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Class 11 Physics · NCERT Chapter 10

Thermal Properties of Matter Class 11 Physics Notes

Complete, exam-ready notes on the thermal properties of matter: temperature and thermometric scales, thermal expansion of solids and liquids, calorimetry and specific heat capacity, change of state and latent heat, heat transfer by conduction, convection and radiation, Newton's law of cooling and the Stefan-Boltzmann law — every NCERT topic with MCQs, mark-wise questions and solved numericals for CBSE, JEE and NEET.

ChapterThermal Properties of MatterClassClass 11SubjectPhysicsBoardCBSEExamsCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is the specific heat capacity of a substance, and what is its SI unit?

Specific heat capacity is the heat required to raise the temperature of unit mass by one degree: Q = mcΔT. Its SI unit is J kg⁻¹ K⁻¹. Water has an unusually high value (≈ 4200 J kg⁻¹ K⁻¹), which is why it is used for cooling and regulates climate.

Temperature and Temperature Scales

Heat and temperature

Heat is the energy transferred between two bodies because of a temperature difference; it flows from the hotter to the colder body. Temperature is a measure of the hotness or coldness of a body. Heat is measured in joules, and temperature in kelvin.

The three common temperature scales are Celsius, Fahrenheit and Kelvin. They are related by C/5 = (F − 32)/9 = (K − 273.15)/5. The Kelvin scale is the absolute thermometric scale; its zero is absolute zero, the lowest possible temperature (−273.15 °C), at which molecular motion is minimal.

C5=F−329=K−273.155\frac{C}{5} = \frac{F - 32}{9} = \frac{K - 273.15}{5}
Relation between Celsius, Fahrenheit and Kelvin scales
  • Boiling point of water: 100 °C = 212 °F = 373.15 K.
  • Freezing point of water: 0 °C = 32 °F = 273.15 K.
  • A change of 1 K equals a change of 1 °C; a change of 1 °F equals 5/9 K.

Kelvin is an absolute scale

Kelvin temperatures are never negative and always begin at absolute zero. Physical formulae for gases and radiation require temperatures in kelvin, not Celsius.

MCQ Questions

1The boiling point of water in the Fahrenheit scale is —

2Absolute zero is approximately —

3A change of 1 kelvin is equal to a change of —

4Heat always flows from —

1 Mark Questions

  1. Distinguish between heat and temperature.
  2. State the SI units of heat and temperature.
  3. What is absolute zero?

2 Mark Questions

  1. Convert 98.6 °F (normal body temperature) to the Celsius scale.
  2. Write the relation between the Celsius, Fahrenheit and Kelvin scales and define each.
  3. Convert 300 K to the Celsius and Fahrenheit scales.

3 Mark Questions

  1. Explain the need for an absolute (Kelvin) scale of temperature and write its relation with the Celsius scale.
  2. At what temperature do the Celsius and Fahrenheit scales read the same numerical value? Derive your answer.

Thermal Expansion of Solids and Liquids

Thermal expansion

Most substances expand on heating. The fractional change in a dimension per degree rise in temperature is the coefficient of expansion. For a solid, α is the linear coefficient, β the superficial (area) coefficient and γ the volume coefficient, with β = 2α and γ = 3α.

ΔL=L0αΔT,β=2α,γ=3α\Delta L = L_0 \alpha \Delta T, \qquad \beta = 2\alpha, \qquad \gamma = 3\alpha
Linear expansion and the relations between expansion coefficients

The volume expansion of a liquid is described by γ_l (its coefficient of volume expansion). Since a liquid is always held in a container that also expands, the apparent expansion observed is slightly less than the real expansion. Water behaves oddly: it contracts on heating from 0 °C to 4 °C (its density is maximum at 4 °C).

  • Steel and concrete in bridges and railway tracks expand on hot days — expansion joints allow for this.
  • A bimetallic strip curves on heating because the two metals expand by different amounts; it is used in thermostats.
  • Water expands on freezing — the reason pipes burst in winter.

Water's strange maximum density

Water is most dense at 4 °C, so it contracts when heated from 0–4 °C and expands above 4 °C. This is why lakes freeze from the top and aquatic life survives the winter beneath.

MCQ Questions

1For a solid, the relation between volume (γ) and linear (α) expansion coefficients is —

2The coefficient of superficial expansion β is related to α by —

3Water has its maximum density at —

4A bimetallic strip bends on heating because —

1 Mark Questions

  1. Define the coefficient of linear expansion.
  2. Write the relations between α, β and γ for a solid.
  3. At what temperature is the density of water maximum?

