Class 11 Physics · Chapter 10
Short answer:
Class 11 Physics Chapter 10 "Thermal Properties of Matter" ke NCERT solutions — temperature scales, thermal expansion, calorimetry, change of state (latent heat), aur heat transfer (conduction, convection, radiation) ke saare 22 exercise questions (10.1 se 10.22) step-by-step solve kiye gaye hain, poori working ke saath. Agar tum class 11 physics ncert solutions poore syllabus ke liye dhoondh rahe ho, to ye chapter Thermodynamics aur Kinetic Theory dono ka base hai.
Chapter 10 "Thermal Properties of Matter" Class 11 Physics ki thermal-physics series ka pehla chapter hai — yahin se temperature, heat aur unke measurement ki proper neev banti hai jo aage Thermodynamics aur Kinetic Theory dono chapters me kaam aayegi. Rationalised 2026-27 syllabus me "Physical World" chapter poori tarah drop ho gaya hai, isliye ab Class 11 Physics sirf 14 chapters ka hai — Units and Measurements se Chapter 1 shuru hokar Waves tak Chapter 14 pe khatam hota hai, aur "Thermal Properties of Matter" ab Chapter 10 hai (pehle Chapter 11 hua karta tha). Agar tumhe class 11 physics all chapters pdf 2026-27 ya class 11 physics deleted syllabus 2026-27 ka poora list chahiye, to school ki latest circular ya ncert.nic.in ka current table of contents hi final source maano — is run me registry-grade cross-check hi mil paya hai, direct NCERT PDF nahi khul saka.
Is chapter me paanch bade ideas hain: (1) temperature scales — Celsius, Fahrenheit aur absolute (Kelvin) scale, (2) thermal expansion — solids, liquids aur gases ka linear/area/volume expansion, (3) specific heat capacity aur calorimetry — mix karne pe heat balance, (4) change of state — melting, boiling, latent heat aur phase diagram (triple point, critical point), aur (5) heat transfer ke teen tareeke — conduction, convection, radiation, jisme Newton's Law of Cooling bhi aata hai. Numericals is chapter me bahut hote hain, isliye unit-consistency (SI vs CGS, °C vs K) sabse zyada marks katne wali galti hai. Agar tumne pehle class 11 physics chapter 2 motion in a straight line ncert solutions ya units and measurements class 11 important questions dekhe honge, to wahi habit — har step me units likhna — yahan bhi kaam aayegi.
Chapter 10 Summary — 5 Minute Revision
Thermal Properties of Matter chapter temperature measurement (Celsius/Fahrenheit/Kelvin scales aur triple point ki idea), thermal expansion (linear α, area β=2α, volume γ=3α, aur paani ka anomalous expansion), calorimetry (Q = mcΔT aur heat balance), change of state (latent heat Q = mL, melting/boiling/sublimation, aur CO2 ka phase diagram — triple point aur critical point) aur heat transfer ke teen modes (conduction — H = KAΔT/x, convection, radiation — Stefan's law aur Newton's Law of Cooling) cover karta hai. Is chapter ki neev pe hi agla Thermodynamics chapter (first law, heat engines) aur Kinetic Theory chapter (gas laws ka microscopic explanation) khada hota hai, isliye concepts yaad rakhna zaroori hai, sirf formula ratna kaafi nahi.
In-Text Questions — Solutions
Example (Temperature scale conversion): Human body temperature 37.0°C hai. Ise Fahrenheit aur Kelvin scale me convert karo.
Fahrenheit conversion formula:
TF = (9/5)TC + 32 = (9/5)(37.0) + 32 = 66.6 + 32 = 98.6°F
Kelvin conversion:
TK = TC + 273.15 = 37.0 + 273.15 = 310.15 K
Isliye body temperature = 98.6°F = 310.15 K.
Example (Linear expansion): Ek iron rod 1.00 m lambi hai 20°C par. Isko 220°C tak garam kiya jaata hai. Agar αiron = 1.2×10-5 K-1 hai, to length me kitna increase hoga?
