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Class 11 Physics · NCERT Chapter 10

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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.

Chapter:Thermal Properties of MatterClass:Class 11Subject:PhysicsBoard:CBSEExams:CBSE · JEE · NEET
8 Key Formulas3 Solved Examples40 Practice MCQs
DWritten byDeep Narayan
Updated
Key Concept Summary

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.

01

Temperature and Temperature Scales

Definition

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.

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

Q1
The boiling point of water in the Fahrenheit scale is —
Q2
Absolute zero is approximately —
Q3
A change of 1 kelvin is equal to a change of —
Q4
Heat always flows from —

1 Mark Questions

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

2 Mark Questions

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

3 Mark Questions

  1. 1
    Explain the need for an absolute (Kelvin) scale of temperature and write its relation with the Celsius scale.
  2. 2
    At what temperature do the Celsius and Fahrenheit scales read the same numerical value? Derive your answer.
02

Thermal Expansion of Solids and Liquids

Definition

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.

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

Q1
For a solid, the relation between volume (γ) and linear (α) expansion coefficients is —
Q2
The coefficient of superficial expansion β is related to α by —
Q3
Water has its maximum density at —
Q4
A bimetallic strip bends on heating because —

1 Mark Questions

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

2 Mark Questions

  1. 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. 2
    Why are gaps left between successive sections of railway tracks?
  3. 3
    Show that for a solid β = 2α and γ = 3α.

3 Mark Questions

  1. 1
    Define α, β and γ for a solid, and derive the relations β = 2α and γ = 3α.
  2. 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.)
03

Calorimetry and Specific Heat

Definition

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.

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

Q1
The heat needed to raise the temperature of mass m by ΔT with specific heat c is —
Q2
The SI unit of specific heat is —
Q3
The substance with an unusually high specific heat is —
Q4
In the method of mixtures, the heat lost by the hotter bodies —

1 Mark Questions

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

2 Mark Questions

  1. 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. 2
    State the principle of the method of mixtures and describe how a calorimeter is used.
  3. 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. 1
    Define specific heat capacity and heat capacity, and state their SI units. Explain the meaning of the method of mixtures.
  2. 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.)
04

Change of State and Latent Heat

Definition

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.

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

Q1
During a change of state at constant pressure, the temperature of the substance —
Q2
The heat required to melt unit mass of a solid is its —
Q3
The latent heat of vaporisation of water is approximately —
Q4
Perspiration cools the body because —

1 Mark Questions

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

2 Mark Questions

  1. 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. 2
    Explain why steam at 100 °C causes more severe burns than water at 100 °C.
  3. 3
    State the latent heat of fusion and vaporisation of water.

3 Mark Questions

  1. 1
    Define latent heat of fusion and vaporisation, and explain why temperature stays constant during a change of state.
  2. 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.)
05

Heat Transfer: Conduction, Convection and Radiation

Definition

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.

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

Q1
The SI unit of thermal conductivity is —
Q2
Heat transfer by the bulk motion of a fluid is called —
Q3
The mode of heat transfer that does not require a medium is —
Q4
Heat from the Sun reaches the Earth by —

1 Mark Questions

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

2 Mark Questions

  1. 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. 2
    Explain how convection produces a sea breeze.
  3. 3
    Why are good insulators used in building walls? Distinguish between conductors and insulators.

3 Mark Questions

  1. 1
    Explain the three modes of heat transfer — conduction, convection and radiation — giving one example of each.
  2. 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.)
06

Black-Body Radiation and Newton's Law of Cooling

Definition

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.

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.

Definition

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

Q1
The power radiated by a black body is proportional to —
Q2
The value of the Stefan-Boltzmann constant σ is approximately —
Q3
A body that absorbs all incident radiation and emits perfectly is called —
Q4
According to Newton's law of cooling, the rate of cooling is proportional to —

1 Mark Questions

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

2 Mark Questions

  1. 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. 2
    Why are radiators painted black while vacuum flasks are silvery?
  3. 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. 1
    State and explain the Stefan-Boltzmann law and Newton's law of cooling, contrasting their temperature dependences.
  2. 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.)
07

NCERT Exercise Questions

From the textbook

All NCERT questions for this chapter — NCERT Physics Part II, Chapter 10 (Thermal Properties of Matter). Questions keep their NCERT numbering so you can match them to the book.