2 Mark Questions

  1. A steel rod of length 1 m and coefficient of linear expansion 1.1 × 10⁻⁵ K⁻¹ is heated from 20 °C to 120 °C. Find its increase in length.
  2. Why are gaps left between successive sections of railway tracks?
  3. Show that for a solid β = 2α and γ = 3α.

3 Mark Questions

  1. Define α, β and γ for a solid, and derive the relations β = 2α and γ = 3α.
  2. A copper plate has a hole of diameter 1 cm at 20 °C. Find the new diameter of the hole when the plate is heated to 220 °C. (α = 1.7 × 10⁻⁵ K⁻¹.)

Calorimetry and Specific Heat

Specific heat capacity

The specific heat capacity c of a substance is the heat required to raise the temperature of unit mass by one degree: Q = mcΔT. Its SI unit is J kg⁻¹ K⁻¹. The molar specific heat is the heat per mole per degree.

Q=mcΔTQ = mc\Delta T
Heat required to change temperature

Calorimetry is the measurement of heat. In a calorimeter, the principle of the method of mixtures states that the heat lost by hotter bodies equals the heat gained by colder ones (assuming no heat loss to the surroundings). Water's high specific heat (≈ 4200 J kg⁻¹ K⁻¹) makes it an ideal coolant and a climate regulator.

  • Heat lost = heat gained in an ideal calorimeter.
  • Water's high specific heat stabilises coastal climates and cools engines.
  • Sand and land heat and cool faster than water because their specific heat is far smaller.

Water as a heat regulator

Because water needs a lot of heat to warm up (high c), oceans and lakes absorb heat in summer and release it slowly in winter, moderating coastal temperatures.

MCQ Questions

1The heat needed to raise the temperature of mass m by ΔT with specific heat c is —

2The SI unit of specific heat is —

3The substance with an unusually high specific heat is —

4In the method of mixtures, the heat lost by the hotter bodies —

1 Mark Questions

  1. Define specific heat capacity. Write its SI unit.
  2. Write the expression Q = mcΔT and identify each symbol.
  3. Why is water used as a coolant?

2 Mark Questions

  1. How much heat is required to raise the temperature of 500 g of water from 20 °C to 40 °C? (c = 4200 J kg⁻¹ K⁻¹.)
  2. State the principle of the method of mixtures and describe how a calorimeter is used.
  3. A block of iron of mass 200 g at 100 °C is dropped into 200 g of water at 20 °C in a calorimeter. Explain why the final temperature is not 60 °C.

3 Mark Questions

  1. Define specific heat capacity and heat capacity, and state their SI units. Explain the meaning of the method of mixtures.
  2. A piece of metal of mass 100 g at 100 °C is placed in 200 g of water at 20 °C in a calorimeter of negligible heat capacity. If the final temperature is 25 °C, find the specific heat of the metal. (c_water = 4200 J kg⁻¹ K⁻¹.)

Change of State and Latent Heat

Latent heat

The latent heat of a substance is the heat absorbed or released when it changes state at constant temperature. The specific latent heat of fusion L_f is the heat to melt unit mass of a solid, and L_v the heat to vaporise unit mass of a liquid: Q = mL.

Q=mL,Lf=3.34×105 J kg−1 (water)Q = mL, \qquad L_f = 3.34\times10^5\ \text{J kg}^{-1}\ (\text{water})
Latent heat and heat of fusion of water

During a change of state the temperature stays constant while heat is absorbed (melting, boiling, sublimation) or released (freezing, condensation, deposition). The latent heat of vaporisation of water is about 2.26 × 10⁶ J kg⁻¹ — over six times its heat of fusion — which is why steam burns are so damaging.

  • Ice at 0 °C absorbs 3.34 × 10⁵ J per kg to melt to water at 0 °C.
  • Water at 100 °C absorbs 2.26 × 10⁶ J per kg to become steam at 100 °C.
  • Perspiration cools the body because the sweat evaporates, taking latent heat from the skin.

Steam burns worse than boiling water

Steam at 100 °C releases its large latent heat of vaporisation (2.26 × 10⁶ J kg⁻¹) when it condenses on skin, delivering far more energy than an equal mass of boiling water — which only cools from 100 °C.

MCQ Questions

1During a change of state at constant pressure, the temperature of the substance —

2The heat required to melt unit mass of a solid is its —

3The latent heat of vaporisation of water is approximately —

4Perspiration cools the body because —

1 Mark Questions

  1. Define latent heat.
  2. Write the expression Q = mL and identify each symbol.
  3. Why is welding done so that heat is applied during melting and boiling?