Formula: ΔL = L0αΔT
ΔL = 1.00 × 1.2×10-5 × (220-20) = 1.2×10-5 × 200 = 2.4×10-3 m
Naya length = 1.00 + 0.0024 = 1.0024 m (increase = 2.4 mm).
Example (Calorimetry — mixture): 200 g paani 20°C par 100 g paani 80°C wale ke saath mix kiya jaata hai. Final mixture temperature nikaalo (heat loss to surroundings ignore karo).
Heat lost by hot water = Heat gained by cold water (principle of calorimetry):
m1c(T-20) = m2c(80-T)
200(T-20) = 100(80-T)
200T - 4000 = 8000 - 100T
300T = 12000 ⟹ T = 40°C
Final mixture temperature = 40°C.
Example (Latent heat): 5 g ice, 0°C par, kitni heat lekar poori tarah paani me convert hoga (0°C par hi)? Lfusion = 336 J/g.
Formula: Q = mL
Q = 5 × 336 = 1680 J
Isliye 1680 J heat chahiye — is process me temperature nahi badhta, sirf phase change hota hai.
Example (Conduction): Ek metal slab, area 0.5 m², thickness 2 cm hai. Ek face 100°C aur doosra 20°C par hai. Agar K = 200 J s-1 m-1 K-1 hai, to steady-state heat flow rate nikaalo.
Formula: H = KAΔT/x
H = (200 × 0.5 × 80) / 0.02 = 8000/0.02 = 4,00,000 W
Heat flow rate = 4×105 W.
Example (Newton's Law of Cooling — conceptual set-up): Ek cup coffee kamre me rakhi hai jahan surrounding temperature 25°C hai. Kya coffee ka cooling rate uniform rahega poore process me? Reason do.
Nahi. Newton's Law of Cooling ke mutabik:
-dT/dt = K(T - T0)
Cooling rate body-surrounding ke temperature difference (T - T0) ke directly proportional hoti hai. Shuru me coffee bahut garam hai to (T-T0) bada hai, isliye cooling fast hoti hai. Jaise-jaise coffee thandi hoti hai, difference chhota hota jaata hai aur cooling rate progressively slow hoti jaati hai — jab tak T = T0 na ho jaaye.
Poore Class 11 Physics ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q22)
10.1 The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Neon:
TC = TK - 273.15 = 24.57 - 273.15 = -248.58°C
TF = (9/5)TC + 32 = (9/5)(-248.58) + 32 = -447.44 + 32 = -415.44°F
Carbon dioxide:
TC = 216.55 - 273.15 = -56.60°C
TF = (9/5)(-56.60) + 32 = -101.88 + 32 = -69.88°F
10.2 Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB?
Triple point of water on Kelvin scale = 273.16 K. Har scale me ek degree ka size, us scale ki triple-point reading se fix hota hai:
1°A = 273.16/200 K, aur 1°B = 273.16/350 K
Kisi bhi actual temperature (Kelvin me T) ko dono scales pe likhne par:
T = TA×(273.16/200) = TB×(273.16/350)
⟹ TA/200 = TB/350 ⟹ TA = (4/7)TB
10.3 The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law R = R0[1 + α(T - T0)]. The resistance is 101.6 Ω at the triple-point of water (273.16 K) and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?
Pehle α nikalte hain lead ke melting point wali reading se:
165.5 = 101.6[1 + α(600.5 - 273.16)]
165.5/101.6 = 1.6289 = 1 + 327.34α ⟹ α = 0.6289/327.34 = 1.922×10-3 K-1
Ab R = 123.4 Ω ke liye T solve karo:
123.4 = 101.6[1 + α(T - 273.16)]
1.2146 - 1 = α(T-273.16) ⟹ T - 273.16 = 0.2146/1.922×10-3 = 111.65
T ≈ 384.8 K
10.4 (a) Why the triple-point of water is a standard fixed point in modern thermometry rather than the ice and steam point? (b) There were two fixed points in the original Celsius scale, which were assigned the number 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale? (c) The absolute temperature (Kelvin scale) T is related to the temperature tc on the Celsius scale by tc = T - 273.15. Why do we have 273.15 in this relation, and not 273.16?
