NCERT Solutions for Class 11 Physics Chapter 10: Thermal Properties of Matter

These Class 11 Physics Chapter 10 solutions cover Thermal Properties of Matter with complete, step-by-step answers to every NCERT exercise (10.1–10.20). Each numerical is solved with full working, correct units and a verified final answer, while the conceptual questions are explained in clear, exam-ready language. The chapter deals with temperature and heat, thermal expansion, specific heat capacity, calorimetry, change of state, latent heat, the three modes of heat transfer and Newton’s law of cooling.

Class: 11 Subject: Physics Chapter: 10 Title: Thermal Properties of Matter Exercises: 10.1–10.20 Session: 2026–27
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Class 11 Physics Chapter 10 Solutions – Overview

Chapter 10, Thermal Properties of Matter, builds a careful physical picture of heat and temperature. It begins by defining heat as energy transferred between bodies due to a temperature difference, and temperature as a measure of hotness, measured on the Celsius, Fahrenheit and Kelvin scales. Using the ideal-gas equation PV = µRT, it introduces absolute temperature and absolute zero (−273.15 °C). The chapter then studies thermal expansion (linear, area and volume, with αv = 3αl) and the anomalous expansion of water, followed by specific heat capacity and calorimetry based on “heat lost = heat gained.” Next come change of state, melting and boiling points, the triple point and latent heat (Q = mL). Finally it explains the three modes of heat transfer — conduction, convection and radiation — including the conduction equation, Stefan–Boltzmann law, Wien’s displacement law and Newton’s law of cooling.

Key Concepts & Definitions

Heat: the form of energy transferred between two systems (or a system and its surroundings) by virtue of a temperature difference. SI unit: joule (J).

Temperature: a measure of the degree of hotness of a body. SI unit: kelvin (K); commonly measured in °C.

Absolute zero: −273.15 °C (= 0 K), the temperature at which an ideal gas would exert zero pressure; the lowest possible temperature.

Thermal expansion: the increase in the dimensions of a body on heating — linear (Δl/l = αlΔT), area (ΔA/A = 2αlΔT) and volume (ΔV/V = αvΔT), with αv = 3αl.

Specific heat capacity (s): the heat required to raise the temperature of unit mass of a substance by one degree; SI unit J kg−1 K−1.

Calorimetry: measurement of heat, based on the principle that in an isolated system heat lost by the hot body equals heat gained by the cold body.

Latent heat (L): heat per unit mass absorbed or released during a change of state at constant temperature; Q = mL (Lf for fusion, Lv for vaporisation).

Modes of heat transfer: conduction (in solids, by molecular collisions), convection (bulk motion of a fluid) and radiation (electromagnetic waves, needs no medium).

Important Formulas

Temperature scales: (tF − 32)/180 = tC/100, i.e. tF = (9/5)tC + 32;   T = tC + 273.15

Ideal-gas equation: PV = µRT, with R = 8.31 J mol−1 K−1

Linear / area / volume expansion: Δl/l = αlΔT;   ΔA/A = 2αlΔT;   ΔV/V = αvΔT;   αv = 3αl

Thermal stress: F/A = Y × (Δl/l) = Y αl ΔT

Heat / specific heat: ΔQ = m s ΔT;   latent heat: Q = mL

Conduction (heat current): H = KA(TC − TD)/L

Stefan–Boltzmann law: H = eσA(T4 − Ts4), σ = 5.67 × 10−8 W m−2 K−4

Wien’s law: λmT = 2.9 × 10−3 m K

Newton’s law of cooling: −dQ/dt ∝ (T2 − T1)

NCERT Exercise Solutions (10.1–10.20)

Questions are reproduced verbatim from the NCERT textbook (Reprint 2026–27). All numerical answers are worked out step by step and cross-checked against the NCERT answer key.

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.

ANSWER Use tC = T − 273.15 and tF = (9/5)tC + 32. Neon: tC = 24.57 − 273.15 = −248.58 °C.   tF = (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. Neon: −248.58 °C = −415.44 °F;   CO2: −56.60 °C = −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?

ANSWER On an absolute scale the size of one degree is fixed by assigning a number to the triple point of water, which is one fixed reference temperature. So the same physical temperature reads 200 on scale A and 350 on scale B. Hence the ratio of the unit sizes gives, for any temperature, TA/200 = TB/350. Therefore TA = (200/350) TB = (4/7) TB.

10.3 The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = Ro [1 + α (T − To)]. 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?

