By the end of this chapter you'll be able to…

  • 1Convert between temperature scales and explain thermal equilibrium
  • 2Apply linear, area, and volume expansion formulae
  • 3Use calorimetry to solve heat-exchange problems
  • 4Calculate heat for change of state using latent heat
  • 5Compare conduction, convection, and radiation and apply their laws
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Why this chapter matters
Thermal properties connect molecular motion to everyday phenomena. Thermal expansion, calorimetry, latent heat, and the three modes of heat transfer explain expansion joints, cooking, climate, and why ice floats -- and form the bridge to thermodynamics.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

Thermal Properties of Matter

'Heat is a form of energy.' — James Prescott Joule

1. Chapter Overview

This chapter explores how MATTER behaves when HEAT is added or removed. Topics include THERMAL EXPANSION (solids, liquids, and gases), CALORIMETRY (heat measurement), CHANGE OF STATE (melting, boiling, latent heat), and HEAT TRANSFER mechanisms — CONDUCTION, CONVECTION, and RADIATION. These concepts bridge macroscopic observations with microscopic molecular motion.


2. Temperature and Heat

  • Temperature: Measure of the AVERAGE kinetic energy of molecules (INTENSIVE property)
  • Heat: Energy transferred BETWEEN systems due to temperature difference (EXTENSIVE property)
  • Thermal Equilibrium: Two systems at the SAME temperature (Zeroth Law of Thermodynamics)

Temperature Scales

ScaleLower Fixed PointUpper Fixed PointFormula
Celsius (°C)0°C (ice point)100°C (steam point)C = K — 273.15
Fahrenheit (°F)32°F212°F°F = (9/5)°C + 32
Kelvin (K)273.15 K373.15 KK = °C + 273.15

3. Thermal Expansion

Linear Expansion (Solids)

  • ΔL = αL₀ΔT (α = coefficient of linear expansion)
  • α: per °C or per K

Area (Superficial) Expansion

  • ΔA = βA₀ΔT (β = 2α for isotropic materials)

Volume Expansion

  • ΔV = γV₀ΔT (γ = 3α for isotropic materials)

Anomalous Expansion of Water

  • Water has MINIMUM volume (maximum density) at 4°C
  • Between 0°C and 4°C, water EXPANDS on cooling (contracts on heating)
  • This is WHY ice floats and ponds freeze from the top down

Thermal Expansion of Liquids

  • Apparent expansion (observed in container): γ_app = γ_real — γ_container
  • Real expansion: γ_real = (1/V)(ΔV/ΔT)

Worked Problem

Q: A steel rail of length 20 m at 20°C. Find expansion at 45°C. (α = 1.2×10⁻⁵ /°C) A: ΔL = αL₀ΔT = 1.2×10⁻⁵ × 20 × (45 — 20) = 1.2×10⁻⁵ × 20 × 25 = 6×10⁻³ m = 6 mm.


4. Calorimetry

  • Heat Capacity C = Q/ΔT (SI: J/K)
  • Specific Heat Capacity c = Q/(mΔT) (SI: J/kg·K)
  • Molar Specific Heat C = Q/(nΔT) (SI: J/mol·K)

Principle of Calorimetry

  • Heat LOST by hot body = Heat GAINED by cold body
  • m₁c₁(T₁ — T) = m₂c₂(T — T₂) (for two-body system)
  • Assumes NO heat loss to surroundings

Worked Problem

Q: 200 g of water at 80°C is mixed with 100 g of water at 30°C. Find final temperature. A: 200×c×(80 — T) = 100×c×(T — 30) → 16000 — 200T = 100T — 3000 → 19000 = 300T → T = 63.3°C.


5. Change of State

Latent Heat

  • Latent Heat of Fusion (L_f): Heat required to change 1 kg from solid to liquid at melting point
    • Ice: L_f = 3.36×10⁵ J/kg
  • Latent Heat of Vaporisation (L_v): Heat required to change 1 kg from liquid to vapour at boiling point
    • Water: L_v = 22.6×10⁵ J/kg

Heating Curve of Water

  • 0°C to 0°C (ice melting): Q = mL_f (temperature CONSTANT during phase change)
  • 0°C to 100°C (water heating): Q = mcΔT
  • 100°C to 100°C (water boiling): Q = mL_v (temperature CONSTANT)

Triple Point

  • The UNIQUE temperature and pressure where all three phases (solid, liquid, vapour) coexist in equilibrium

Worked Problem

Q: How much heat is needed to convert 100 g of ice at -10°C to steam at 100°C? (c_ice = 2100 J/kg·K, L_f = 3.36×10⁵ J/kg, L_v = 22.6×10⁵ J/kg) A: Q₁ = mcΔT_ice = 0.1×2100×10 = 2100 J. Q₂ = mL_f = 0.1×3.36×10⁵ = 33600 J. Q₃ = mcΔT_water = 0.1×4200×100 = 42000 J. Q₄ = mL_v = 0.1×22.6×10⁵ = 226000 J. Total = 303700 J.