(a) Ice-point aur steam-point pressure ke saath badalte hain aur exactly reproduce karna mushkil hai (impurities, pressure-dependence). Triple-point sirf ek unique pressure aur temperature pe hota hai (solid, liquid, vapour teeno equilibrium me) — isliye ye ek zyada precise, reproducible fixed point hai jo pressure pe depend nahi karta.
(b) Doosra fixed point absolute zero (0 K) hai — jahan ideal gas ka pressure theoretically zero ho jaata hai.
(c) 273.16 K sirf water ke triple point ki value hai. Water ka normal freezing/melting point (1 atm pressure par) triple-point se thoda kam, 273.15 K hota hai (kyunki pressure zyada hone se freezing point thoda gir jaata hai). Isliye °C scale me 0°C ko normal ice-point se define kiya jaata hai, jo 273.15 K ke barabar hai — na ki triple-point (273.16 K) ke.
10.5 Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made — Pressure at triple-point of water: A = 1.250×105 Pa, B = 0.200×105 Pa. Pressure at normal melting point of sulphur: A = 1.797×105 Pa, B = 0.287×105 Pa. (a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B? (b) What do you think is the reason for slightly different answers from A and B?
Constant-volume gas thermometer formula:
T = 273.16 × (P/Ptriple)
Thermometer A:
TA = 273.16 × (1.797×105/1.250×105) = 273.16 × 1.4376 = 392.69 K
Thermometer B:
TB = 273.16 × (0.287×105/0.200×105) = 273.16 × 1.435 = 391.98 K
(b) Oxygen aur hydrogen dono perfectly ideal gas nahi hain, isliye alag readings aati hain. Agar dono thermometers ke readings ko zero-pressure limit tak extrapolate karein, to dono ki values same aa jaayengi (ideal-gas limit).
10.6 A steel tape 1 m long is correctly calibrated for a temperature of 27.0°C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0°C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when temperature is 27.0°C? (αsteel = 1.20×10-5 K-1).
45°C par tape khud bhi expand ho chuki hai, isliye har "1 cm" mark actually thoda zyada lambi ho gayi hai:
Fractional expansion = αΔT = 1.20×10-5 × 18 = 2.16×10-4
Actual length at 45.0°C:
L = 63.0 × (1 + 2.16×10-4) = 63.0136 cm ≈ 63.0136 cm
Ab rod khud bhi steel ka hai, isliye 45°C se 27°C tak thanda hone par usi fraction se sikudega:
L27°C = 63.0136 / (1 + 2.16×10-4) ≈ 63.0000 cm
Length at 27.0°C = 63.0 cm — same as calibration temperature, kyunki tape aur rod dono steel ke hone se expansion effect self-consistent ho jaata hai.
10.7 A large steel wheel is to be fitted onto a shaft of the same material. At 27°C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using a 'dry ice' (solid CO2) bath. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of steel is constant over the required temperature range: αsteel = 1.20×10-5 K-1.
Shaft (8.70 cm) ko cool karke wheel ke hole (8.69 cm) jitna sikudna hai, yaani diameter 0.01 cm ghatana hai:
ΔD = D0αΔT ⟹ ΔT = ΔD/(D0α) = 0.01/(8.70 × 1.20×10-5)
ΔT = 0.01/(1.044×10-4) ≈ 95.8 K
Shaft ko isliye 27°C se lagbhag 95.8 K neeche thanda karna hoga:
T = 27 - 95.8 ≈ -68.8°C ≈ -69°C
Dry ice se ye temperature achieve karna practically possible hai.
