Thermal Properties of Matter Class 11 Notes | CBSE Physics Chapter 10

Chapter summary

Thermal Properties of Matter explains how heat and temperature differ and what happens when matter is heated, covering temperature scales, thermal expansion of solids and liquids, calorimetry and latent heat during phase change, and the three modes of heat transfer. It also builds the radiation laws that decide how hot bodies glow and cool, namely the Stefan-Boltzmann law, Wien’s displacement law and Newton’s law of cooling. For NEET this chapter is a steady source of formula-based and conceptual questions, especially expansion, calorimetry and black body radiation.

Chapter notes

Table of Contents


Key Concepts

1. Temperature and Heat

Heat is energy that flows from a hotter body to a colder body because of their temperature difference. SI unit: joule (J). Heat is energy in transit - once it is absorbed, we call it internal energy, not heat.

Temperature is the measure of the degree of hotness or coldness of a body. It decides the direction of heat flow: energy always moves from high to low temperature until both reach the same value (thermal equilibrium). SI unit: kelvin (K).

Key idea: A bucket of warm water can hold more heat than a spark, even though the spark is at a far higher temperature. Heat depends on mass; temperature does not.


2. Thermometry and Temperature Scales

A thermometer measures temperature using a property that changes regularly with it - usually the expansion of mercury or alcohol in a glass capillary.

The Three Common Scales

  • Celsius (°C): ice point 0 °C, steam point 100 °C.
  • Fahrenheit (°F): ice point 32 °F, steam point 212 °F.
  • Kelvin (K): the absolute/SI scale; 0 K is absolute zero. T(K) = t(°C) + 273.15.

Conversion between Celsius and Fahrenheit:

(C − 0)/100 = (F − 32)/180, i.e. C/5 = (F − 32)/9

The ideal-gas (absolute) scale is built on the fact that, at constant volume, the pressure of a gas falls linearly with temperature and would reach zero at −273.15 °C - this defines absolute zero.


3. Thermal Expansion

Most substances expand on heating because the molecules vibrate more vigorously and their average separation increases. There are three kinds, each with its own coefficient.

TypeFormulaCoefficient
Linear (length)ΔL = α L₀ ΔTα = coefficient of linear expansion
Area (superficial)ΔA = β A₀ ΔTβ = coefficient of area expansion
Volume (cubical)ΔV = γ V₀ ΔTγ = coefficient of volume expansion

Relation between the coefficients: β = 2α and γ = 3α, so α : β : γ = 1 : 2 : 3.

Units: each coefficient has units of K⁻¹ (or °C⁻¹).

Anomalous Expansion of Water

Water behaves strangely between 0 °C and 4 °C: it contracts on heating and expands on cooling, reaching its maximum density at 4 °C. This is why ice forms on top of a pond while fish survive in the 4 °C water below.


4. Specific Heat Capacity

The specific heat capacity (c) is the heat needed to raise the temperature of 1 kg of a substance by 1 K (or 1 °C).

Q = m c ΔT

  • SI unit: J kg⁻¹ K⁻¹.
  • Water has an unusually large specific heat (4186 J kg⁻¹ K⁻¹), which is why it is used as a coolant and why coastal climates are mild.

The heat capacity of a body is C = m c (unit J K⁻¹) - the heat needed to raise the whole body’s temperature by 1 K. The molar specific heat is the heat per mole per kelvin.


5. Calorimetry

Calorimetry is the measurement of heat. It rests on the principle of mixtures: when bodies at different temperatures are mixed, heat lost by the hotter body equals heat gained by the colder body (assuming no loss to surroundings).

Heat lost = Heat gained

This single equation lets you find an unknown specific heat, mass, or final temperature. The apparatus used is a calorimeter, usually a copper vessel because copper has a low specific heat and conducts well.


6. Change of State and Latent Heat

When a substance changes state (solid ⇄ liquid ⇄ gas), its temperature stays constant even though heat is being added or removed. This “hidden” heat is the latent heat.

