale

Socho tum thermometer se bukhar naapte ho — 98°F, 37°C. Scale alag hai par garam to same hai. Temperature = hotness ka measure, Heat = temperature difference se transfer hone wali energy. Heat hamesha garam se thande ki taraf jaati hai.

3 scales: Celsius mein ice 0°C steam 100°C, Fahrenheit mein ice 32°F steam 212°F, Kelvin mein ice 273.15K steam 373.15K. Relation C/100 = (F-32)/180 = (K-273.15)/100. Kelvin absolute scale hai, iska zero -273.15°C jahan sab molecular motion band theoretically.

Gas ka law: PV = μRT, μ moles, R=8.31 J/mol·K gas constant. Constant pressure pe volume ∝ temperature, constant volume pe pressure ∝ temperature. Isi se absolute zero ka idea aaya — graph ko extend karo toh -273.15 pe zero volume.

Thermal equilibrium matlab do bodies touch mein raho aur heat exchange band ho jaye toh same temperature. Zeroth law ka basis.

Temp hotness, Heat energy transfer due to ΔT hot→cold. Scales: 0C=32F=273.15K, 100C=212F=373.15K. Conversion C/100=(F-32)/180=(K-273.15)/100. Ideal gas PV=μRT R=8.31, absolute zero -273.15C where P or V→0. Thermal equilibrium → same T zeroth law.
Temperature Scales & Ideal Gas
Revision Points — After Diagram 1:
• Heat = energy in transit due to ΔT, Temperature = hotness measure, equilibrium = no net heat flow.
• Scales: C/100=(F-32)/180=(K-273.15)/100, K = C+273.15, absolute zero -273.15°C.
• Ideal gas: PV=μRT, at low P high T real gases approximate ideal, absolute zero from extrapolation.
• Zeroth law: if A equilibrium with B and B with C, then A with C → defines temperature.

2. Thermal Expansion - Garam Karo Toh Failo

Socho ek lohe ka gate garmi mein jam ho jata hai, sardi mein khul jaata hai. Kyun? Garam karne pe atoms zyada hilte hain, average distance badhta hai → length badhti hai. Linear: ΔL = α L ΔT. α coefficient of linear expansion small ~10⁻⁵ /K. Steel α≈12e-6, aluminium 23e-6.

Area ka ΔA = β A ΔT, β≈2α, Volume ka ΔV = γ V ΔT, γ≈3α for solids. Liquids ka γ zyada ~10⁻⁴, gases ka ~3.7e-3 (1/273). Isliye railway tracks mein gap chhodte hain, bridges mein rollers.

Bimetallic strip: do metals alag α wale jod do — brass aur iron. Garam karo toh zyada α wala zyada failta hai → strip bend hoti hai. Fire alarm, thermostat, auto indicator isi se kaam karta hai. Thanda garam se circuit on off.

Paani ka alag natak — 0°C se 4°C tak garam karo toh sikudta hai, 4°C ke baad failta hai. 4°C pe density max. Isliye ice upar tairta hai, neeche paani 4°C rehta hai jisme machhli jee leti hai sardi mein. Bottle mein paani bhar ke freezer mein rakho toh phat jaati hai kyunki baraf ka volume zyada.

ΔL=αLΔT α~1e-5 /K, β≈2α ΔA=βAΔT, γ≈3α ΔV=γVΔT. Liquids γ~1e-4, gases 1/273 per K. Bimetallic bend due to α diff thermostat. Water anomalous 0-4C contracts expands after, max density 4C ice floats bottle bursts.
Thermal Expansion α β γ & Bimetal
Revision Points — After Diagram 2:
• Linear ΔL=αLΔT, area β≈2α, volume γ≈3α for isotropic solid.
• α steel 12e-6, Al 23e-6 /K, liquids ~1e-4, gases ~3.3e-3.
• Bimetallic: Δα causes bending → thermostat, fire alarm, gap in rails.
• Water anomaly: 0→4C density increases, 4C max density, ice volume more than water → floats.

