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Class 11 Physics Chapter 10 Thermal Properties of Matter — Formulas & Key Points
Chapter 10 Thermal Properties of Matter is crucial for CBSE Class 11 Physics board exams and competitive tests like JEE and NEET. This chapter deals with how matter responds to heat—expansion, temperature change, phase transitions, and heat transfer mechanisms. The 2025 CBSE board typically allocates 4-6 marks to questions from this chapter. This formula sheet organizes every equation, constant, and definition you need for quick revision and problem-solving.
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Key takeaways
- ✓Linear thermal expansion: ΔL = αL₀ΔT where α is coefficient of linear expansion with unit K⁻¹ or °C⁻¹
- ✓Heat gained or lost: Q = mcΔT for temperature change without phase change; specific heat c has units J kg⁻¹ K⁻¹
- ✓Latent heat formula: Q = mL where L is latent heat (J/kg) for phase transitions at constant temperature
- ✓Newton's law of cooling: dT/dt ∝ (T - T₀) describes exponential temperature decay in cooling bodies
- ✓Stefan-Boltzmann law: E = σAT⁴ governs radiant heat emission with σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴
- ✓Thermal conductivity K links heat flow rate: (Q/t) = KA(T₁ - T₂)/x for steady-state conduction
- ✓Wien's displacement law: λₘT = b (constant) relates peak emission wavelength to absolute temperature
Core Formulas: Thermal Expansion
Thermal expansion describes dimensional changes in matter when temperature varies. Three types exist: linear (length), area (surface), and volume (bulk). NCERT Class 11 Physics emphasizes the relationship between these coefficients. For solids, linear expansion dominates rod and rail problems; area expansion matters in sheet metal; volume expansion applies to liquids and gases. The coefficient α, β, γ represent linear, area, and volume expansion respectively, with the fundamental relation γ = 3α and β = 2α holding for isotropic solids. These formulas appear frequently in numerical problems worth 2-3 marks each in CBSE exams.
Thermal Expansion Formula Table
This table consolidates all thermal expansion relationships for quick reference during problem-solving. Pay special attention to units—temperature change ΔT can be in Kelvin or Celsius (numerically identical for differences). The final length/area/volume formulas use binomial approximation (1 + x) where x is small, so terms with α², α³ are negligible. In CBSE board exams, questions often provide initial dimensions and ask for final dimensions after heating or the temperature change needed for a specific expansion. Always verify whether the question asks for change (Δ) or final value.
- Type of Expansion | Formula | Application Context
- Linear expansion (length) | ΔL = αL₀ΔT or L = L₀(1 + αΔT) | Rods, rails, bridges; α in K⁻¹
- Area expansion (surface) | ΔA = βA₀ΔT or A = A₀(1 + βΔT) | Metal sheets, plates; β = 2α
- Volume expansion (bulk) | ΔV = γV₀ΔT or V = V₀(1 + γΔT) | Liquids, gases, solids; γ = 3α
- Apparent expansion of liquid | γₐₚₚ = γₗᵢᵩᵤᵢ𝒹 - γ_container | When liquid in container expands
- Relation between coefficients | γ = 3α, β = 2α | Valid for isotropic solids only
Calorimetry: Heat Transfer Without Phase Change
Calorimetry measures heat exchange in systems. The fundamental principle states heat lost by hot body equals heat gained by cold body in an isolated system. Specific heat capacity c is substance-dependent: water has c = 4186 J kg⁻¹ K⁻¹, copper 385 J kg⁻¹ K⁻¹, aluminium 900 J kg⁻¹ K⁻¹. Molar heat capacity C relates to specific heat by C = Mc where M is molar mass. Water equivalent W = mc is the mass of water that would absorb the same heat as the calorimeter for the same temperature rise. Class 11 Physics solutions often involve mixing problems where two substances reach thermal equilibrium.