Q = m L

  • Latent heat of fusion (L_f): heat per kg to convert a solid to liquid at its melting point. For ice, L_f ≈ 3.34 × 10⁵ J kg⁻¹.
  • Latent heat of vaporization (L_v): heat per kg to convert a liquid to vapour at its boiling point. For water, L_v ≈ 2.26 × 10⁶ J kg⁻¹.
  • SI unit: J kg⁻¹.

[DIAGRAM: A temperature-vs-heat graph for ice → water → steam, showing two flat plateaus at 0 °C (melting) and 100 °C (boiling) where temperature stays constant.]

This is why steam at 100 °C scalds far worse than water at 100 °C - the steam releases its large latent heat of vaporization on your skin.


7. Heat Transfer: Conduction

Conduction is the transfer of heat through a material without bulk movement of the material itself - energy is passed from molecule to molecule. It is the main mode in solids, especially metals.

For a rod of length L and cross-section A with ends at temperatures T₁ and T₂, the rate of heat flow is:

Q/t = kA(T₁ − T₂)/L

  • k = coefficient of thermal conductivity (unit W m⁻¹ K⁻¹); high for metals, low for wood and air.
  • (T₁ − T₂)/L is the temperature gradient.

8. Heat Transfer: Convection

Convection is the transfer of heat by the actual movement of the heated material itself. It occurs in fluids (liquids and gases) where warm, less-dense regions rise and cooler regions sink, setting up convection currents.

  • Natural convection: driven by density differences (sea breeze, boiling water).
  • Forced convection: the fluid is pushed by a pump or fan (a room heater’s blower, blood circulation).

9. Heat Transfer: Radiation

Radiation is the transfer of heat by electromagnetic waves and needs no medium - this is how the Sun’s energy reaches the Earth through empty space.

Every body above 0 K emits thermal radiation. Dark, rough surfaces are good absorbers and good emitters; shiny, polished surfaces are poor absorbers and good reflectors.


10. Blackbody Radiation and Stefan-Boltzmann Law

A perfectly black body absorbs all radiation falling on it and is also the best possible emitter. Its emissivity e = 1.

The Stefan-Boltzmann law states that the energy radiated per unit area per unit time by a black body is proportional to the fourth power of its absolute temperature:

E = σT⁴ (for a real body, E = eσT⁴)

  • σ = Stefan’s constant = 5.67 × 10⁻⁸ W m⁻² K⁻⁴.
  • The net rate of loss by a body at temperature T in surroundings at T₀ is E = eσ(T⁴ − T₀⁴).

11. Wien’s Displacement Law

A hot body radiates over many wavelengths, but the wavelength at which it radiates most strongly shifts with temperature. Wien’s law says this peak wavelength is inversely proportional to absolute temperature:

λ_m T = b

  • b = Wien’s constant = 2.9 × 10⁻³ m·K.
  • This is why a heated iron glows dull red first, then orange, then white as it gets hotter - the peak moves to shorter wavelengths.

12. Newton’s Law of Cooling

Newton’s law of cooling states that for a small temperature difference, the rate of loss of heat of a body is directly proportional to the difference between its temperature and that of its surroundings.

−dT/dt = K(T − T₀)

  • It is an approximation of the Stefan-Boltzmann law, valid only for small temperature differences.
  • A graph of temperature against time is an exponential decay curve approaching the surrounding temperature T₀.

This is why a cup of hot tea cools quickly at first and then more slowly as it nears room temperature.


Read the rest of the chapter →Hide the rest ↑

Weightage in Board & Entrance Exams

ExamTypical WeightageMost-Tested Areas
CBSE Board (Class 11)4–5 marksThermal expansion, calorimetry, latent heat, conduction formula
JEE Main / Advanced1–2 questionsExpansion, conduction in series/parallel, Stefan-Boltzmann, Wien’s law
NEET1–2 questionsSpecific & latent heat, modes of heat transfer, Newton’s law of cooling

[TABLE: Question-type split - VSA (1 mark): scales, coefficients, definitions; SA (2–3 marks): expansion & calorimetry numericals, conduction; LA (5 marks): radiation laws, Newton’s law of cooling derivation.]