10.8 A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C? (αcopper = 1.70×10-5 K-1)
Hole solid material jaisa hi expand karta hai (hole "missing material" nahi, balki wahi expand hota hai jo sheet ka material karta):
ΔD = D0αΔT = 4.24 × 1.70×10-5 × 200
ΔD = 4.24 × 3.4×10-3 = 1.44×10-2 cm (hole bada hoga).
10.9 A brass wire 1.8 m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of -39°C, what is the tension developed in the wire, if its diameter is 2.0 mm? (αbrass = 2.0×10-5 K-1, Ybrass = 0.91×1011 Pa)
Wire cool hokar sikudna chahta hai but rigid supports rokte hain — isliye thermal strain hi mechanical strain ban jaata hai:
Strain = αΔT = 2.0×10-5 × 66 = 1.32×10-3
Stress = Y × strain = 0.91×1011 × 1.32×10-3 = 1.20×108 Pa
Area = πr² = π×(1.0×10-3)² = 3.14×10-6 m²
Tension F = Stress × Area = 1.20×108 × 3.14×10-6 ≈ 377 N
10.10 A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C? Is there a thermal stress developed at the junction? The ends of the rod are free to expand. (αbrass = 2.0×10-5 K-1, αsteel = 1.2×10-5 K-1)
ΔT = 250 - 40 = 210°C
ΔLbrass = 50 × 2.0×10-5 × 210 = 0.21 cm
ΔLsteel = 50 × 1.2×10-5 × 210 = 0.126 cm
Total ΔL = 0.21 + 0.126 = 0.336 cm
Kyunki dono rods ke ends free hain expand karne ke liye, koi bhi rod doosre ki expansion ko rokta nahi — isliye junction pe koi thermal stress develop nahi hota.
10.11 The coefficient of volume expansion of glycerine is 49×10-5 K-1. What is the fractional change in its density for a 30°C rise in temperature?
Density aur volume inversely related hain, mass constant. Volume expansion se density ka fractional decrease approx γΔT hota hai:
Δρ/ρ ≈ -γΔT = -49×10-5 × 30 = -1.47×10-2
Yaani density lagbhag 1.47% kam ho jaati hai.
10.12 A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g-1 K-1.
Total energy delivered by machine in 2.5 min = 150 s:
Etotal = 10,000 W × 150 s = 1.5×106 J
50% block ko milta hai:
Q = 0.5 × 1.5×106 = 7.5×105 J
Mass = 8.0 kg = 8000 g. Q = mcΔT se:
ΔT = Q/(mc) = 7.5×105/(8000 × 0.91) = 7.5×105/7280 ≈ 103°C
10.13 A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g-1 K-1; heat of fusion of water = 335 J g-1)
Copper 500°C se 0°C tak thanda hote hue heat deta hai:
Q = mcΔT = 2500 g × 0.39 × 500 = 4,87,500 J
Ye heat ice ko melt karne me use hoti hai (Q = mL):
mice = Q/L = 4,87,500/335 ≈ 1455 g ≈ 1.45 kg
10.14 In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm³ of water at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value?
Heat lost by metal = heat gained by (water + calorimeter):
mmetalcmetal(150-40) = (mwater + w)cwater(40-27)
0.20 × cmetal × 110 = (0.150 + 0.025) × 4186 × 13
22 cmetal = 0.175 × 4186 × 13 = 9523 J
cmetal = 9523/22 ≈ 433 J kg-1 K-1
Agar surroundings me bhi heat loss ho raha hai, to metal ne actually usse zyada heat kho di hai jitni sirf paani+calorimeter ne li — matlab actual heat loss zyada hai. Fixed ΔT (110°C) ke saath zyada heat ke liye zyada cmetal chahiye. Isliye is calculation se mila jawaab actual value se chhota (underestimate) hoga.