- Heat gained/lost | Q = mcΔT | Temperature change without phase transition; Q in joules, m in kg
- Heat capacity (body) | S = mc | Total heat needed to raise body temperature by 1 K; unit J K⁻¹
- Specific heat capacity | c = Q/(mΔT) | Heat per unit mass per degree; unit J kg⁻¹ K⁻¹
- Molar heat capacity | C = Mc | Heat per mole per degree; unit J mol⁻¹ K⁻¹; M is molar mass
- Water equivalent | W = mc | Effective mass of water with same heat capacity; unit kg
- Principle of calorimetry | Heat lost = Heat gained | Σ(mcΔT)hot = Σ(mcΔT)cold in isolated system
Latent Heat and Phase Transitions
Latent heat is energy absorbed or released during phase change at constant temperature. For water, latent heat of fusion Lf = 3.34 × 10⁵ J/kg (ice to water at 0°C) and latent heat of vaporization Lv = 2.26 × 10⁶ J/kg (water to steam at 100°C). These values appear in NCERT Class 11 Physics solved examples. During melting or boiling, all absorbed heat breaks intermolecular bonds rather than increasing kinetic energy (temperature). CBSE exam problems often combine sensible heat (mcΔT) and latent heat (mL) in multi-step processes like heating ice from -10°C to steam at 110°C.
- Latent heat (general) | Q = mL | Heat for phase change at constant T; L in J/kg
- Latent heat of fusion | Q = mLf | Solid ↔ liquid; Lf(ice) = 3.34 × 10⁵ J/kg
- Latent heat of vaporization | Q = mLv | Liquid ↔ gas; Lv(water) = 2.26 × 10⁶ J/kg
- Latent heat of sublimation | Q = mLs | Solid ↔ gas directly; Ls = Lf + Lv
Heat Transfer: Conduction Formulas
Heat conduction occurs through direct molecular contact in solids. Fourier's law governs steady-state conduction where temperature gradient drives heat flow. Thermal conductivity K is material property: metals like copper (K = 385 W m⁻¹ K⁻¹) conduct well, insulators like wood (K ≈ 0.1 W m⁻¹ K⁻¹) poorly. In series arrangement (composite slabs), total thermal resistance R = Σ(x/KA); in parallel, conductances add. CBSE 11 Physics notes emphasize the analogy with electrical resistance: temperature difference analogous to voltage, heat flow rate to current. The 2024-25 board paper included a 3-mark numerical on composite wall conduction.
- Fourier's law (heat flow rate) | Q/t = KA(T₁ - T₂)/x | K = thermal conductivity (W m⁻¹ K⁻¹), x = thickness
- Thermal resistance | R = x/(KA) | Analogous to electrical resistance; unit K/W
- Series combination (slabs) | R_total = R₁ + R₂ +... | (Q/t) = (T₁ - Tₙ)/(ΣRᵢ)
- Parallel combination | 1/R_total = 1/R₁ + 1/R₂ +... | Equal temperature difference across each path
- Temperature at interface (2 slabs) | T = (K₁T₁/x₁ + K₂T₂/x₂)/(K₁/x₁ + K₂/x₂) | Steady state, no heat accumulation
Heat Transfer: Convection and Radiation
Convection transfers heat through bulk fluid motion (natural or forced); no simple formula like conduction, but heat transfer rate depends on surface area, temperature difference, and convection coefficient h. Radiation needs no medium—electromagnetic waves carry energy. Stefan-Boltzmann law applies to ideal black bodies (emissivity ε = 1); real objects have ε < 1. Newton's law of cooling approximates convective and radiative cooling when temperature difference is small. Wien's displacement law explains why hot objects glow red (longer λ) while hotter ones appear blue-white (shorter λ). These concepts appear in CBSE 11 Physics theory questions and JEE MCQs.
- Stefan-Boltzmann law | E = σAT⁴ | Power radiated by black body; σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴
- Radiated power (real body) | P = εσAT⁴ | ε = emissivity (0 ≤ ε ≤ 1); T in Kelvin only
- Net radiation exchange | P_net = εσA(T⁴ - T₀⁴) | Between body at T and surroundings at T₀
- Wien's displacement law | λₘT = b | b = 2.9 × 10⁻³ m·K; λₘ = peak wavelength
- Newton's law of cooling | dT/dt = -k(T - T₀) | Rate of cooling proportional to excess temperature
- Newton's law (integrated) | T - T₀ = (Tᵢ - T₀)e⁻ᵏᵗ | Exponential decay; k depends on surface properties
Key Definitions and Physical Quantities
Understanding precise definitions prevents conceptual errors in CBSE exams. Temperature measures average kinetic energy of molecules; heat is energy in transit due to temperature difference. Thermal equilibrium means no net heat flow between systems at same temperature (Zeroth Law). Specific heat is intensive property (independent of mass); heat capacity is extensive (proportional to mass). Black body is idealized perfect absorber and emitter (ε = 1, α = 1). Emissivity ε and absorptivity α are equal for any body at thermal equilibrium (Kirchhoff's law). Thermal conductivity K quantifies conduction ability; insulators have low K, conductors high K. These terms appear in 1-mark definition questions regularly.