Important Definitions

TermDefinition
HeatEnergy in transit due to a temperature difference; SI unit joule
TemperatureMeasure of degree of hotness; decides the direction of heat flow; SI unit kelvin
Coefficient of linear expansion (α)Fractional change in length per unit rise in temperature: α = ΔL/(L₀ΔT)
Specific heat capacity (c)Heat to raise 1 kg of a substance by 1 K: Q = mcΔT
Heat capacity (C)Heat to raise the whole body by 1 K: C = mc
Latent heat (L)Heat per unit mass to change state at constant temperature: Q = mL
Thermal conductivity (k)Measure of how readily a material conducts heat: Q/t = kAΔT/L
Black bodyA body that absorbs and emits all radiation (e = 1)
Stefan-Boltzmann lawEnergy radiated per unit area: E = σT⁴
Newton’s law of coolingRate of cooling ∝ temperature difference: −dT/dt = K(T − T₀)

Solved Examples

Example 1

Convert 37 °C (normal body temperature) to the Fahrenheit and Kelvin scales.

Answer: F = (9/5)C + 32 = (9/5)(37) + 32 = 66.6 + 32 = 98.6 °F. T = 37 + 273.15 = 310.15 K.

Example 2

A steel rod 1 m long is heated from 20 °C to 120 °C. Find the increase in length. (α = 1.2 × 10⁻⁵ K⁻¹)

Answer: ΔL = αL₀ΔT = (1.2 × 10⁻⁵)(1)(100) = 1.2 × 10⁻³ m = 1.2 mm.

Example 3

How much heat is needed to raise the temperature of 2 kg of water from 25 °C to 75 °C? (c = 4186 J kg⁻¹ K⁻¹)

Answer: Q = mcΔT = 2 × 4186 × 50 = 4.186 × 10⁵ J.

Example 4

Calculate the heat required to convert 0.5 kg of ice at 0 °C completely into water at 0 °C. (L_f = 3.34 × 10⁵ J kg⁻¹)

Answer: Q = mL_f = 0.5 × 3.34 × 10⁵ = 1.67 × 10⁵ J.

Example 5

One face of a copper slab of area 0.2 m² and thickness 2 cm is kept at 100 °C and the other at 0 °C. Find the rate of heat flow. (k = 400 W m⁻¹ K⁻¹)

Answer: Q/t = kA(T₁ − T₂)/L = (400 × 0.2 × 100)/0.02 = 8000/0.02 = 4 × 10⁵ W.

Example 6

The surface temperature of a star is 6000 K. Using Wien’s law, find the wavelength at which it radiates most strongly. (b = 2.9 × 10⁻³ m·K)

Answer: λ_m = b/T = (2.9 × 10⁻³)/6000 = 4.83 × 10⁻⁷ m (≈ 483 nm, blue-green light).


Important Questions for Board Exams

1-Mark Questions (VSA)

  1. Define the coefficient of linear expansion and give its SI unit.
  2. Why does water have an anomalous expansion, and at what temperature is its density maximum?
  3. Name the mode of heat transfer that does not require a material medium.
  4. Why is steam at 100 °C more dangerous than water at 100 °C?
  5. What is the value and SI unit of Stefan’s constant?

2–3-Mark Questions (SA)

  1. Derive the relation γ = 3α between the coefficients of volume and linear expansion.
  2. State the principle of calorimetry and explain how it is used to find an unknown specific heat.
  3. Distinguish between conduction, convection, and radiation with one example of each.
  4. State Wien’s displacement law and use it to explain the colour change of a heated iron piece.

5-Mark Questions (LA)

  1. State and explain Newton’s law of cooling. Show how it follows from the Stefan-Boltzmann law for small temperature differences, and sketch the cooling curve.
  2. Derive the expression for the rate of heat conduction through a rod and define thermal conductivity. Discuss conductors in series.
  3. Explain blackbody radiation and state the Stefan-Boltzmann law. Find the net rate of heat loss of a body at temperature T placed in surroundings at T₀.