10.15 Given below are observations on molar specific heats at room temperature of some gases. Predict the molar specific heat of a monatomic gas from kinetic theory (Cv = (3/2)R) and compare with experimental values. Also comment on the agreement for diatomic and triatomic gases.
Kinetic theory (equipartition of energy) se monatomic gas ke degrees of freedom = 3 (sirf translational):
Cv = (3/2)R = (3/2)(8.31) = 12.47 J mol-1 K-1 — ye He, Ne, Ar jaise monatomic gases ke experimental values se bahut acchi tarah match karta hai.
Diatomic gas me 2 rotational DOF extra jud jaate hain (total 5):
Cv = (5/2)R ≈ 20.8 J mol-1 K-1
— ye O2, N2 jaise gases se reasonably match karta hai.
Triatomic (non-linear) gas me 6 DOF (3 trans + 3 rot):
Cv = 3R ≈ 24.9 J mol-1 K-1
— par experimental values isse aksar zyada milte hain, kyunki simple equipartition vibrational degrees of freedom ko account nahi karta, jo polyatomic molecules me room temperature pe bhi kaafi active ho jaate hain, isliye disagreement dikhta hai.
10.16 Answer the following questions based on the P-T phase diagram of CO2: (a) At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in equilibrium? (b) What is the effect of a decrease of pressure on the fusion and boiling point of CO2? (c) What are the critical temperature and pressure for CO2? What is the significance of the critical point? (d) Is CO2 solid, liquid or gas at (i) -70°C under 1 atm, (ii) -60°C under 10 atm, (iii) 15°C under 56 atm?
(a) Teeno phases sirf triple point par coexist karte hain: T ≈ -56.6°C, P ≈ 5.11 atm.
(b) CO2 ki fusion curve ka slope (water ke ulta) positive hota hai — isliye pressure kam karne se dono fusion aur boiling point kam ho jaate hain.
(c) Critical temperature ≈ 31.1°C, critical pressure ≈ 73.0 atm. Is point ke upar liquid aur gas phase me koi distinction nahi rehta — sirf pressure badhaakar gas ko liquefy nahi kar sakte.
(d) (i) -70°C, 1 atm → vapour (triple point se neeche temperature, low pressure). (ii) -60°C, 10 atm → solid (high pressure, low temp region). (iii) 15°C, 56 atm → liquid (triple point aur critical point ke beech ka region).
10.17 Answer the following questions based on the P-T phase diagram of CO2: (a) CO2 at 1 atm pressure and temperature -60°C is compressed isothermally. Does it go through a liquid phase? (b) What happens when CO2 at 4 atm pressure is cooled from room temperature at constant pressure? (c) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm pressure and temperature -65°C as it is heated up to room temperature at constant pressure. (d) CO2 is heated to a temperature 70°C and compressed isothermally. What changes in its properties do you expect to observe?
(a) Nahi. -60°C, triple-point temperature (-56.6°C) se neeche hai, isliye is isotherm par liquid phase kabhi accessible nahi hota — compression seedha vapour se solid me le jaayega (sublimation curve cross karke).
(b) 4 atm, triple-point pressure (5.11 atm) se kam hai, isliye is isobar par cooling se gas seedha solid ban jaayega (liquid phase bypass hoga — sublimation).
(c) 10 atm, triple-point pressure se zyada hai. Heating karne par solid pehle melt hoga (fusion curve cross) liquid banega, phir aur heat karne par boil hoga (vaporisation curve cross) aur vapour ban jaayega — normal solid→liquid→vapour sequence.
(d) 70°C, critical temperature (31.1°C) se upar hai — CO2 ab supercritical fluid state me hai. Isothermal compression karne par koi phase transition/liquefaction nahi hoga, koi meniscus (liquid-gas boundary) nahi dikhega — density continuously badhegi bina kisi sharp phase-change ke.