ANSWER Take To = 273.16 K, Ro = 101.6 Ω. From the law, R = Ro[1 + α(T − To)]. First find α using the lead point: 165.5 = 101.6 [1 + α(600.5 − 273.16)]. 165.5/101.6 = 1 + α(327.34) ⇒ 1.6289 − 1 = 327.34α ⇒ α = 0.6289/327.34 = 1.9213 × 10−3 K−1. Now for R = 123.4 Ω: 123.4 = 101.6 [1 + α(T − 273.16)] ⇒ 123.4/101.6 = 1 + α(T − 273.16). 1.2146 − 1 = (1.9213 × 10−3)(T − 273.16) ⇒ T − 273.16 = 0.2146/1.9213 × 10−3 = 111.69 K. T = 273.16 + 111.69 = 384.8 K (approx.).

10.4 Answer the following: (a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)? (b) There were two fixed points in the original Celsius scale as mentioned above 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 absolute 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? (d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

ANSWER (a) The triple point of water (the unique temperature and pressure at which ice, water and vapour coexist) occurs at one definite temperature, 273.16 K, independent of external conditions. The melting point of ice and the boiling point of water both change with pressure, so they are not unique and cannot serve as truly fixed reference points. (b) The other fixed point on the Kelvin scale is absolute zero (0 K) itself. (c) On the Celsius scale, 0 °C is taken as the melting point of ice, which is 273.15 K. The triple point of water is 273.16 K, i.e. 0.01 °C. Since the relation connects the Celsius zero (ice point) to Kelvin, we use 273.15, not 273.16. (d) The Fahrenheit interval is 5/9 of the Celsius (Kelvin) interval, so 1 K = (9/5) Fahrenheit-sized degrees. Triple point on this scale = 273.16 × (9/5) = 491.69.

10.5 Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

TemperaturePressure thermometer APressure thermometer B
Triple-point of water1.250 × 105 Pa0.200 × 105 Pa
Normal melting point of sulphur1.797 × 105 Pa0.287 × 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 behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

ANSWER (a) For a constant-volume gas thermometer, T ∝ P, so T = 273.16 × (P/Ptr). 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) The small difference arises because oxygen and hydrogen are not perfectly ideal gases at the pressures used. To reduce the discrepancy, the measurements should be repeated at lower and lower pressures, and the temperature plotted against the triple-point pressure should be extrapolated to the limit of zero pressure, where all gases behave ideally and give the same reading.

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 the temperature is 27.0 °C? Coefficient of linear expansion of steel = 1.20 × 10−5 K−1.

ANSWER At 45 °C the tape itself has expanded, so each centimetre mark is now longer than 1 cm. ΔT = 45 − 27 = 18 °C. Length of 1 tape-division at 45 °C = 1 cm × (1 + αΔT) = 1 × (1 + 1.20 × 10−5 × 18) = 1.000216 cm. Actual length of the rod = (reading) × (length per division) = 63.0 × 1.000216 = 63.0136 cm. Actual length at 45 °C = 63.0 + 0.0136 = 63.0136 cm (to 3 significant figures, the change is 0.0136 cm and the total reads 63.0 cm). At 27 °C the tape is correctly calibrated, so the steel rod reads its true length: length at 27 °C = 63.0 cm.

10.7 A large steel wheel is to be fitted on to 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 ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: αsteel = 1.20 × 10−5 K−1.

ANSWER The shaft must shrink until its diameter equals the wheel-hole diameter, 8.69 cm. Let T2 be the required temperature; T1 = 27 °C = 300 K. d2 = d1[1 + α(T2 − T1)] ⇒ 8.69 = 8.70[1 + 1.20 × 10−5(T2 − 27)]. 8.69/8.70 = 1 + 1.20 × 10−5(ΔT) ⇒ 0.998851 − 1 = 1.20 × 10−5(ΔT). ΔT = (−1.1494 × 10−3)/(1.20 × 10−5) = −95.8 °C. T2 = 27 − 95.8 = −68.8 °C ≈ −69 °C. The wheel slips on when the shaft is cooled to about −69 °C.

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? Coefficient of linear expansion of copper = 1.70 × 10−5 K−1.

ANSWER A hole expands just like a solid disc of the same material, so Δd = d αl ΔT. ΔT = 227 − 27 = 200 °C. Δd = 4.24 × 1.70 × 10−5 × 200 = 4.24 × 3.40 × 10−3 = 0.0144 cm. The diameter increases by 1.44 × 10−2 cm.