Quick Revision Points

  • Heat is energy in transit; temperature decides the direction of heat flow
  • Scale conversion: C/5 = (F − 32)/9; T(K) = t(°C) + 273.15
  • Expansion: ΔL = αL₀ΔT, ΔA = βA₀ΔT, ΔV = γV₀ΔT with α : β : γ = 1 : 2 : 3
  • Water has maximum density at 4 °C (anomalous expansion)
  • Specific heat: Q = mcΔT; water c = 4186 J kg⁻¹ K⁻¹
  • Calorimetry: heat lost = heat gained
  • Latent heat: Q = mL; ice L_f ≈ 3.34 × 10⁵, water L_v ≈ 2.26 × 10⁶ J kg⁻¹
  • Conduction: Q/t = kA(T₁ − T₂)/L; convection needs fluid movement; radiation needs no medium
  • Stefan-Boltzmann: E = σT⁴; σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴
  • Wien’s law: λ_m T = b; b = 2.9 × 10⁻³ m·K
  • Newton’s law of cooling: −dT/dt = K(T − T₀); exponential decay toward T₀

Next Chapter: Chapter 11 - Thermodynamics

🃏 Flash Cards: Thermal Properties of Matter

Class 11 Physics · Chapter 10 – swipe through all 8 cards to understand the whole chapter.

🌡️Start here1/8

Heat, Temperature & Scales

Heat is energy in transit between bodies; temperature decides which way it flows.

C/5 = (F − 32)/9 = (K − 273.15)/5

Heat in joules (1 cal ≈ 4.186 J); temperature in kelvin.

  • Heat always flows hot → cold; equal temperature = thermal equilibrium (Zeroth Law).
  • A step of 1 °C = 1 K, but 1 °C ≠ 1 °F; absolute zero = 0 K = −273.15 °C.
  • −40° reads the same on both Celsius and Fahrenheit (classic exam point).
📏Core law2/8

Thermal Expansion

Heating widens average atomic spacing, so solids and liquids expand.

ΔL = α L0 ΔT · α : β : γ = 1 : 2 : 3

β = 2α, γ = 3α for an isotropic solid; coefficients per K.

  • Linear ΔL = αL0ΔT, areal ΔA = βA0ΔT, volume ΔV = γV0ΔT.
  • A hole expands like the surrounding material — it gets bigger, not smaller.
  • Liquids have only γ, and it is much larger than for solids.
💧Key fact3/8

Water Anomaly & Thermal Stress

Water misbehaves near 0 °C, and a clamped rod that cannot expand builds up stress.

stress = Y α ΔT · density max at 4 °C

Y is Young’s modulus; anomaly lets lakes freeze top-down.

  • Between 0 °C and 4 °C water contracts on heating; density is maximum at 4 °C.
  • Ice floats and ponds freeze from the surface down because of this anomaly.
  • A bimetallic strip bends because two metals have different α.
🔥Core law4/8

Calorimetry & Specific Heat

Raising temperature needs heat set by mass and specific heat; an isolated system conserves it.

Q = m c ΔT · heat lost = heat gained

Water c ≈ 4186 J kg⁻1 K⁻1 (1 cal g⁻1 °C⁻1) — unusually high.

  • Specific heat c = heat to raise 1 kg by 1 K; heat capacity = mc (J/K).
  • In an isolated system heat lost = heat gained (principle of calorimetry).
  • Final temperature always lies between the two starting temperatures.
❄️Key process5/8

Latent Heat & Phase Change

During melting or boiling, heat goes into breaking bonds at constant temperature.

Q = m L · L_fusion(ice) ≈ 3.34 × 105 J kg⁻1

L_vaporisation(water) ≈ 22.6 × 105 J kg⁻1.

  • Temperature stays flat on the heating curve through a phase change.
  • Energy breaks bonds, it does not raise kinetic energy, so T is constant.
  • Steam at 100 °C burns worse than water at 100 °C — it carries latent heat.
♨️Three modes6/8

Conduction, Convection & Radiation

Heat travels three ways; steady-state conduction behaves like an electric circuit.

Q/t = k A (T1 − T2) / L · R = L / (k A)

k = thermal conductivity (W m⁻1 K⁻1); only radiation needs no medium.