- Temperature: Measure of average kinetic energy of molecules; determines direction of heat flow; SI unit Kelvin (K)
- Heat: Energy transferred between systems due to temperature difference; not a state function; unit joule (J)
- Thermal equilibrium: State where two systems in contact have no net heat exchange; same temperature
- Specific heat capacity: Heat required per unit mass per unit temperature rise; intensive property
- Latent heat: Heat per unit mass for phase change at constant temperature; breaks intermolecular bonds
- Thermal conductivity: Material property indicating heat conduction rate; metals high, insulators low
- Black body: Ideal body that absorbs all incident radiation; perfect emitter (ε = 1)
- Emissivity: Ratio of emission by real body to black body at same temperature; 0 ≤ ε ≤ 1
Important Physical Constants and Values
Memorize these standard values from NCERT Class 11 Physics for quick substitution in numerical problems. The Stefan-Boltzmann constant σ and Wien's constant b appear in radiation calculations. Specific heats and latent heats of common substances are frequently needed. Always use absolute temperature (Kelvin) in Stefan-Boltzmann and Wien's laws—adding 273 to Celsius values. In calorimetry problems, assume specific heat of water as 4200 J kg⁻¹ K⁻¹ (or 1 cal g⁻¹ °C⁻¹) unless stated otherwise. Coefficients of linear expansion α are typically in the range 10⁻⁵ to 10⁻⁶ K⁻¹ for common solids. Keep units consistent: convert grams to kilograms, centimeters to meters before calculation.
- Stefan-Boltzmann constant: σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴
- Wien's displacement constant: b = 2.9 × 10⁻³ m·K (or 0.29 cm·K)
- Specific heat of water: c = 4186 J kg⁻¹ K⁻¹ ≈ 4200 J kg⁻¹ K⁻¹ = 1 cal g⁻¹ °C⁻¹
- Latent heat of fusion of ice: Lf = 3.34 × 10⁵ J/kg = 80 cal/g at 0°C
- Latent heat of vaporization of water: Lv = 2.26 × 10⁶ J/kg = 540 cal/g at 100°C
- Coefficient of linear expansion (typical): Steel α ≈ 1.2 × 10⁻⁵ K⁻¹; Copper α ≈ 1.7 × 10⁻⁵ K⁻¹; Glass α ≈ 9 × 10⁻⁶ K⁻¹
- Thermal conductivity examples: Copper K = 385 W m⁻¹ K⁻¹; Aluminium K = 205 W m⁻¹ K⁻¹; Steel K ≈ 50 W m⁻¹ K⁻¹; Glass K ≈ 0.8 W m⁻¹ K⁻¹
Common Mistakes, Units, and Sign Conventions
Many Class 11 Physics solutions lose marks due to unit errors and sign mismanagement. Always use SI units in formulas: mass in kg (not g), length in m (not cm), temperature in K for Stefan-Boltzmann and Wien's laws but K or °C interchangeable for ΔT. In calorimetry, heat gained is positive, heat lost is negative—but magnitudes are equal in principle of calorimetry. Never write temperature as negative in Stefan-Boltzmann law (must be absolute). When calculating thermal expansion, ensure L₀ is initial length, not final. In Newton's law of cooling, T - T₀ must remain positive (body hotter than surroundings). Thermal resistance R has unit K/W, not °C/W (numerically same but conceptually Kelvin is absolute scale).
