Thermal Properties of Matter Notes for NEET — Class 11 NCERT

Introduction

Heat is energy in transit because of a temperature difference. Thermal physics is the study of how heat moves and how matter responds — through expansion, phase changes, and energy balance.

For NEET 2027 expect 1 to 2 questions. Calorimetry, conduction (especially composite slabs), Stefan-Boltzmann and Newton's law of cooling are the favourites. Six formulas in the cheat sheet cover most of it.

Temperature and temperature scales

Three scales appear in NEET:

T_{K} = T_{C} + 273.15, T_{F} = \frac{9}{5} T_{C} + 32.

Absolute zero is $0 K = - 273.15° C$ . The size of one Celsius degree equals the size of one Kelvin. One Fahrenheit degree is $5/9$ of a Celsius degree.

Temperature in °C

Quick presets (Celsius)

Celsius

25.00 °C

Fahrenheit

77.00 °F

Kelvin

298.15 K

Thermal expansion

Heat a solid and it expands. Three coefficients describe the expansion:

Δ L = L_{0} α Δ T, Δ A = A_{0} β Δ T, Δ V = V_{0} γ Δ T .

For an isotropic solid:

β = 2 α, γ = 3 α (α : β : γ = 1 : 2 : 3) .

Typical values: brass $α \approx 19 \times 1 0^{- 6} K^{- 1}$ , steel $\approx 11 \times 1 0^{- 6}$ , glass $\approx 9 \times 1 0^{- 6}$ , copper $\approx 17 \times 1 0^{- 6}$ .

Linear coefficient α: 1.10e-5 K⁻¹

Initial L₀: 1.00 m

ΔT: 50 K

Effective coefficient

α = 1.10e-5 K⁻¹

Change

Δ = 5.500e-4 m (0.055%)

Final L

1.000550 m

Δ L = L_{0} α Δ T = 1 \times 1.10 e - 5 \times 50 = 5.500 e - 4 m

Anomalous expansion of water

Most substances expand on heating. Water between 0 °C and 4 °C contracts on heating — a unique anomaly. Density is maximum at 4 °C. This is why ice floats and lakes freeze top down, allowing aquatic life to survive winter.

Specific heat capacity

Heat needed to raise unit mass by one kelvin:

Q = m c Δ T .

SI unit J/(kg·K). Water has the highest of common substances: $c_{w} = 4186 J/(kg\cdotK)$ . Compare metals: copper $\approx 390$ , aluminium $\approx 900$ , iron $\approx 460$ .

Molar specific heat

Heat per mole per kelvin. For ideal gases, molar specific heat at constant volume $C_{V}$ and at constant pressure $C_{P}$ differ by $C_{P} - C_{V} = R$ (Mayer's relation).

Calorimetry — heat exchange

When two bodies at different temperatures are placed in thermal contact (in an insulated container), heat flows from the hotter body to the cooler until they reach a common temperature. Energy conservation:

m_{1} c_{1} (T_{1} - T_{f}) = m_{2} c_{2} (T_{f} - T_{2}) .

Solving:

T_{f} = \frac{m _{1} c _{1} T _{1} + m _{2} c _{2} T _{2}}{m _{1} c _{1} + m _{2} c _{2}} .

Hot body

Mass m₁: 200 g

c₁: 460 J/(kg·K)

T₁: 100 °C

Cold body

Mass m₂: 500 g

c₂: 4186 J/(kg·K)

T₂: 20 °C

Equilibrium temperature

23.37 °C

T_{f} = \frac{m _{1} c _{1} T _{1} + m _{2} c _{2} T _{2}}{m _{1} c _{1} + m _{2} c _{2}}

Heat lost by hot

7050 J

Heat gained by cold

7050 J

Latent heat and phase changes

At a phase transition, heat is absorbed or released without a change in temperature:

Q = m L .