  • Conduction: atom-to-atom through solids; metals best via free electrons.
  • Convection: bulk movement of fluid; radiation: EM waves through vacuum.
  • Rods in series add resistances; in parallel add conductances.
Core law7/8

Black Body Radiation Laws

A black body is the best absorber and emitter; its glow encodes its temperature.

E = σ T4 · λ_max T = b

σ = 5.67 × 10⁻8 W m⁻2 K⁻4; b = 2.9 × 10⁻3 m·K; T in kelvin.

  • Stefan–Boltzmann: net power P = e σ A (T4 − T04), e = emissivity (0–1).
  • Wien’s law: hotter bodies peak at shorter λ (red → white → blue).
  • Kirchhoff: a good absorber is a good emitter at a given temperature.
Final law8/8

Newton’s Law of Cooling

A body only moderately hotter than its surroundings loses heat in proportion to that excess.

−dT/dt = k (T − T0)

Valid for small (T − T0); an approximation of Stefan’s law.

  • The body cools fastest when hottest; cooling slows as T nears T0.
  • ln(T − T0) vs t is a straight line; T decays exponentially toward T0.
  • Numericals use (T1 − T2)/t = k[(T1 + T2)/2 − T0].
Swipe Click a card to focus 8 cards
📝 Practice Thermal Properties of Matter — 10 NEET PYQs
Real previous-year questions · with answers & solutions
Start →Close ✕
Tap an option to check your answer and see the worked solution. Every question is a real NEET previous-year question.
Q1NEET 2021
A cup of coffee cools from 90 °C to 80 °C in t minutes when the room temperature is 20 °C. The time taken by a similar cup to cool from 80 °C to 60 °C at the same room temperature is:
Correct answer: B. Average-temperature form: (Δ T)/(t)=K(T_(avg)-T₀). First: (10)/(t)=K(85-20)=65K⇒ K=(2)/(13t). Second: (20)/(t’)=K(70-20)=50K=(100)/(13t)⇒ t’=(20×13t)/(100)=(13)/(5)t.
🔎 See the full step-by-step solution in the app →
Q2NEET 2020
The quantities of heat required to raise the temperatures of two solid copper spheres of radii r₁ and r₂ (with r₁=1.5 r₂) through 1 K are in the ratio:
Correct answer: D. Q=mcΔ T with m=ρ·(4)/(3)π r³. For equal material and Δ T, Q∝ r³. So (Q₁)/(Q₂)=((r₁)/(r₂))³=(1.5)³=(27)/(8).
🔎 See the full step-by-step solution in the app →
Q3NEET 2019
A copper rod of length 88 cm and an aluminium rod of unknown length have their increase in length independent of the rise in temperature. Given α_(Cu)=1.7×10⁻⁵ K⁻¹ and α_(Al)=2.2×10⁻⁵ K⁻¹, the length of the aluminium rod is:
Correct answer: C. For the increase in length to be the same for any Δ T, α_(Cu)L_(Cu)=α_(Al)L_(Al). So L_(Al)=(α_(Cu)L_(Cu))/(α_(Al))=(1.7×10⁻⁵×88)/(2.2×10⁻⁵)=68 cm.
🔎 See the full step-by-step solution in the app →
Q4NEET 2018
A black body radiates power P and emits maximum energy at wavelength λ₀. If its temperature changes so that it now emits maximum energy at wavelength (3)/(4)λ₀, the power radiated becomes nP. The value of n is:
Correct answer: A. By Wien’s law λₘ T= const, so (T₂)/(T₁)=(λ₀)/((3/4)λ₀)=(4)/(3). By Stefan’s law P∝ T⁴, so n=((4)/(3))⁴=(256)/(81).
🔎 See the full step-by-step solution in the app →
Q5NEET 2016
Coefficients of linear expansion of brass and steel rods are α₁ and α₂; their lengths are l₁ and l₂. If (l₂-l₁) is to remain the same at all temperatures, the required condition is:
Correct answer: C. (l₂-l₁) constant means both rods expand by the same amount for any Δ T: Δ l₁=Δ l₂, i.e. α₁ l₁Δ T=α₂ l₂Δ T. Hence α₁ l₁=α₂ l₂.
🔎 See the full step-by-step solution in the app →
Q6NEET 2016
A piece of ice falls from a height h so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice, and all of this energy goes into melting. Taking latent heat of ice =3.4×10⁵ J kg⁻¹ and g=10 m s⁻², the value of h is:
Correct answer: B. Energy available to melt =(1)/(4)mgh=mL. So h=(4L)/(g)=(4×3.4×10⁵)/(10)=1.36×10⁵ m=136 km.
🔎 See the full step-by-step solution in the app →
Q7NEET 2015
The two ends of a metal rod are maintained at 100 °C and 110 °C, and the rate of heat flow through it is 4.0 J/s. If instead the ends are maintained at 200 °C and 210 °C, the rate of heat flow will be:
Correct answer: D. Steady-state conduction rate (dQ)/(dt)=(kA Δ T)/(L)∝Δ T. The temperature difference is 10 °C in both cases (110-100=210-200), so the rate is unchanged at 4.0 J/s.
🔎 See the full step-by-step solution in the app →
Q8NEET 2010
A cylindrical metallic rod in thermal contact with two heat reservoirs at its two ends conducts an amount of heat Q in time t. The rod is then melted and recast into a rod of half the original radius. The amount of heat the new rod conducts (same reservoirs, same time t) is:
Correct answer: B. Volume is conserved: π r₁² l₁=π r₂² l₂ with r₂=r₁/2 gives l₂=4l₁. Conduction Q∝(A)/(l)∝(r²)/(l). So (Q₂)/(Q₁)=(r₂²/l₂)/(r₁²/l₁)=((1/4))/(4)=(1)/(16), giving Q₂=Q/16.
🔎 See the full step-by-step solution in the app →
Q9NEET 2008
On a new linear temperature scale, the W scale, the freezing and boiling points of water are 39 °W and 239 °W respectively. The temperature corresponding to 39 °C on the Celsius scale is:
Correct answer: B. Equal fractions between fixed points: (C-0)/(100-0)=(W-39)/(239-39). So (39)/(100)=(W-39)/(200)⇒ W-39=78⇒ W=117 °W.
🔎 See the full step-by-step solution in the app →
Q10NEET 1992
A mercury-in-glass thermometer can conveniently be used to measure temperatures up to about:
Correct answer: C. Mercury boils near 357 °C, so a mercury thermometer is usable up to about 360 °C; beyond this the mercury vaporises.
🔎 See the full step-by-step solution in the app →
View 20+ more practice questions, gamified →
Free · no signup · works in your browser
Studying this chapter? Track it - saved on this device, no login.