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? Co-efficient of linear expansion of brass = 2.0 × 10−5 K−1; Young’s modulus of brass = 0.91 × 1011 Pa.

ANSWER On cooling, the wire tries to contract but the rigid supports prevent it, so a tensile (thermal) stress develops. ΔT = (−39) − 27 = −66 °C; magnitude 66 °C. Thermal strain = α|ΔT| = 2.0 × 10−5 × 66 = 1.32 × 10−3. Cross-sectional area A = πr2; r = 1.0 mm = 1.0 × 10−3 m, so A = π(1.0 × 10−3)2 = 3.142 × 10−6 m2. Tension F = Y × strain × A = (0.91 × 1011)(1.32 × 10−3)(3.142 × 10−6). F = 0.91 × 1.32 × 3.142 × 1011−3−6 = 3.774 × 102 N. Tension developed ≈ 3.8 × 102 N (about 380 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 (Co-efficient of linear expansion of brass = 2.0 × 10−5 K−1, steel = 1.2 × 10−5 K−1).

ANSWER ΔT = 250 − 40 = 210 °C; each rod length L = 50 cm = 0.50 m. Brass: Δlbrass = L αbrass ΔT = 0.50 × 2.0 × 10−5 × 210 = 2.1 × 10−3 m = 0.21 cm. Steel: Δlsteel = L αsteel ΔT = 0.50 × 1.2 × 10−5 × 210 = 1.26 × 10−3 m = 0.126 cm ≈ 0.13 cm. Total change in length = 0.21 + 0.126 = 0.34 cm (an expansion). Since the ends are free to expand, each rod expands without restriction, so no thermal stress is developed at the junction.

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?

ANSWER Density ρ = m/V. As temperature rises, mass is constant but volume increases by ΔV = V αv ΔT, so density decreases. Fractional change in density |Δρ/ρ| = αv ΔT (for small changes). |Δρ/ρ| = 49 × 10−5 × 30 = 1470 × 10−5 = 0.0147. Fractional change in density = 0.0147 ≈ 1.5 × 10−2 (a decrease).

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.

ANSWER Power P = 10 kW = 104 W; time t = 2.5 min = 150 s. Total energy = P × t = 104 × 150 = 1.5 × 106 J. Only 50% heats the block: Q = 0.5 × 1.5 × 106 = 7.5 × 105 J. s = 0.91 J g−1 K−1 = 910 J kg−1 K−1; m = 8.0 kg. ΔT = Q/(ms) = (7.5 × 105)/(8.0 × 910) = (7.5 × 105)/7280 = 103 °C. Rise in temperature ≈ 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).

ANSWER Maximum ice melts when the copper cools all the way to 0 °C, giving up all its heat to the ice. Heat released by copper Q = m s ΔT = 2.5 kg × 390 J kg−1 K−1 × (500 − 0) °C. Q = 2.5 × 390 × 500 = 4.875 × 105 J. Ice melted mice = Q/Lf = (4.875 × 105 J)/(335 × 103 J kg−1) = 1.455 kg. Maximum mass of ice that can melt ≈ 1.5 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 for specific heat of the metal?

ANSWER Mass of water = 150 cm3 × 1 g cm−3 = 0.150 kg; water equivalent of calorimeter = 0.025 kg, so total water-equivalent = 0.150 + 0.025 = 0.175 kg. swater = 4186 J kg−1 K−1. Heat lost by metal = m s (150 − 40) = 0.20 × s × 110. Heat gained by water + calorimeter = 0.175 × 4186 × (40 − 27) = 0.175 × 4186 × 13 = 9523.15 J. Heat lost = heat gained: 0.20 × s × 110 = 9523.15 ⇒ 22s = 9523.15. s = 9523.15/22 = 432.9 J kg−1 K−1 = 0.43 J g−1 K−1. If heat losses to the surroundings are not negligible, some of the metal’s heat escapes instead of warming the water, so the calculated specific heat would be smaller than the actual value.