2 Mark Questions

  1. How much heat is needed to convert 100 g of ice at 0 °C to water at 0 °C? (L_f = 3.34 × 10⁵ J kg⁻¹.)
  2. Explain why steam at 100 °C causes more severe burns than water at 100 °C.
  3. State the latent heat of fusion and vaporisation of water.

3 Mark Questions

  1. Define latent heat of fusion and vaporisation, and explain why temperature stays constant during a change of state.
  2. Find the heat required to convert 200 g of ice at −10 °C to steam at 100 °C. (c_ice = 2100, c_water = 4200, L_f = 3.34×10⁵, L_v = 2.26×10⁶ J kg⁻¹, all in J kg⁻¹ K⁻¹.)

Heat Transfer: Conduction, Convection and Radiation

Conduction

Conduction is the transfer of heat through a material from its hotter to its colder parts without any bulk motion of the material itself. It obeys Fourier's law: the rate of heat flow H = kA(ΔT/L), where k is the thermal conductivity and A the cross-section.

H=Qt=kA T1−T2LH = \frac{Q}{t} = kA\,\frac{T_1 - T_2}{L}
Rate of heat flow by conduction

Metals are good conductors (high k); air and insulating materials like glass wool and wood are poor conductors (low k). Convection is heat transfer by the bulk motion of a fluid — warm fluid rises and cool fluid sinks, setting up currents. Radiation is the emission of energy in the form of electromagnetic waves; it needs no medium and can travel through vacuum (solar heat).

  • Thermal conductivity k has SI unit W m⁻¹ K⁻¹.
  • Convection explains land and sea breezes and the circulation of hot water in a geyser.
  • Radiation is the only mode of heat transfer that can travel through a vacuum.

Sea breeze and land breeze

By day, land heats faster than the sea, warm air rises over land and cool air flows in from the sea — a sea breeze. At night the land cools faster and the breeze reverses — a land breeze. Both are convection currents.

MCQ Questions

1The SI unit of thermal conductivity is —

2Heat transfer by the bulk motion of a fluid is called —

3The mode of heat transfer that does not require a medium is —

4Heat from the Sun reaches the Earth by —

1 Mark Questions

  1. Define thermal conductivity.
  2. Name the three modes of heat transfer.
  3. Which mode of heat transfer can operate in a vacuum?

2 Mark Questions

  1. A metal rod of length 0.5 m and cross-section 10⁻⁴ m² has a temperature difference of 100 °C across its ends. If k = 200 W m⁻¹ K⁻¹, find the rate of heat flow.
  2. Explain how convection produces a sea breeze.
  3. Why are good insulators used in building walls? Distinguish between conductors and insulators.

3 Mark Questions

  1. Explain the three modes of heat transfer — conduction, convection and radiation — giving one example of each.
  2. A window pane is 1 cm thick, 1 m × 1 m in area, with a temperature difference of 20 °C between its faces. Find the heat conducted per second. (k_glass = 0.8 W m⁻¹ K⁻¹.)

Black-Body Radiation and Newton's Law of Cooling

Black body and its radiation

A black body is an ideal body that absorbs all radiation incident on it and emits radiation at all wavelengths. The Stefan-Boltzmann law states that the power radiated per unit area is proportional to the fourth power of the absolute temperature: P = σAT⁴, with σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴.

P=σAT4,net rate=σA(T4−T04)P = \sigma A T^4, \qquad \text{net rate} = \sigma A (T^4 - T_0^4)
Stefan-Boltzmann law for black-body radiation

Good absorbers are also good emitters; dark surfaces absorb and radiate heat well, while shiny (polished) surfaces reflect radiation and radiate poorly. This is why radiators are painted black and solar collectors use dark surfaces.

Newton's law of cooling

When a body cools, the rate of loss of heat is proportional to the excess of its temperature over that of the surroundings (for small temperature differences): dT/dt ∝ (T − T₀). The cooling curve is exponential.

  • The rate of radiation depends on T⁴, so a tiny rise in temperature greatly increases radiated power.
  • Thermometers and kettles are often polished to reduce heat loss by radiation.
  • A black body is the best absorber and the best emitter of radiation.

Why radiators are black

A black surface absorbs and emits radiation best, so household radiators are painted black to maximise heat loss, while silver vacuum flasks use shiny surfaces to minimise it.