6. Heat Transfer

Conduction

  • Fourier's Law: Q/t = kA(ΔT/d) (k = thermal conductivity, W/m·K)
  • Thermal Resistance: R = d/kA
  • Good conductors: Metals (k ≈ 200-400 W/m·K)
  • Insulators: Wood, air, glass wool (k ≈ 0.01-0.1 W/m·K)

Convection

  • Heat transfer by the ACTUAL MOVEMENT of heated material
  • Natural convection: Warm fluid rises (density decrease)
  • Forced convection: Fan or pump circulates fluid
  • Examples: Sea breeze, land breeze, household heating

Radiation

  • Heat transfer by ELECTROMAGNETIC WAVES (does NOT require medium)
  • Stefan-Boltzmann Law: P = εσAT⁴ (σ = 5.67×10⁻⁸ W/m²·K⁴)
  • Wien's Displacement Law: λ_m T = b (b = 2.9×10⁻³ m·K)
    • Higher temperature → shorter peak wavelength
    • Sun (~5800 K) emits mostly visible; Earth (~300 K) emits infrared

Newton's Law of Cooling

  • Rate of cooling ∝ Temperature difference with surroundings
  • dT/dt = -k(T — T₀)
  • Approximate for moderate temperature differences

7. Common Mistakes

  1. Temperature and heat are NOT the same: Heat is energy transfer; temperature measures average KE
  2. Phase changes occur at CONSTANT temperature: Adding heat during melting/boiling does NOT raise temperature
  3. Water expansion anomaly: Water expands when cooled from 4°C to 0°C (unlike most substances)
  4. Radiation does NOT require a medium: The Sun's heat reaches Earth through the vacuum of space
  5. Newton's law of cooling is approximate: It's accurate only when the temperature difference is small

8. CBSE Exam Focus

  1. Thermal expansion numericals — gaps in rails, bimetallic strips
  2. Calorimetry — mixture problems (3/5-mark)
  3. Latent heat — phase change calculations (5-mark)
  4. Heat conduction — thermal resistance and composite walls
  5. Stefan-Boltzmann law and Wien's displacement law
  6. Newton's law of cooling — numericals

9. Key Formulas

  • ΔL = αL₀ΔT, ΔA = βA₀ΔT, ΔV = γV₀ΔT (β = 2α, γ = 3α)
  • Q = mcΔT (sensible heat)
  • Q = mL (latent heat)
  • Q/t = kA(ΔT/d) (conduction)
  • P = εσAT⁴ (radiation), σ = 5.67×10⁻⁸ W/m²·K⁴
  • λ_m T = constant (2.9×10⁻³ m·K)
  • dT/dt = -k(T — T₀) (Newton's cooling)

10. Self-Test (5+ Q&A)

Q1: A 100 g copper block (c = 390 J/kg·K) at 200°C is placed in 200 g water at 20°C. Find final temperature. A: Heat lost = Heat gained: 0.1×390×(200 — T) = 0.2×4200×(T — 20) → 39(200 — T) = 840(T — 20) → 7800 — 39T = 840T — 16800 → 24600 = 879T → T = 28°C.

Q2: Why does a brass disc pass through a brass ring when both are heated equally? A: Both expand equally (same α). The hole expands as if it were made of the same material, so the disc still fits.

Q3: At what temperature is the Celsius reading equal to Fahrenheit? A: C = (9/5)C + 32 → -40°C = -40°F.

Q4: Calculate the cooling rate of a body at 60°C in surroundings at 30°C if it cools from 50°C to 40°C in 5 minutes. A: Newton's law: Rate = -kΔT_avg. This requires experimental data or the constant k.

Q5: Why is the bottom of a lake warmer than the top in winter? A: Anomalous expansion of water: Water is densest at 4°C. In winter, surface water cools and sinks until the whole lake reaches 4°C. Further cooling makes surface water less dense (0-4°C), so it stays on top and freezes.