Frequently Asked Questions

What is the difference between heat and temperature?

Heat is energy that flows between two bodies because of a temperature difference, measured in joules. Temperature measures the degree of hotness and decides the direction of heat flow, which is always from the hotter body to the colder one.

What are the formulas for linear, areal and volume thermal expansion?

Linear expansion is given by change in length equals alpha times original length times change in temperature, areal by beta times original area times change in temperature, and volume by gamma times original volume times change in temperature. For an isotropic solid the coefficients are related as alpha : beta : gamma equals 1 : 2 : 3, so beta is 2 alpha and gamma is 3 alpha.

What is the principle of calorimetry and the formula for heat?

The heat needed to change a body’s temperature is Q equals mass times specific heat times change in temperature. The principle of calorimetry states that in an isolated system the heat lost by the hotter body equals the heat gained by the colder one, so the final temperature lies between the two starting temperatures.

Is Thermal Properties of Matter important for NEET and what carries the most weight?

Yes, it is part of the NEET Physics syllabus and usually contributes a question or two each year. The most frequently tested areas are thermal expansion and thermal stress, calorimetry and latent heat, and the radiation laws including Stefan-Boltzmann, Wien’s displacement law and Newton’s law of cooling.

What is the difference between Stefan’s law and Newton’s law of cooling?

Stefan’s law says the power radiated per unit area of a black body is proportional to the fourth power of its absolute temperature and applies at any temperature. Newton’s law of cooling says the rate of heat loss is proportional to the temperature excess over the surroundings, and it is only an approximation of Stefan’s law valid when the body is moderately hotter than its surroundings.

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top