CBSE Class 11 Physics Chapter 11 Notes: Thermal Properties of Matter

Chapter Introduction

Why does a metal spoon feel colder than a wooden spoon at the same temperature? Why do railway tracks have gaps between sections? How does a thermometer work? Chapter 11 of NCERT Physics for Class 11 deals with the thermal properties of matter — how bodies respond to heat. From thermal expansion and calorimetry to the three modes of heat transfer (conduction, convection, radiation), this chapter connects everyday phenomena to fundamental physics. The radiation laws (Stefan-Boltzmann, Wien's displacement) also serve as a bridge to modern physics.

Key Definitions

Detailed Concepts

Temperature Scales

Conversion formulas:

• T(K) = T(°C) + 273.15
• T(°F) = (9/5) T(°C) + 32

Thermal Expansion

When a body is heated, its molecules vibrate more vigorously and occupy more space. The three types of expansion are:

Why these relations hold: Consider a cube of side L. On heating, each side expands by factor (1 + α_l ΔT). The new volume = L³(1 + α_l ΔT)³ ≈ L³(1 + 3α_l ΔT) for small ΔT. Hence α_v ≈ 3α_l. Similarly, the new area of a face ≈ L²(1 + 2α_l ΔT), giving α_a ≈ 2α_l.

Anomalous expansion of water: Water contracts from 0 °C to 4 °C and expands above 4 °C. This is why lakes freeze from the top — water at 4 °C (maximum density) sinks to the bottom, preserving aquatic life.

Specific Heat Capacity and Calorimetry

Heat required: Q = msΔT, where m = mass, s = specific heat, ΔT = temperature change.

Calorimetry (principle of mixtures): In an insulated system, heat lost by hot body = heat gained by cold body.

m₁s₁(T₁ – T) = m₂s₂(T – T₂), where T is the final equilibrium temperature.

Molar specific heats for gases:

• At constant volume: C_v
• At constant pressure: C_p
• Mayer's relation: C_p – C_v = R (for ideal gas)

Change of State and Latent Heat

During a phase change, temperature remains constant even though heat is being supplied/removed.

Heat during phase change: Q = mL, where L is the appropriate latent heat.

• L_f for ice = 3.36 × 10⁵ J/kg
• L_v for water = 2.26 × 10⁶ J/kg

Note: L_v >> L_f because vaporisation requires overcoming intermolecular forces AND doing work against atmospheric pressure to create volume expansion.

Heat Transfer: Conduction

Fourier's law of heat conduction: The rate of heat flow through a slab is:

H = dQ/dt = KA(ΔT/Δx)

where K = thermal conductivity, A = cross-sectional area, ΔT/Δx = temperature gradient.

For a slab of thickness L with temperatures T₁ and T₂ on either side:

H = KA(T₁ – T₂) / L

Thermal resistance: R_th = L / KA (analogous to electrical resistance R = ρl/A)

Thermal conductivity values (in W m⁻¹ K⁻¹):

Good conductors (metals) have K >> 1; insulators (wood, air) have K << 1.

Heat Transfer: Convection

Convection is heat transfer by the actual motion of the fluid itself. Hot fluid rises (lower density), cool fluid sinks (higher density), setting up a convection current.

Types:

• Natural convection: Driven by buoyancy (e.g., sea breeze, atmospheric circulation)
• Forced convection: Driven by a pump or fan (e.g., car radiator, fan-forced heating)

Heat Transfer: Radiation

Radiation is heat transfer by electromagnetic waves — no medium is required.

Stefan-Boltzmann law: Total power radiated by a body at temperature T:

P = σeAT⁴

where σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ (Stefan-Boltzmann constant), e = emissivity (0 ≤ e ≤ 1), A = surface area.

For a black body (e = 1): P = σAT⁴

Net radiated power (body at T in surroundings at T₀): P_net = σeA(T⁴ – T₀⁴)

Wien's displacement law: The wavelength at which a black body emits maximum radiation:

λ_m T = b = 2.898 × 10⁻³ m·K

Hotter bodies radiate at shorter wavelengths (blue-shift). This is why a heated iron goes from dark red → orange → yellow as temperature increases.

Newton's law of cooling (approximation for small ΔT):

dT/dt = –k(T – T₀)

where T₀ is the surrounding temperature and k is a positive constant. This gives an exponential cooling curve: T(t) = T₀ + (T_initial – T₀)e⁻ᵏᵗ

Important Derivations

1. Relation α_v ≈ 3α_l

Consider a cube of side L₀ at temperature T₀. When heated to T₀ + ΔT:

New side = L₀(1 + α_l ΔT)

New volume = L₀³(1 + α_l ΔT)³ ≈ L₀³(1 + 3α_l ΔT) (neglecting higher-order terms)

But also: New volume = V₀(1 + α_v ΔT)

Comparing: α_v ≈ 3α_l

2. Newton's Law of Cooling from Stefan-Boltzmann Law

Net power: P = σeA(T⁴ – T₀⁴) = σeA(T² + T₀²)(T + T₀)(T – T₀)