10.18 A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 minutes, what is the average rate of extra evaporation caused by the drug? Assume evaporation is the only heat-loss mechanism. Mass of child = 30 kg. Specific heat of body ≈ that of water; latent heat of evaporation of water at that temperature ≈ 580 cal g-1.
Temperature fall:
ΔT = 101°F - 98°F = 3°F = 3×(5/9)°C = 5/3°C
Heat lost by body:
Q = mcΔT = 30,000 g × 1 cal g-1°C-1 × (5/3) = 50,000 cal
Mass of sweat evaporated:
m = Q/L = 50,000/580 ≈ 86.2 g
Average rate = 86.2 g / 20 min ≈ 4.31 g/min
10.19 A 'thermacole' icebox is a cheap and efficient method for storing small quantities of food in hot climates. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 hours. The outside temperature is 45°C, and the coefficient of thermal conductivity of thermacole = 0.01 J s-1 m-1 K-1. (Heat of fusion of water = 335×103 J kg-1)
Total surface area of cube (6 faces):
A = 6 × (0.30)² = 0.54 m²
Heat flow rate (conduction, ΔT = 45-0 = 45°C, x = 0.05 m):
H = KAΔT/x = (0.01 × 0.54 × 45)/0.05 = 4.86 W
Total heat entering in 6 h = 6×3600 = 21,600 s:
Q = 4.86 × 21,600 ≈ 1.05×105 J
Ice melted:
m = Q/L = 1.05×105/3.35×105 ≈ 0.313 kg
Ice remaining = 4.0 - 0.313 ≈ 3.69 kg
10.20 A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. (Thermal conductivity of brass = 109 J s-1 m-1 K-1; heat of vaporisation of water = 2256×103 J kg-1)
Rate of heat needed to vaporise water:
dQ/dt = (6.0/60) kg/s × 2256×103 J/kg = 0.1 × 2256000 = 2,25,600 W
Conduction equation (Twater side = 100°C):
dQ/dt = KA(T1-100)/x
2,25,600 = (109 × 0.15 × (T1-100))/0.01
2,25,600 = 1635 × (T1-100) ⟹ T1-100 = 138.0
T1 ≈ 238°C
10.21 Explain why: (a) a body with large reflectivity is a poor emitter, (b) a brass tumbler feels much colder than a wooden tray on a chilly day, (c) an optical pyrometer calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives correct value when the piece is in the furnace, (d) the earth without its atmosphere would be inhospitably cold, (e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.
(a) Good reflector = poor absorber (kyunki opaque body me reflectivity+absorptivity ≈ 1). Kirchhoff's law ke mutabik achha absorber hi achha emitter hota hai — isliye poor absorber, poor emitter bhi hoga.
(b) Brass ek achha thermal conductor hai, isliye haath ki garmi jaldi kheech leta hai (fast conduction away se cold feel hota hai). Wood poor conductor hai, isliye haath ki garmi utni jaldi nahi kheenchti — dono actually same ambient temperature par hain, sirf conductivity ka farak feel hota hai.
(c) Pyrometer ideal blackbody ke liye calibrated hai. Open me red-hot iron ki emissivity 1 se kam hai (perfect black body nahi), isliye ye kam radiate karta hai aur pyrometer temperature underestimate karta hai. Furnace ke andar cavity multiple reflections se effectively blackbody jaisi radiation deti hai, isliye reading sahi aati hai.
(d) Atmosphere (greenhouse gases jaise CO2, water vapour) outgoing infrared radiation ko trap karti hai, jisse surface garam rehta hai. Atmosphere ke bina saari absorbed heat raat me freely escape ho jaayegi, jaise Moon par extreme din-raat temperature variation hoti hai.
(e) Steam, radiator me condense hote waqt apni bahut badi latent heat of vaporisation release karti hai (sirf temperature difference se milne wali heat se kahin zyada), isliye per kg steam se hot water ke comparison me zyada heat delivered hoti hai — heating zyada efficient hota hai.