10.15 Given below are observations on molar specific heats at room temperature of some common gases.

GasMolar specific heat (Cv) (cal mol−1 K−1)
Hydrogen4.87
Nitrogen4.97
Oxygen5.02
Nitric oxide4.99
Carbon monoxide5.01
Chlorine6.17

The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

ANSWER All the gases listed are diatomic. Besides the three translational degrees of freedom that a monatomic gas has, a diatomic molecule also has rotational degrees of freedom. To raise its temperature, heat must be shared among more modes of motion, so its molar specific heat is greater than that of a monatomic gas. Considering only translational and rotational modes, a diatomic gas has Cv ≈ (5/2)R ≈ 4.97 cal mol−1 K−1, which agrees well with the values for all the gases except chlorine. The higher value for chlorine shows that, in addition to rotational modes, vibrational modes of motion are also active in chlorine at room temperature, contributing extra to its specific heat.

10.16 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 the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g−1.

ANSWER Temperature drop = 101 − 98 = 3 °F. Convert to Celsius: ΔT = 3 × (5/9) = 1.667 °C (= 1.667 K). Heat lost by body Q = m s ΔT = 30000 g × 1 cal g−1 K−1 × 1.667 = 30000 × 1.667 = 5.0 × 104 cal. Mass of sweat evaporated m = Q/L = (5.0 × 104 cal)/(580 cal g−1) = 86.2 g. Average rate of extra evaporation = m/time = 86.2 g / 20 min = 4.3 g min−1. Average rate of extra evaporation ≈ 4.3 g/min.

10.17 A ‘thermacole’ icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. 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 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is 0.01 J s−1 m−1 K−1. [Heat of fusion of water = 335 × 103 J kg−1]

ANSWER Total surface area of the cube A = 6 × (side)2 = 6 × (0.30)2 = 6 × 0.09 = 0.54 m2. Thickness L = 5.0 cm = 0.05 m; ΔT = 45 − 0 = 45 °C; t = 6 h = 6 × 3600 = 21600 s. Heat conducted in Q = KAΔT·t/L = (0.01 × 0.54 × 45 × 21600)/0.05. Numerator = 0.01 × 0.54 × 45 × 21600 = 5248.8; Q = 5248.8/0.05 = 1.0498 × 105 J. Ice melted m = Q/Lf = (1.0498 × 105)/(335 × 103) = 0.313 kg. Ice remaining = 4.0 − 0.313 = 3.687 kg ≈ 3.7 kg.

10.18 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.

ANSWER Rate of vaporisation = 6.0 kg/min = 6.0/60 = 0.10 kg s−1. Heat current needed H = (dm/dt) × Lv = 0.10 × 2256 × 103 = 2.256 × 105 J s−1. Conduction: H = KA(T − 100)/L, with A = 0.15 m2, L = 0.01 m, K = 109. 2.256 × 105 = (109 × 0.15 × (T − 100))/0.01 = 1635 × (T − 100). T − 100 = (2.256 × 105)/1635 = 137.98 °C. T = 100 + 138 = 237.98 ≈ 238 °C.

10.19 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 (for measuring high temperatures) 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 a correct value for the temperature when the same 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

ANSWER (a) A good emitter is also a good absorber (Kirchhoff’s law). A body with large reflectivity absorbs very little of the radiation falling on it, so it is a poor absorber and therefore a poor emitter. (b) Brass is a good conductor of heat while wood is a poor conductor. On a cold day, the brass tumbler quickly conducts heat away from the hand, making it feel much colder; the wooden tray conducts heat very slowly, so it does not feel as cold. (c) An optical pyrometer is calibrated assuming black-body radiation, H ∝ T4. In the open, a red-hot iron piece (emissivity < 1) radiates less energy than a black body at the same temperature, so the pyrometer reads too low. Inside a furnace, the piece is surrounded by walls at the same temperature and the radiation effectively becomes black-body radiation, so the reading is correct. (d) The atmosphere (especially CO2 and water vapour) traps the infrared radiation re-emitted by the earth (greenhouse effect), keeping it warm. Without the atmosphere this heat would escape directly into space, leaving the earth inhospitably cold. (e) Steam at 100 °C carries a large amount of latent heat of vaporisation (2.26 × 106 J kg−1) in addition to its sensible heat. When it condenses in the radiator it releases this large latent heat, so steam heating warms a building far more effectively than hot water of equal mass.

10.20 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.

ANSWER By Newton’s law of cooling, (change in temperature)/time = K × (mean temperature excess over surroundings). First stage: 80 → 50 °C in 5 min. Δθ = 30 °C; mean temp = (80 + 50)/2 = 65 °C; excess = 65 − 20 = 45 °C. So 30/5 = K(45) ⇒ 6 = 45K ⇒ K = 6/45 = 0.1333 min−1. Second stage: 60 → 30 °C in time t. Δθ = 30 °C; mean temp = (60 + 30)/2 = 45 °C; excess = 45 − 20 = 25 °C. 30/t = K(25) = 0.1333 × 25 = 3.333 ⇒ t = 30/3.333 = 9 min. Time to cool from 60 °C to 30 °C = 9 minutes.