3. Specific Heat & Calorimetry & Latent Heat - Kitni Heat Lagegi?

Socho ek kilo paani aur ek kilo loha same burner pe rakho — loha jaldi garam, paani dheere. Kyun? Specific heat alag. Q = m c ΔT. c specific heat capacity J/kg·K, paani ka 4186 sabse zyada, isliye samundar dheere garam thanda hota hai. Lohe ka 450.

Molar specific heat C = M c. Gases ke liye do: Cp constant pressure pe, Cv constant volume pe, Cp = Cv + R.

Ab do cheezein mix karo garam thandi — garam wala heat dega thanda lega jab tak same temperature na ho. Principle of calorimetry: Heat lost = Heat gained. Calorimeter mein experiment karte hain, uska water equivalent bhi add karte hain. Jaise chai mein thanda doodh milao toh final temp beech mein.

Change of state: Ice ko 0°C pe garam karo toh 0 pe hi pighalta hai, temp nahi badhta jab tak pura pighle. Kyunki heat bonds todne mein ja rahi hai. Q = m L, L latent heat. Fusion Lf ice 3.33e5 J/kg, vaporisation Lv water 22.6e5 J/kg. Graph mein temperature vs heat mein plateaus at 0 and 100°C. Isi se steam se zyada jalan hoti hai paani se, kyunki Lv zyada.

Q=mcΔT c specific heat water 4186 J/kgK high. Molar C=Mc Cp=Cv+R for ideal gas. Calorimetry: heat lost=heat gained + calorimeter. Latent Q=mL temp constant during change, Lf 3.33e5 ice, Lv 22.6e5 water J/kg, steam burn worse due to large Lv.
Specific Heat & Calorimetry & Latent
Revision Points — After Diagram 3:
• Q=mcΔT, c high → slow temp change, water 4186 highest.
• Calorimetry: Σ heat lost = Σ heat gained, include calorimeter water equivalent.
• Latent: Q=mL, Lf ice 3.33e5, Lv 22.6e5 J/kg, temp constant during melting/boiling.
• Graph T vs Q: slope 1/mc, flat plateau at melting/boiling = latent.

4. Conduction - Lohe Ka Karchhi Garam Kyun?

Steel ki karchhi se sabzi hilao toh handle bhi garam ho jaata hai, par lakdi ka nahi. Kyun? Conduction. Atoms apne jagah pe hilte hain garam side pe zyada, thand side pe kam ko energy dete hain collision se. Solid mein hota hai mainly.

Fourier's law: Heat flow rate Q/t = K A ΔT / L. K thermal conductivity W/m·K, A area, L thickness, ΔT temp difference. K metals high copper 385, aluminium 237, steel 40-50, wood 0.12, air 0.026. Isliye copper bottom wale bartan jaldi garam, steel body durable.

Thermal resistance Rth = L/KA K/W, jaise electrical resistance. Series mein R add: Rtotal=R1+R2. Composite wall ka heat flow U=1/Rtotal. Ghar ki deewar mein brick + insulation ka R add hota hai.

Daily trick: Thandi mein metal ka darwaza haath se zyada thanda lagta hai lakdi se, kyunki metal ka K high → haath se heat jaldi kheench leta hai, halanki temp same.

Conduction via atomic vibrations collisions, Fourier Q/t=K A ΔT/L, K high metals Cu 385 Al 237 Steel 50 low insulators wood 0.12 air 0.026 W/mK. Thermal resistance Rth=L/KA series add Rtotal. Same temp metal feels colder due to high K drains heat fast.
Conduction K = KAΔT/L
Revision Points — After Diagram 4:
• Conduction: heat via vibrations no bulk motion, steady state Fourier law Q/t=KAΔT/L.
• K values: Cu ~385, Al ~237, Steel ~50, Glass ~0.8, Wood ~0.12, Air ~0.026 W/mK.
• Resistance Rth=L/KA, series Rtotal=R1+R2+R3, overall U=1/Rtotal Q/t=U A ΔT.
• Good conductor feels colder/hotter quickly due to fast heat drain/source.