11. Conclusion

Thermal properties connect the MICROSCOPIC (molecular kinetic energy) to the MACROSCOPIC (expansion, heat transfer). Understanding thermal expansion is crucial in engineering (expansion joints, thermostats). Calorimetry is the practical measurement of heat. Phase changes involve LARGE amounts of energy at constant temperature. Heat transfer mechanisms explain everything from cooking to climate. This chapter forms the BRIDGE to thermodynamics.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Thermal expansion
delta-L = alpha L0 delta-T; beta = 2 alpha; gamma = 3 alpha
Linear, area, and volume expansion coefficients for isotropic solids.
Calorimetry
Q = m c delta-T (sensible); Q = m L (latent)
Heat lost = heat gained in an isolated mixture.
Conduction (Fourier's law)
Q/t = k A (delta-T / d)
k is thermal conductivity; thermal resistance R = d/(kA).
Radiation laws
P = e sigma A T^4; lambda_m T = b
Stefan-Boltzmann and Wien's displacement laws; sigma = 5.67e-8.
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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Treating heat and temperature as the same
Heat is energy in transit due to a temperature difference; temperature measures the average kinetic energy of molecules.
WATCH OUT
Expecting temperature to rise during a phase change
During melting or boiling, added heat goes into latent heat and the temperature stays constant.
WATCH OUT
Assuming all substances contract on cooling
Water shows anomalous expansion -- between 4 C and 0 C it expands on cooling, so ice floats and ponds freeze top-down.
WATCH OUT
Thinking radiation needs a medium
Radiation travels as electromagnetic waves and crosses a vacuum, which is how the Sun's heat reaches Earth.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Expansion
A 20 m steel rail at 20 C is heated to 45 C (alpha = 1.2e-5 /C). Find the expansion.
Show solution
delta-L = alpha L0 delta-T = 1.2e-5 x 20 x 25 = 6e-3 m = 6 mm.
Q2MEDIUM· Calorimetry
A 100 g copper block (c = 390 J/kg K) at 200 C is placed in 200 g water at 20 C. Find the final temperature.
Show solution
0.1 x 390 x (200 - T) = 0.2 x 4200 x (T - 20). 39(200 - T) = 840(T - 20). 7800 - 39T = 840T - 16800, so 24600 = 879T, T = 28 C.
Q3HARD· Latent Heat
Find the heat to convert 100 g ice at -10 C to steam at 100 C (c_ice = 2100, L_f = 3.36e5, c_water = 4200, L_v = 22.6e5 J/kg).
Show solution
Q1 = 0.1 x 2100 x 10 = 2100 J; Q2 = 0.1 x 3.36e5 = 33600 J; Q3 = 0.1 x 4200 x 100 = 42000 J; Q4 = 0.1 x 22.6e5 = 226000 J. Total = 303700 J.
Q4EASY· Scales
At what temperature is the Celsius reading equal to the Fahrenheit reading?
Show solution
C = (9/5)C + 32 gives C = -40, so -40 C = -40 F.
Q5MEDIUM· Concept
Why is the bottom of a lake warmer than the top in winter?
Show solution
Water is densest at 4 C. As the surface cools it sinks until the lake reaches 4 C; further cooled surface water (0-4 C) is less dense, so it stays on top and freezes, leaving the bottom water at about 4 C.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • Temperature measures average molecular KE; heat is transferred energy.
  • Expansion: delta-L = alpha L0 delta-T; beta = 2 alpha, gamma = 3 alpha.
  • Water has maximum density at 4 C (anomalous expansion).
  • Calorimetry: heat lost = heat gained; Q = m c delta-T.
  • Phase change: Q = m L at constant temperature; L_f(ice) = 3.36e5, L_v(water) = 22.6e5 J/kg.
  • Conduction Q/t = kA(delta-T/d); convection moves the medium; radiation needs none.
  • Stefan-Boltzmann P = e sigma A T^4; Wien's law lambda_m T = constant; Newton's cooling dT/dt = -k(T - T0).

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 5-7 marks across the chapter

Question typeMarks eachTypical countWhat it tests
Latent heat / change of state3-51Heating-curve calculations
Calorimetry31Mixture problems
Heat transfer / expansion2-31Conduction, radiation laws, thermal expansion
Prep strategy
  • Memorise expansion relations (beta = 2 alpha, gamma = 3 alpha)
  • Practise calorimetry and heating-curve numericals
  • Learn the three heat-transfer mechanisms and their laws
  • Note phase changes occur at constant temperature

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Engineering expansion joints

Gaps in rails, bridges, and pipelines accommodate thermal expansion and prevent buckling.

Thermostats and sensors

Bimetallic strips use differential expansion to switch heaters and appliances on and off.

Climate and weather

Convection currents and radiation drive sea breezes, ocean currents, and Earth's energy balance.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

  1. Break heating-curve problems into sensible and latent stages
  2. Set heat lost equal to heat gained for mixtures
  3. Use the right heat-transfer law for the mode asked
  4. Remember phase changes happen at constant temperature

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

  • Solve composite-wall conduction problems using thermal resistances in series and parallel.
  • Derive Newton's law of cooling as an approximation of the Stefan-Boltzmann law.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Physics examHigh
JEE Main and Advanced (Thermal Physics)Medium
NEET PhysicsMedium

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

During melting, the added heat is used to break the bonds that hold the rigid solid structure together, converting it into liquid, rather than to increase the molecules' kinetic energy. Since temperature reflects average kinetic energy, it remains constant at the melting point until all the ice has melted. This energy is the latent heat of fusion.

Metals have a much higher thermal conductivity than wood. When you touch a metal handle it conducts heat away from your hand rapidly, so your skin cools quickly and feels cold. Wood conducts heat poorly, so it removes heat from your hand slowly and feels warmer, even though both objects are at the same temperature.
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Last reviewed on 29 May 2026. Written and reviewed by subject-matter experts — read about our process.
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