Extra Practice Questions

Short Answer Type Questions

Q1. Why does water have an anomalous expansion, and why is this important for aquatic life?

ANSWERBetween 0 °C and 4 °C water contracts on heating and expands on cooling, so it has maximum density at 4 °C. In winter the colder, less dense water (below 4 °C) stays on top and freezes first, while the denser 4 °C water sinks to the bottom. This keeps lakes liquid at the bottom, allowing fish and plants to survive the cold.

Q2. State Stefan–Boltzmann law and write the expression for the net rate of heat loss by a body.

ANSWERThe energy radiated per unit time by a perfect radiator is H = σAT4, where σ = 5.67 × 10−8 W m−2 K−4. For a body of emissivity e at temperature T with surroundings at Ts, the net rate of heat loss is H = eσA(T4 − Ts4).

Q3. A blacksmith heats an iron ring before fitting it onto a wooden wheel. Explain the physics.

ANSWEROn heating, the iron ring expands and its diameter becomes larger than the wooden rim, so it slips on easily. As it cools, it contracts and grips the rim tightly. This is an application of linear thermal expansion (d2 = d1[1 + αΔT]).

Q4. Why does the temperature of a substance remain constant during a change of state even though heat is supplied?

ANSWERDuring melting or boiling, the supplied heat (latent heat) is used to overcome the intermolecular forces and change the arrangement of molecules, not to increase their kinetic energy. Since temperature depends on average kinetic energy, it stays constant until the change of state is complete.

Q5. Distinguish between conduction, convection and radiation in one line each.

ANSWERConduction: heat flows between adjacent parts of a body by molecular collisions without bulk motion (mainly in solids). Convection: heat is carried by the actual bulk movement of a heated fluid. Radiation: heat is transferred by electromagnetic waves and needs no material medium.

Long Answer Type Questions

Q1. Derive the relation αv = 3αl between the coefficients of volume and linear expansion.

ANSWERConsider a cube of side l. On heating by ΔT, each side increases by Δl = αl l ΔT. New volume = (l + Δl)3 = l3 + 3l2Δl + 3l(Δl)2 + (Δl)3. Since Δl is very small, the squared and cubed terms are neglected, so ΔV = 3l2Δl = 3l2(αl l ΔT) = 3αl V ΔT. But by definition ΔV = αv V ΔT. Comparing the two, αv = 3αl. (Similarly the coefficient of area expansion is 2αl.)

Q2. Describe how heat is transferred by conduction and obtain the expression for the rate of heat flow through a uniform bar.

ANSWERIn conduction, the hotter molecules vibrate more vigorously and pass energy to neighbouring slower molecules through collisions, so heat flows without any net movement of matter. Consider a bar of length L and uniform cross-section A with its ends maintained at temperatures TC and TD (TC > TD), the sides insulated. In the steady state the rate of heat flow (heat current) H is found to be proportional to the area A and the temperature difference (TC − TD) and inversely proportional to L: H = KA(TC − TD)/L, where K is the thermal conductivity of the material (SI unit W m−1 K−1). A larger K means the material conducts heat faster.

Q3. State Newton’s law of cooling, write its mathematical form and explain how it can be verified experimentally.

ANSWERNewton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the temperature difference between the body and its surroundings, provided this difference is small: −dQ/dt = k(T2 − T1). Writing dQ = ms dT2, we get dT2/(T2 − T1) = −K dt, which on integration gives loge(T2 − T1) = −Kt + c, i.e. T2 = T1 + C′e−Kt. To verify it, hot water in a calorimeter is allowed to cool and its temperature noted at regular intervals. A graph of loge(T2 − T1) against time t is a straight line with a negative slope, confirming the law.