5. Convection & Radiation & Newton's Cooling - Hawa Aur Bina Hawa Ke

Garam paani mein thanda paani dalo toh upar neeche ghumta hai — convection. Hot fluid less dense upar jaata hai, cold dense neeche aata hai loop banta hai. Sea breeze din mein land garam → hawa upar → sea se thandi hawa aati hai. Raat mein ulta land breeze. AC ka thanda hawa upar lagate hain taaki convection se poora room thanda ho.

Radiation mein medium nahi chahiye. Sab bodies EM waves emit karti hain temperature se. Stefan-Boltzmann: E = σ A e T⁴. σ=5.67e-8 W/m²K⁴, e emissivity 0-1 blackbody e=1 ideal emitter absorber. T⁴ se high temp pe bahut radiation — suraj ~5800K bahut radiate.

Newton's law of cooling: Garam chai room mein rakho toh thandi hone ki speed temp difference pe depend. dT/dt = -k (T - Ts). Ts surrounding temp, k constant depends on material environment. Chhota ΔT pe valid, solution T(t)=Ts+(T0-Ts)e^{-kt} exponential decay.

Kaala kapda garmi mein zyada garam kyunki e absorptivity high black absorbs more, safed reflect karta hai isliye thanda. Thermos mein shiny surface radiation rokta hai, vacuum conduction convection rokta hai.

Convection: hot less dense rises cold denser sinks buoyancy loop sea breeze day land→sea night reverse. Radiation no medium EM waves, Stefan E=σA e T⁴ T⁴ dependence, e=1 blackbody ideal. Newton cooling dT/dt=-k(T-Ts) rate∝ΔT exponential T→Ts, small ΔT valid, k empirical.
Convection & Radiation & Cooling
Revision Points — After Diagram 5:
• Convection needs fluid motion: hot ↑ less dense, cold ↓ denser, sea breeze day, land breeze night.
• Radiation: all bodies emit, E=σAeT⁴, σ=5.67e-8, black e=1 best emitter/absorber.
• Newton cooling: dT/dt=-k(T-Ts) → exponential decay T(t)=Ts+(T0-Ts)e^{-kt}, small ΔT only.
• Practical: black absorbs more heats more, white reflects, thermos uses shiny + vacuum.

6. Blackbody Laws & Greenhouse - Suraj Ka Rang

Ek ideal blackbody saari aane wali radiation absorb kar le aur sab wavelength pe emit kare. Real black nahi hota par kaala dibba chhed wala close. Uska spectrum temperature pe depend.

Wien's displacement: λm T = 2.9×10⁻³ m·K. λm peak wavelength jahan intensity max. T badhe toh λm kam → blue shift. Suraj 5800K → λm~500 nm green-yellow visible, isliye white dikhta hai. Thanda star laal, garam neela. Isi se remote temperature nikalte hain star ka colour se.

Stefan's law se total power ∝T⁴. Agar do stars same size par ek double hot toh 16 times zyada radiate. Earth ka energy balance isi se.

Greenhouse effect: Atmosphere visible ko aane deta hai, par Earth se nikalne wali IR ko CO2, water vapour absorb kar lete hain → atmosphere garam → Earth garam rehta hai warna -18°C hota ab 15°C hai. Zyada CO2 se global warming. Kaanch ke ghar mein bhi same — glass visible aane de IR roke.