Exercises (NCERT Q10.1–Q10.20)

  • 10.1) The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
  • 10.2) Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T_A and T_B?
  • 10.3) The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: R = R₀[1 + α(T − T₀)]. The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?
  • 10.4) Answer the following: (a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)? (b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale? (c) The absolute temperature (Kelvin scale) T is related to the temperature t_c on the Celsius scale by t_c = T − 273.15. Why do we have 273.15 in this relation, and not 273.16? (d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?
  • 10.5) Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made: (temperature, P_A, P_B) triple-point of water: 1.250 × 10⁵ Pa, 0.200 × 10⁵ Pa; normal melting point of sulphur: 1.797 × 10⁵ Pa, 0.287 × 10⁵ Pa. (a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B? (b) What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?
  • 10.6) A steel tape 1 m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0 °C? Coefficient of linear expansion of steel = 1.20 × 10⁻⁵ K⁻¹.
  • 10.7) A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using 'dry ice'. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: α_steel = 1.20 × 10⁻⁵ K⁻¹.
  • 10.8) A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10⁻⁵ K⁻¹.
  • 10.9) A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of −39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm? Co-efficient of linear expansion of brass = 2.0 × 10⁻⁵ K⁻¹; Young's modulus of brass = 0.91 × 10¹¹ Pa.
  • 10.10) A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a 'thermal stress' developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10⁻⁵ K⁻¹, steel = 1.2 × 10⁻⁵ K⁻¹).
  • 10.11) The coefficient of volume expansion of glycerine is 49 × 10⁻⁵ K⁻¹. What is the fractional change in its density for a 30 °C rise in temperature?
  • 10.12) A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g⁻¹ K⁻¹.
  • 10.13) A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g⁻¹ K⁻¹; heat of fusion of water = 335 J g⁻¹).
  • 10.14) In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm³ of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?
  • 10.15) Given below are observations on molar specific heats at room temperature of some common gases. Gas, molar specific heat (C_v in cal mol⁻¹ K⁻¹): Hydrogen 4.87, Nitrogen 4.97, Oxygen 5.02, Nitric oxide 4.99, Carbon monoxide 5.01, Chlorine 6.17. The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?
  • 10.16) A child running a temperature of 101 °F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 minutes, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g⁻¹.
  • 10.17) A 'thermacole' icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is 0.01 J s⁻¹ m⁻¹ K⁻¹. [Heat of fusion of water = 335 × 10³ J kg⁻¹].
  • 10.18) A brass boiler has a base area of 0.15 m² and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s⁻¹ m⁻¹ K⁻¹; Heat of vaporisation of water = 2256 × 10³ J kg⁻¹.
  • 10.19) Explain why: (a) a body with large reflectivity is a poor emitter (b) a brass tumbler feels much colder than a wooden tray on a chilly day (c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace (d) the earth without its atmosphere would be inhospitably cold (e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.
  • 10.20) A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.

Answers at a glance

  1. 1
    Q10.1 → Neon: −248.58 °C, −415.44 °F; carbon dioxide: −56.60 °C, −69.88 °F.
  2. 2
    Q10.2 → T_A = (4/7) T_B, i.e. T_A : T_B = 4 : 7.
  3. 3
    Q10.3 → The temperature is about 385 K.
  4. 4
    Q10.4 → (a) the triple point is unique, ice/steam points vary with pressure (b) absolute zero, 0 K (c) 0 °C is at 273.15 K, the melting point, not the triple point (d) 491.7.
  5. 5
    Q10.5 → (a) A reads 392.69 K, B reads 391.98 K (b) non-ideal gas behaviour; use low pressures and extrapolate to zero pressure.
  6. 6
    Q10.6 → Actual length at 45 °C ≈ 63.014 cm; length at 27 °C = 63.0 cm.
  7. 7
    Q10.7 → The wheel slips on when the shaft cools to about −69 °C.
  8. 8
    Q10.8 → The hole diameter increases by about 0.0144 cm (1.44 × 10⁻² cm).
  9. 9
    Q10.9 → A tension of 3.8 × 10² N develops in the wire.
  10. 10
    Q10.10 → Combined change in length ≈ 0.35 cm; no thermal stress since the ends are free.
  11. 11
    Q10.11 → Density decreases by a fraction 1.47 × 10⁻² (1.47%).
  12. 12
    Q10.12 → The block temperature rises by 103 °C.
  13. 13
    Q10.13 → About 1.45 kg of ice can melt.
  14. 14
    Q10.14 → c ≈ 0.43 × 10³ J kg⁻¹ K⁻¹; with heat losses the computed value is smaller than the actual one.
  15. 15
    Q10.15 → Diatomic gases store extra energy in rotational (≈5/2 R) and for chlorine also vibrational modes; chlorine's large value shows its vibrations are active at room temperature.
  16. 16
    Q10.16 → The average extra evaporation rate is 4.3 g min⁻¹.
  17. 17
    Q10.17 → About 3.7 kg of ice remains after 6 hours.
  18. 18
    Q10.18 → The flame part in contact with the boiler is at about 238 °C.
  19. 19
    Q10.19 → (a) reflectors absorb (and so emit) little (b) brass conducts heat from the hand, wood does not (c) open-air iron is a poor radiator, furnace iron approaches a black body (d) the atmosphere traps heat (e) steam releases latent heat on condensing.
  20. 20
    Q10.20 → It takes 10 minutes to cool from 60 °C to 30 °C.
08

CBSE Pattern Questions

How CBSE asks this chapter

Physics board paper structure: Section A (MCQ/VSA 1M), Section B (2M), Section C (3M), Section D (5M). Thermal Properties of Matter typically carries ~7 marks: 1-mark MCQs on specific heat or thermal expansion, a 2-mark numerical on heat exchange or expansion, a 3-mark derivation of the conduction law or Newton's law of cooling, and a 5-mark calorimetry/radiation question.