MCQs & Assertion–Reason

1. The SI unit of coefficient of thermal conductivity is:

(a) J s−1 m−1 K−1    (b) J kg−1 K−1    (c) W m−2 K−4    (d) J K−1

2. The relation between the coefficients of volume and linear expansion is:

(a) αv = αl    (b) αv = 2αl    (c) αv = 3αl    (d) αv = αl/3

3. Water has its maximum density at:

(a) 0 °C    (b) 4 °C    (c) 100 °C    (d) −4 °C

4. Absolute zero on the Celsius scale is approximately:

(a) 0 °C    (b) −100 °C    (c) −273.15 °C    (d) 273.16 °C

5. The amount of heat required to change unit mass of a substance from solid to liquid at constant temperature is called:

(a) specific heat capacity    (b) latent heat of fusion    (c) latent heat of vaporisation    (d) heat capacity

6. The mode of heat transfer that does not require any material medium is:

(a) conduction    (b) convection    (c) radiation    (d) all of these

7. According to Wien’s displacement law, as the temperature of a body increases, the wavelength of maximum emission:

(a) increases    (b) decreases    (c) stays constant    (d) becomes zero

8. The substance with the highest specific heat capacity among the following is:

(a) copper    (b) iron    (c) mercury    (d) water

9. The triple point of water corresponds to a temperature of:

(a) 273.15 K    (b) 273.16 K    (c) 0 K    (d) 373 K

10. According to the Stefan–Boltzmann law, the energy radiated per unit time by a black body is proportional to:

(a) T    (b) T2    (c) T3    (d) T4

Answer key: 1-(a), 2-(c), 3-(b), 4-(c), 5-(b), 6-(c), 7-(b), 8-(d), 9-(b), 10-(d).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: A hole in a metal plate expands when the plate is heated.

Reason: A hole expands in the same way as a solid piece of the same material would.

A-R 2. Assertion: Steam at 100 °C causes more severe burns than boiling water at 100 °C.

Reason: Steam carries an additional latent heat of vaporisation that it releases on condensing on the skin.

A-R 3. Assertion: A good absorber of radiation is also a good emitter.

Reason: A body with high reflectivity is an excellent emitter of radiation.

A-R 4. Assertion: The temperature of a substance does not change during a change of state.

Reason: During a change of state the supplied heat is used to change the molecular arrangement rather than to raise kinetic energy.

A-R 5. Assertion: On a chilly day a metal handle feels colder than a wooden one at the same temperature.

Reason: Metal is a much better conductor of heat than wood, so it draws heat away from the hand faster.

Answer key: 1-(A), 2-(A), 3-(C), 4-(A), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Forgetting that a temperature difference is the same in °C and K (so ΔT = 18 °C = 18 K), but an absolute temperature must use T = tC + 273.15.
  • Mixing up αl, 2αl and 3αl for length, area and volume expansion.
  • Using the wrong sign in expansion problems — cooling means contraction (negative ΔT) and develops tensile stress.
  • Leaving specific heat in J g−1 K−1 when mass is in kg; convert to J kg−1 K−1 (multiply by 1000).
  • Forgetting the latent-heat term Q = mL when a change of state occurs — not all the heat goes into a temperature change.
  • In calorimetry, missing the water equivalent of the calorimeter when applying “heat lost = heat gained.”
  • Using end temperatures instead of the mean temperature excess in Newton’s law of cooling problems.

How to score full marks in this chapter

Always write the formula first, substitute values with units, and box the final answer with its unit. Convert all quantities to SI before substituting. In calorimetry, clearly write “heat lost = heat gained” and include the calorimeter’s water equivalent. For expansion problems, decide whether the question needs linear, area or volume expansion. In Newton’s law of cooling, use the average of the initial and final temperatures as the body’s temperature. For conceptual questions on radiation, quote Kirchhoff’s law, Stefan–Boltzmann law (H ∝ T4) or Wien’s law where relevant.

Frequently Asked Questions

What is Class 11 Physics Chapter 10 about?

Chapter 10, Thermal Properties of Matter, deals with heat and temperature, temperature scales and the ideal-gas equation, thermal expansion (linear, area and volume), specific heat capacity, calorimetry, change of state and latent heat, the three modes of heat transfer (conduction, convection, radiation) and Newton’s law of cooling.

How many exercise questions are there in Class 11 Physics Chapter 10?

There are 20 NCERT exercise questions, numbered 10.1 to 10.20. This page solves every one of them step by step, with all numericals verified against the NCERT answer key.

What is the relation between αv and αl?

The coefficient of volume expansion is three times the coefficient of linear expansion: αv = 3αl. This is derived by expanding a cube and neglecting the very small higher-order terms in Δl.

Are these Class 11 Physics Chapter 10 solutions free?

Yes. All solutions are free and follow the official NCERT Physics textbook (Reprint 2026–27), with original step-by-step answers.

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