Blackbody ideal e=1 perfect absorber emitter. Wien λm T=2.9e-3 mK peak shifts blue with T hot blue red cold used for star temp. Stefan total power ∝T⁴. Greenhouse atmosphere transmits visible traps IR via CO2 H2O → Earth warm 15C not -18C extra CO2 warming.
Wien λmT=2.9e-3 & Stefan & Greenhouse
Revision Points — After Diagram 6:
• Blackbody e=1 ideal, spectrum shape depends only on T.
• Wien: λm T = 2.9×10⁻³ mK, hot → peak short λ blue, cold → long λ red.
• Stefan: total E=σAeT⁴, T double → 16× power.
• Greenhouse: visible in IR trapped by CO2/H2O → warms Earth, extra CO2 → global warming, glass house same.
Gold Reference - Chapter 10 Thermal
ConceptFormula (Correct)Daily Example
Temp ScalesC/100=(F-32)/180=(K-273.15)/100 K=C+273.15 absolute zero -273.15CFever 37C=98.6F=310K
Ideal GasPV=μRT R=8.31 J/molK, PV=NkT, absolute zero extrapolation P/V→0Balloon in cold shrinks
ExpansionΔL=αLΔT β≈2α ΔA=βAΔT γ≈3α ΔV=γVΔT α steel 12e-6 Al 23e-6 liquids ~1e-4 gases 1/273, water anomaly 0-4C density max 4CRail gap, bimetal thermostat, bottle burst
Specific HeatQ=mcΔT c water 4186 J/kgK highest, C=M c, Cp=Cv+R, calorimetry lost=gained, Q=mL Lf 3.33e5 Lv 22.6e5 J/kgWater slow heat, steam burn worse
ConductionQ/t=K A ΔT/L K Cu 385 Al 237 Steel 50 Wood 0.12 Air 0.026 W/mK Rth=L/KA series addSteel spoon hot handle wood not
ConvectionHot less dense rises cold denser sinks buoyancy, sea breeze day sea→land night reverseAC top placement, sea breeze
RadiationE=σ A e T⁴ σ=5.67e-8 e 0-1 black 1, Wien λmT=2.9e-3 mK blue hot red cold, Newton dT/dt=-k(T-Ts) exponentialBlack cloth hot, thermos shiny, chai cooling
GreenhouseAtmosphere visible in IR trapped CO2 H2O → Earth 15C vs -18C without, extra CO2 warmingGlass house, global warming
Final Revision — Precise & Exam-Ready

1. Temperature & Gas: Heat = energy due to ΔT, Temp = hotness. Scales C/100=(F-32)/180=(K-273.15)/100, K=C+273.15. Absolute zero -273.15C where gas P/V→0. Ideal gas PV=μRT R=8.31. Zeroth law defines T equilibrium.

2. Expansion: ΔL=αLΔT, ΔA≈2αAΔT, ΔV≈3αVΔT for solid isotropic. α steel 12e-6 Al 23e-6 /K liquids ~1e-4 gases ~1/273. Bimetal bend due to α diff → thermostat fire alarm. Water 0→4C contracts density max 4C ice floats.

3. Heat Capacity & Phase: Q=mcΔT c water 4186 highest. Molar Cp=Cv+R. Calorimetry: heat lost = heat gained including calorimeter. Latent Q=mL temp constant during change Lf ice 3.33e5 Lv water 22.6e5 J/kg T vs Q plateau.

4. Conduction: Q/t=KAΔT/L K Cu 385 Al 237 Steel 50 Wood 0.12 Air 0.026 W/mK. Rth=L/KA series adds Rtotal=R1+R2. Metal feels colder same T due to high K drains heat.

5. Convection Radiation Cooling: Convection hot less dense ↑ cold ↓ buoyancy sea breeze day sea→land. Radiation no medium E=σAeT⁴ T⁴ strong, e=1 blackbody ideal. Wien λmT=2.9e-3 peak blue hot red cold star temp. Newton dT/dt=-k(T-Ts) small ΔT exponential T(t)=Ts+(T0-Ts)e^{-kt}. Greenhouse visible in IR trapped by CO2 H2O → Earth warm.

Numerical Checklist: C↔F↔K conversion, PV=μRT, ΔL=αLΔT, β≈2α γ≈3α, Q=mcΔT, lost=gained, Q=mL, Q/t=KAΔT/L, E=σAeT⁴, λmT=2.9e-3, dT/dt=-k(T-Ts). Units J, J/kgK, W/mK, K.
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