MCQ Questions

Q1
The heat required to raise the temperature of unit mass of a substance by 1 K is its
Q2
The coefficient of area expansion of a solid is related to its linear expansion coefficient α by
Q3
The heat required to convert 1 kg of ice at 0°C to water at 0°C is
Q4
A body that absorbs all radiation incident on it is called

1 Mark Questions

  1. 1
    Define specific heat capacity and give its SI unit.
  2. 2
    Define latent heat of fusion and latent heat of vaporisation.
  3. 3
    Define the coefficient of linear expansion of a solid.
  4. 4
    State Newton's law of cooling.

2 Mark Questions

  1. 1
    A copper rod of length 1 m is heated from 20°C to 100°C. If the coefficient of linear expansion is 1.7 × 10⁻⁵ K⁻¹, find the increase in length. (Ans: Δl = αlΔT = 1.36 mm)
  2. 2
    How much heat is needed to raise the temperature of 200 g of water from 25°C to 75°C? (c = 4.18 × 10³ J/kg K) (Ans: Q = mcΔT = 4.18 × 10⁴ J)
  3. 3
    5 kg of water at 60°C is mixed with 5 kg of water at 20°C. Find the final temperature assuming no heat loss. (Ans: 40°C)

3 Mark Questions

  1. 1
    State and explain the law of thermal conduction (Fourier's law) and derive the heat-current formula for a rod of length l and area A. (Ans: H = kA(θ₁ − θ₂)/l)
  2. 2
    Explain the principle of calorimetry and describe how the specific heat of a solid is measured by the method of mixtures.
  3. 3
    State Stefan's law of black body radiation and explain the Newton's law of cooling as its consequence. (Ans: E ∝ T⁴)

5 Mark Questions

  1. 1
    A copper calorimeter of mass 100 g contains 150 g of water at 25°C. Steam at 100°C is passed until the temperature reaches 30°C. Find the mass of steam condensed (L = 2.26 × 10⁶ J/kg, c = 4180 J/kg K).
  2. 2
    Explain heat transfer by conduction, convection and radiation with examples, describe the green-house effect, and state Stefan's law and Wien's displacement law.
09

PDF: Assignments & Tests

Printable study material

Free downloadable PDFs for Thermal Properties of Matter — two topic-wise assignments covering every concept of the chapter, and two full CBSE-pattern tests (each 90 minutes, 40 marks) mirroring the board paper. No signup required.

Free PDF Downloads & Assignments

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Practice path: complete Assignment 1 first (topic-by-topic), then Assignment 2 (mixed difficulty with full solutions), then attempt the timed tests.

Thermal Properties of Matter Assignment 1 — Topic-wise Questions

Covers all six topics: temperature and temperature scales; thermal expansion of solids and liquids; calorimetry and specific heat; change of state and latent heat; heat transfer — conduction, convection and radiation; and black-body radiation and Newton's law of cooling. Arranged topic by topic.

thermal-properties-of-matter-class-11-notes-assignment-1.pdf
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Thermal Properties of Matter Assignment 2 — Challenge + Integrated Problems

Higher-difficulty set with multi-topic integration problems (scale conversions, expansion, calorimetry, latent heat, conduction, black-body radiation, Newton's cooling) and full worked answers. Includes an answer key.

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Thermal Properties of Matter Test 1 — CBSE Pattern (90 min, 40 marks)

Full board-pattern test: Section A (MCQ/VSA 1M × 10), Section B (2M × 4), Section C (3M × 4), Section D (5M × 2) = 40 marks. Time: 90 minutes. Includes answer key.

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Thermal Properties of Matter Test 2 — CBSE Pattern (90 min, 40 marks)

Second full paper with different questions, same 40-mark / 90-minute blueprint. Includes answer key.

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Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Temperature scales

Linear expansion

Expansion relations

Specific heat

Latent heat

Conduction

Stefan-Boltzmann

Newton's cooling

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • 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).

Practice Numericals

Step-by-step worked examples

Standard exam-level problems solved with clear conceptual breakdowns.

1

Heat to warm water

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

Step-by-Step Solution

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

Final Answer

5.04 × 10⁵ J

2

Heat to melt ice then warm it

Problem StatementHow 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.)

Step-by-Step Solution

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

Final Answer

4.18 × 10⁴ J

3

Power radiated by a black body

Problem StatementA 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.)

Step-by-Step Solution

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

Final Answer

≈ 2.84 × 10³ W

FAQ

Frequently asked 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.

Interactive Quiz

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Active Recall Practice

Chapter MCQ Mock Test

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