10.22 A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C.
Newton's Law of Cooling (average-temperature form):
(T1-T2)/t = K[(T1+T2)/2 - T0]
First case (80°C → 50°C, 5 min, T0=20°C):
(80-50)/5 = K[(80+50)/2 - 20] ⟹ 6 = K×45 ⟹ K = 0.1333 min-1
Second case (60°C → 30°C):
(60-30)/t = K[(60+30)/2 - 20] = 0.1333×25 = 3.333
t = 30/3.333 ≈ 9.0 minutes
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| Celsius ↔ Fahrenheit | TF = (9/5)TC + 32 |
| Celsius ↔ Kelvin | TK = TC + 273.15 |
| Ideal gas thermometer | T = 273.16 × (P/Ptriple) |
| Linear expansion | ΔL = L0αΔT (α = coefficient of linear expansion) |
| Area (superficial) expansion | ΔA = A0βΔT, jahan β = 2α |
| Volume (cubical) expansion | ΔV = V0γΔT, jahan γ = 3α |
| Fractional density change | Δρ/ρ ≈ -γΔT |
| Specific heat capacity | Q = mcΔT |
| Molar specific heat | C = Q/(nΔT); monatomic Cv=(3/2)R, diatomic Cv=(5/2)R |
| Latent heat | Q = mL (L = latent heat of fusion/vaporisation) |
| Calorimetry (heat balance) | Heat lost = Heat gained (no external loss) |
| Conduction (steady state) | H = dQ/dt = KAΔT/x (K = thermal conductivity) |
| Thermal stress (constrained rod) | Stress = Y × αΔT (Y = Young's modulus) |
| Newton's Law of Cooling | -dT/dt = K(T - T0); average form: (T1-T2)/t = K[(T1+T2)/2 - T0] |
| Stefan's Law (radiation) | E = σT4 (per unit area, ideal blackbody) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Temperature ko Celsius aur Kelvin me mix kar dena — ΔT expansion/heat formulas me Celsius aur Kelvin difference same hota hai, par absolute temperature (jaise gas thermometer formula T = 273.16×P/Ptriple) hamesha Kelvin me hi lena hai, Celsius me nahi.
- Area expansion me β = 2α aur volume expansion me γ = 3α bhool jaana — students seedha α use kar dete hain jabki question area ya volume expansion maang raha ho.
- Thermal stress wale sawaal (jaise 10.9) me sirf strain (αΔT) likh kar chhod dena — stress nikalne ke liye Young's modulus se multiply karna zaroori hai, aur phir area se multiply karke hi tension/force milega.
- Calorimetry problems me calorimeter ka 'water equivalent' add karna bhool jaana — sirf paani ki heat balance likh dena galat answer dega jab question me calorimeter ka water equivalent diya ho.
- Latent heat wale sawaal me phase-change ke dauran temperature change bhi jod dena — jab tak phase change chal raha hai (melting/boiling), temperature constant rehta hai, sirf Q = mL lagta hai, Q = mcΔT nahi.
- Newton's Law of Cooling me directly (T1-T2)/t = K(T-T0) me single T daal dena instead of average temperature (T1+T2)/2 — is chapter ke numerical (jaise 10.22) me ye sabse common calculation error hai.
Board-Style Important Questions
- Assertion-reason ya MCQ style sawaal aksar poochta hai ki area expansion coefficient (β) aur volume expansion coefficient (γ) ka linear expansion coefficient (α) se kya ratio hai — β=2α, γ=3α yaad rakhna zaroori hai.
- Reasoning-based sawaal — 'brass tumbler chilly day pe wooden tray se zyada thanda kyun lagta hai' jaisa conceptual conduction-based explanation (Q10.21(b) type) frequently repeat hota hai.
- Numerical based on calorimetry (heat lost = heat gained) with calorimeter's water equivalent included — Q10.14 jaisa pattern bahut common hai.