- Unit error: Always convert g → kg, cm → m before substituting in formulas; check final unit matches expected
- Temperature scale: Use Kelvin (not Celsius) in T⁴ radiation formulas; for ΔT either scale works (numerically equal)
- Sign in calorimetry: Heat gained (+), heat lost (−); equate magnitudes: |Q_lost| = |Q_gained|
- Expansion formula: ΔL = αL₀ΔT uses initial length L₀, not average or final length
- Latent heat direction: Melting/boiling absorbs heat (+Q); freezing/condensation releases heat (−Q)
- Thermal resistance: R = x/(KA) increases with thickness x, decreases with area A and conductivity K
- Newton's cooling: Valid only for small ΔT (≈ 30 K difference); exponential decay form used for larger differences
- Emissivity range: 0 ≤ ε ≤ 1; black body ε = 1, perfect reflector ε = 0; real objects between
Memory Tricks and Mnemonics
Smart mnemonics help retain formulas under exam pressure. For expansion coefficients, remember 'LAG': Linear-Area-Gamma with ratio 1:2:3. Calorimetry problems: 'Hot Loses, Cold Gains' (HL=CG). Stefan-Boltzmann T⁴ dependence: 'Stefan's Tea For Tuesday' (T⁴). Wien's law: 'Wavelength Maximum Times Temperature' = constant (WMT). Specific heat water is 4200 ≈ 'four-two-hundred' J kg⁻¹ K⁻¹. Latent heat of vaporization is roughly 6-7 times latent heat of fusion. For series thermal resistance: resistances add like series electrical resistors; for parallel, conductances add. These tricks work best when paired with practice—solving 20-30 numericals from Class 11 Physics notes cements recall.
- LAG rule: Linear (α), Area (β = 2α), Gamma (γ = 3α)—ratio 1:2:3
- Calorimetry sign: 'HL = CG' → Heat Lost equals (Cold) Heat Gained
- Stefan's T⁴: 'Stefan's Tea For Tuesday'—temperature to fourth power
- Wien's peak: 'Wavelength Maximum × Temperature = constant'—inverse relation
- Water specific heat: '4-2-hundred' → 4200 J kg⁻¹ K⁻¹ (or 1 cal g⁻¹ °C⁻¹)
- Latent heat ratio: Lv/Lf ≈ 6.7 for water (2.26/0.334 × 10⁶/10⁵)
- Thermal resistance: Series add (like electrical R); parallel conductances add (like 1/R)
Last-Minute Revision Box
This one-page summary is your 15-minute review before the CBSE Class 11 Physics exam. Cover the page, recall each formula, then check. Focus on units and when each applies. Stefan-Boltzmann and Wien's laws need Kelvin; calorimetry and expansion can use Celsius for ΔT. Practice dimensional analysis: [Q/t] = [KA(ΔT)/x] = W confirms Fourier's law. Remember phase-change problems involve multiple steps: heat ice, melt ice, heat water, boil water, heat steam—each with distinct formula. Thermal Properties of Matter typically yields 4-6 marks in CBSE board; master these formulas and you secure them. For deeper conceptual clarity and personalized doubt-solving with photo upload of any problem, students across India now rely on CBSETUTOR.ai—offering 24×7 AI tutoring for Classes 6-12 at just ₹999/month (one flat fee for all subjects). Start your 3-day free trial and transform chapter revision into confident problem-solving.
- Expansion: ΔL = αL₀ΔT (linear), ΔA = 2αA₀ΔT (area), ΔV = 3αV₀ΔT (volume); γ = 3α
- Calorimetry: Q = mcΔT (no phase change); Q = mL (phase change at constant T); Heat lost = Heat gained
- Conduction: (Q/t) = KA(T₁ - T₂)/x; series R_total = ΣRᵢ where Rᵢ = xᵢ/(KᵢAᵢ)
- Radiation: P = εσAT⁴ (T in Kelvin); Net: P = εσA(T⁴ - T₀⁴); Wien: λₘT = 2.9 × 10⁻³ m·K
- Newton cooling: dT/dt = -k(T - T₀) linear form; T(t) = T₀ + (Tᵢ - T₀)e⁻ᵏᵗ exponential form
- Constants: σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴; c_water = 4200 J kg⁻¹ K⁻¹; Lf(ice) = 3.34 × 10⁵ J/kg; Lv(water) = 2.26 × 10⁶ J/kg
- Check units: mass kg, length m, T in K (for T⁴), ΔT in K or °C (same numerically), power in W, heat in J
- Common traps: Use L₀ (initial) not L (final) in expansion; T⁴ needs absolute temp; ε ≤ 1; phase change at constant T
Frequently asked questions
What is the difference between heat and temperature in Chapter 10?+
Temperature measures average kinetic energy of molecules and determines direction of heat flow. Heat is energy transferred between systems due to temperature difference—it is not a property of a body but energy in transit. In CBSE exams, remember: temperature is measured in K or °C, heat in joules or calories.