MCQ Questions

1The power radiated by a black body is proportional to —

2The value of the Stefan-Boltzmann constant σ is approximately —

3A body that absorbs all incident radiation and emits perfectly is called —

4According to Newton's law of cooling, the rate of cooling is proportional to —

1 Mark Questions

  1. State the Stefan-Boltzmann law.
  2. Define a black body.
  3. State Newton's law of cooling.

2 Mark Questions

  1. A body of surface area 0.1 m² at temperature 500 K radiates as a black body. Find the power radiated. (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴.)
  2. Why are radiators painted black while vacuum flasks are silvery?
  3. A body cools from 80 °C to 70 °C in 5 minutes when the room is at 20 °C. Estimate how long it takes to cool from 70 °C to 60 °C using Newton's law of cooling.

3 Mark Questions

  1. State and explain the Stefan-Boltzmann law and Newton's law of cooling, contrasting their temperature dependences.
  2. A black-body sphere of radius 5 cm at 1000 K radiates energy. Find the rate of energy radiated. (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴.)

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Temperature scales

C5=F−329=K−273.155\frac{C}{5} = \frac{F-32}{9} = \frac{K-273.15}{5}

Linear expansion

ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T

Expansion relations

β=2α,γ=3α\beta = 2\alpha, \quad \gamma = 3\alpha

Specific heat

Q=mcΔTQ = mc\Delta T

Latent heat

Q=mLQ = mL

Conduction

H=kA T1−T2LH = kA\,\frac{T_1 - T_2}{L}

Stefan-Boltzmann

P=σAT4P = \sigma A T^4

Newton's cooling

dTdt∝(T−T0)\frac{dT}{dt} \propto (T - T_0)

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • Kelvin temperatures are used in all gas and radiation formulae.
  • For a solid: β = 2α, γ = 3α.
  • Water has its maximum density at 4 °C.
  • Q = mcΔT; water c ≈ 4200 J kg⁻¹ K⁻¹.
  • Latent heat: ice L_f ≈ 3.34 × 10⁵, steam L_v ≈ 2.26 × 10⁶ J kg⁻¹.
  • Conduction needs a medium; radiation does not.
  • Stefan-Boltzmann: P ∝ T⁴; Newton's cooling: rate ∝ (T − T₀).

Solved problems

Worked examples

JEE / NEET-style numericals, solved step by step.

1

Heat to warm water

How much heat is required to raise 2 kg of water from 20 °C to 80 °C? (c_water = 4200 J kg⁻¹ K⁻¹.)

  1. Q = mcΔT = 2 × 4200 × (80 − 20) = 2 × 4200 × 60.
  2. Q = 504000 J = 5.04 × 10⁵ J.

Answer

5.04 × 10⁵ J

2

Heat to melt ice then warm it

How much heat is needed to melt 100 g of ice at 0 °C and raise the resulting water to 20 °C? (L_f = 3.34 × 10⁵ J kg⁻¹, c = 4200 J kg⁻¹ K⁻¹.)

  1. Heat to melt: Q₁ = mL = 0.1 × 3.34 × 10⁵ = 3.34 × 10⁴ J.
  2. Heat to warm: Q₂ = mcΔT = 0.1 × 4200 × 20 = 8400 J.
  3. Total = 33400 + 8400 = 41800 J.

Answer

4.18 × 10⁴ J

3

Power radiated by a black body

A black body of surface area 0.05 m² is at 1000 K. Find the power radiated. (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴.)

  1. P = σAT⁴ = 5.67 × 10⁻⁸ × 0.05 × (1000)⁴.
  2. P = 5.67 × 10⁻⁸ × 0.05 × 10¹² = 2835 W.

Answer

≈ 2.84 × 10³ W

FAQ

Common questions

What is the difference between heat and temperature?

Heat is the energy transferred between bodies because of a temperature difference; it flows from the hotter to the colder body and is measured in joules. Temperature is the degree of hotness of a body, measured in kelvin.

Why is water used as a coolant?

Water has a very high specific heat capacity (about 4200 J kg⁻¹ K⁻¹), so it absorbs a large amount of heat for only a small rise in temperature. This makes it ideal for cooling engines and regulating climate.

Why does the temperature stay constant during a change of state?

The heat absorbed (or released) during melting, boiling or condensation is used to change the potential energy of the molecules rather than their kinetic energy — so the temperature, which measures average kinetic energy, does not change. The heat involved is the latent heat.

What are the three modes of heat transfer?

Conduction (transfer through a material by molecular vibration, no bulk motion), convection (transfer by the bulk motion of a fluid carrying heat), and radiation (transfer by electromagnetic waves, which needs no medium and can travel through vacuum).

Why does a body radiate more as its temperature rises sharply?

By the Stefan-Boltzmann law the radiated power is proportional to the fourth power of the absolute temperature (P = σAT⁴). A small rise in temperature therefore produces a very large increase in the rate of radiated energy.

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