- Newton's Law of Cooling numerical jahan do temperature-drop cases diye jaate hain aur second case ka time nikaalna hota hai — Q10.22 jaisa pattern.
- Long-answer/derivation type — thermal expansion (linear/area/volume relation derive karna) ya conduction ka steady-state heat flow formula derive karke numerical solve karna, jaisa 10.19/10.20 pattern.
- CO2 ka P-T phase diagram diya jaakar triple point, critical point, aur alag-alag conditions me phase identify karne wala multi-part sawaal (Q10.16/10.17 jaisa) — is chapter ka sabse conceptually heavy pattern hai.
Aksar Poochhe Jaane Wale Sawaal
Class 11 Physics ke saare chapters ek jagah (all chapters pdf 2026-27) kahan se milenge?
NCERT ki official site ncert.nic.in par 2026-27 session ke liye current textbook PDF milta hai — wahi final source hai class 11 physics all chapters pdf 2026-27 ke liye. Rationalised syllabus me ab Physics Part 1 me Units and Measurements se Gravitation (Ch.1-7, mechanics) aur Part 2 me Mechanical Properties of Solids se Waves (Ch.8-14, thermal/oscillations/waves) — dono parts milaakar 14 chapters hain. Ye sirf printing/binding split hai, syllabus unit ka distinction nahi.
Is chapter ka board exam me weightage kitna hota hai?
Exact chapter-wise weightage har saal thodi vary kar sakti hai aur ye 11th ki school-level exams par depend karta hai (11th boards nahi karwata, sirf 12th karwata hai). Isliye koi specific number claim karne se better hai ki current session ki official curriculum document ya apne school ke exam pattern circular se confirm kar lo.
Thermal Properties of Matter ka Thermodynamics aur Kinetic Theory se kya connection hai?
Ye teeno chapters ek series hain. Thermal Properties me temperature scales, expansion, calorimetry aur latent heat cover hota hai; agla chapter thermodynamics class 11 physics notes wale topics — first law, heat engines, entropy jaisi macroscopic ideas — cover karta hai; aur uske baad kinetic theory of gases class 11 derivation wala chapter isi heat aur temperature ko molecule-level pe explain karta hai (pressure, temperature ka microscopic origin).
Specific heat capacity aur latent heat me kya farak hai — dono formulas confuse ho jaate hain?
Specific heat capacity (Q = mcΔT) tab use hoti hai jab temperature change ho raha ho lekin phase same rahe (jaise paani 20°C se 40°C tak garam hona). Latent heat (Q = mL) tab use hoti hai jab phase change ho raha ho (ice se water, ya water se steam) — is dauran temperature constant rehta hai, sirf state badalta hai. Dono ko ek hi sawaal me mix karna hai to alag-alag steps me calculate karna padta hai.
NCERT Exemplar is chapter ke liye kaafi hai ya extra practice chahiye?
NCERT textbook ke 22 exercise questions concept-clarity ke liye kaafi solid base dete hain, lekin higher-order numericals (jaise phase-diagram based multi-part questions) ke liye NCERT Exemplar practice karna helpful hota hai. Agar tumne pehle laws of motion class 11 ncert exemplar ya gravitation class 11 numericals with solutions jaise chapters exemplar se practice kiye hain, to wahi habit yahan bhi continue karo.
Chapter shuru karne se pehle konse pichhle chapters revise karne chahiye?
Units and Measurements (SI units, significant figures) aur Mechanical Properties of Solids (stress-strain, Young's modulus) dono directly is chapter me use hote hain — jaise thermal stress wale sawaal (10.9, 10.10) me Young's modulus chahiye hi hoga. Agar tumne pehle class 11 physics chapter 2 motion in a straight line ncert solutions type numericals practice kiye hain, to unit-consistency ki wahi habit yahan calorimetry aur conduction ke numericals me bhi kaam aayegi.
Class 11 Physics — Saare Chapters
Board exam tak sirf revision karna hai?
Class 11 Physics ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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