Why do we use γ = 3α for volume expansion but not always exact?+
The relation γ = 3α holds strictly for isotropic solids (properties same in all directions). For anisotropic materials like crystals, expansion coefficients differ along axes so γ ≠ 3α. NCERT Class 11 Physics assumes isotropic materials unless stated otherwise, so use γ = 3α in numerical problems.
How to solve calorimetry problems with multiple substances and phase changes?+
List all processes step-by-step: heating ice (mcΔT), melting (mLf), heating water (mcΔT), boiling (mLv), heating steam. Sum heat for each step. Equate total heat lost by hot body to total heat gained by cold body. Units must be consistent. Water equivalent simplifies: treat calorimeter as equivalent mass of water.
When do I use Stefan-Boltzmann law versus Newton's law of cooling?+
Stefan-Boltzmann law (P = σAT⁴) calculates radiated power from hot bodies; used when radiation dominates. Newton's law of cooling (rate ∝ ΔT) is an approximation for convective and radiative cooling when temperature excess is small (< 30 K). For large ΔT or pure radiation, use Stefan-Boltzmann with net power formula P = εσA(T⁴ - T₀⁴).
What are common mistakes students make in thermal expansion numericals?+
Using final length instead of initial length L₀; mixing units (cm and m, g and kg); forgetting to convert Celsius to Kelvin where needed (though ΔT is same in both); applying γ = 3α to liquids (correct only for isotropic solids); confusing coefficient of apparent expansion with real expansion in liquid-in-container problems.
How to remember the values of latent heats for water?+
Latent heat of fusion Lf ≈ 334 kJ/kg (think '3-3-4' or 80 cal/g). Latent heat of vaporization Lv ≈ 2260 kJ/kg (think '2-2-6-0' or 540 cal/g). Mnemonic: 'Vaporization is roughly 7 times fusion'—this ratio helps cross-check. These appear in 90 percent of Class 11 Physics solutions involving phase change.
Why must I use Kelvin in Stefan-Boltzmann law but Celsius is okay in calorimetry?+
Stefan-Boltzmann law involves T⁴—using Celsius (which can be negative or zero) gives wrong results. Absolute temperature (Kelvin) is mandatory. In calorimetry Q = mcΔT, only temperature difference matters: ΔT in Kelvin equals ΔT in Celsius numerically (a 10 K rise = 10 °C rise), so either scale works for ΔT.
How does CBSETUTOR.ai help with Thermal Properties of Matter numericals?+
CBSETUTOR.ai offers 24×7 AI tutoring for CBSE Class 11 Physics. Upload a photo of any numerical from Chapter 10, get step-by-step solutions with formula explanations. Practice unlimited calorimetry, expansion, conduction problems with instant feedback. All subjects for Classes 6-12 at ₹999/month flat fee. Start a free 3-day trial and see your scores improve within weeks.
What is thermal resistance and how is it analogous to electrical resistance?+
Thermal resistance R = x/(KA) opposes heat flow just as electrical resistance opposes current. Temperature difference ΔT is analogous to voltage, heat flow rate Q/t to current. In series, R_total = ΣRᵢ (like series resistors); in parallel, conductances 1/R add (like parallel resistors). This analogy simplifies composite slab problems in CBSE exams.
How many marks does Chapter 10 carry in CBSE Class 11 Physics board exam?+
Thermal Properties of Matter typically carries 4-6 marks in the CBSE Class 11 Physics annual exam—usually one 3-mark numerical and one or two 1-2 mark questions (definition, derivation, conceptual). JEE Main allocates 1-2 questions from this chapter. Mastering formulas and solving 20-30 NCERT and exemplar problems ensures full marks.
Related resources
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