For T close to T₀: T + T₀ ≈ 2T₀, T² + T₀² ≈ 2T₀²

So P ≈ σeA × 4T₀³ × (T – T₀) = k(T – T₀), where k = 4σeAT₀³

This gives: dT/dt = –k(T – T₀) → Newton's law of cooling

Common Mistakes

1. Confusing heat and temperature: Heat is energy in transit (J); temperature is a measure of hotness (K). A large iceberg at 0 °C has enormous heat content but low temperature. 2. Using Celsius in gas laws and radiation laws: Always convert to Kelvin before using ideal gas equation, Stefan-Boltzmann law, or Wien's law. 3. Forgetting latent heat during phase change: During melting or boiling, temperature does NOT change. Heat goes into breaking intermolecular bonds, not raising temperature. Q = mL, not Q = msΔT. 4. Applying linear expansion formula to area or volume: Use α_a ≈ 2α_l for area and α_v ≈ 3α_l for volume, not α_l directly. 5. Wrong Stefan-Boltzmann formula: Power is proportional to T⁴ (absolute temperature in Kelvin), not ΔT⁴. Using Celsius temperature gives completely wrong results. 6. Confusing thermal conductors and insulators: High K = good conductor (metals); low K = good insulator (wood, air). Do not mix them up in practical applications.

Numerical Practice

Example 1: Thermal Expansion

A steel rod of length 5 m is heated from 20 °C to 70 °C. If α_l = 1.2 × 10⁻⁵ /°C, find the increase in length.

ΔL = α_l L₀ ΔT = 1.2 × 10⁻⁵ × 5 × (70 – 20) = 1.2 × 10⁻⁵ × 5 × 50 = 3.0 × 10⁻³ m = 3 mm

Example 2: Calorimetry

50 g of ice at 0 °C is mixed with 200 g of water at 40 °C. Find the final temperature. (s_water = 4186 J/kg·K, L_f = 3.36 × 10⁵ J/kg)

Heat to melt ice: Q₁ = mL_f = 0.05 × 3.36 × 10⁵ = 16800 J

Heat to raise melted ice to T: Q₂ = 0.05 × 4186 × (T – 0) = 209.3T

Heat lost by warm water: Q₃ = 0.2 × 4186 × (40 – T) = 837.2(40 – T)

Heat gained = Heat lost: 16800 + 209.3T = 837.2(40 – T)

16800 + 209.3T = 33488 – 837.2T

1046.5T = 16688 → T ≈ 15.9 °C

Example 3: Stefan-Boltzmann Law

A black body at 1000 K has a surface area of 0.01 m². Find the total power radiated.

P = σAT⁴ = 5.67 × 10⁻⁸ × 0.01 × (1000)⁴ = 5.67 × 10⁻⁸ × 0.01 × 10¹² = 567 W

Exam Tips

1. Temperature conversions are common 1-mark questions: Know °C ↔ K and °C ↔ °F. 2. Thermal expansion relations (α_a ≈ 2α_l, α_v ≈ 3α_l): Be ready to derive or justify these. 3. Calorimetry numericals: Set up the heat balance equation carefully. If a phase change is involved, include mL for the transition AND msΔT for subsequent heating. 4. Three modes of heat transfer: Know the mechanism, formula, and at least one example for each (conduction, convection, radiation). 5. Newton's law of cooling: The derivation from Stefan-Boltzmann law is a 3-mark favourite. Also be prepared to solve exponential cooling problems. 6. Wien's law: Usually a 1-mark formula recall or a quick numerical (find λ_m given T or vice versa). 7. Draw a heating curve: Temperature vs heat supplied graph showing plateaus during phase changes — this is a frequently tested diagram.

NCERT Exercise Focus Areas

The NCERT exercises for this chapter test a well-defined set of problem types. Knowing the pattern helps you allocate preparation time wisely.

Exam strategy: Calorimetry-with-phase-change and conduction-through-slabs problems carry the most marks and are where students lose points. Practise at least 5 problems of each type. Wien's law and temperature-conversion questions are quick scoring opportunities — do not skip them.

Conceptual Questions for Practice

Test your understanding with these short-answer questions. They require no calculation — just clear reasoning from the concepts above.

1. Why are metallic cooking utensils provided with wooden or plastic handles? Think about thermal conductivity and which material resists heat flow.

2. Why does a thick glass tumbler crack more easily than a thin one when hot water is poured into it? Consider differential expansion between the inner and outer surfaces and how thickness affects the temperature gradient.

3. Two bodies at the same temperature are brought into contact. Will heat flow between them? Justify your answer using the definition of heat.

4. A hole in a metal sheet is heated. Does the hole expand or contract? Apply the concept of area expansion to the material around the hole.

5. Why does a piece of steel feel colder than a piece of wood at the same room temperature? Explain using thermal conductivity and the rate of heat transfer away from your skin.

Quick Revision Notes

Key constants: σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, b = 2.898 × 10⁻³ m·K

Anomalous water: Maximum density at 4 °C, minimum volume at 4 °C