For water:

Latent heat of fusion (ice → water at 0 °C): $L_{f} = 334 kJ/kg$ .
Latent heat of vaporization (water → steam at 100 °C): $L_{v} = 2260 kJ/kg$ .

The heating curve of water (heat in vs temperature out) shows two horizontal plateaus at 0 °C and 100 °C where heat goes into phase change instead of temperature.

Mass m: 100 g

Initial temperature T_i: -20 °C

Final temperature T_f: 120 °C

Total heat needed

Q = 309.46 kJ

Ice heating

4.18 kJ

Ice → water (fusion)

33.40 kJ

Water heating

41.86 kJ

Water → steam (vaporization)

226.00 kJ

Steam heating

4.02 kJ

c_ice = 2090 J/(kg·K) · c_water = 4186 · c_steam = 2010

L_fusion = 334 kJ/kg · L_vaporization = 2260 kJ/kg

Q = \sum m c Δ T + \sum m L

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Conduction (Fourier's law)

Heat flows through a solid because hotter molecules pass kinetic energy to colder neighbours. Steady-state rate of flow through a slab of thickness $L$ , area $A$ , between hot face $T_{1}$ and cold face $T_{2}$ :

\frac{d Q}{d t} = \frac{k A ( T _{1} - T _{2} )}{L} .

$k$ is thermal conductivity, units W/(m·K). Copper $\approx 400$ , aluminium $\approx 200$ , steel $\approx 50$ , glass $\approx 1$ , water $\approx 0.6$ , air $\approx 0.024$ .

Composite slabs in series

For two slabs joined face to face, heat flow is the same through both. Total thermal resistance $R = L / (k A)$ adds:

R_{eq} = R_{1} + R_{2} \Rightarrow \frac{L _{eq}}{k _{eq} A} = \frac{L _{1}}{k _{1} A} + \frac{L _{2}}{k _{2} A} .

Composite slabs in parallel

\frac{1}{R _{eq}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} .

k₁: 400 W/(m·K)

L₁: 1.00 m

Cross-section A: 1.00 cm²

T_hot: 100 °C

T_cold: 0 °C

Heat current

dQ/dt = 4.000 W

\frac{d Q}{d t} = \frac{k A Δ T}{L} = \frac{400 \times 0.0001 \times 100}{1} = 4.000 W

Convection

Heat transfer in fluids by bulk motion: hotter (less dense) regions rise, cooler regions sink. Examples: boiling water, sea breezes, atmospheric circulation. Newton's law of cooling (for convective heat loss from a body to the surroundings) follows: rate of heat loss is proportional to temperature difference.

Thermal radiation

All bodies emit electromagnetic radiation depending on their temperature. Radiation does not need a medium and travels at the speed of light.

Stefan-Boltzmann's law

Power radiated per unit area by a black body:

\frac{P}{A} = σ T^{4} .

$σ = 5.67 \times 1 0^{- 8} W/(m^{2} \cdotK^{4})$ . For a real body multiply by emissivity $e$ (between 0 and 1). Doubling $T$ multiplies emitted power by $2^{4} = 16$ .

Wien's displacement law

The wavelength at which a black body radiates most strongly:

λ_{m a x} T = b, b = 2.898 \times 1 0^{- 3} m\cdotK .

Sun ( $T \approx 5800$ K) peaks at $λ \approx 500 nm$ — visible. A red star at 3000 K peaks in the infrared. Rising temperature shortens the peak wavelength (red → yellow → blue).

Temperature T: 5800 K

Emissivity e: 1.00 (1.0 = black body)

Quick presets

Radiated power per m² (Stefan-Boltzmann)

6.416e+7 W/m²

\frac{P}{A} = e σ T^{4} = 1.00 \times 5.67 \times 1 0^{- 8} \times 580 0^{4}

Peak wavelength (Wien)

λ_peak = 500 nm

Visible

λ_{m a x} = \frac{b}{T} = \frac{2.898 \times 1 0 ^{- 3}}{5800}

Newton's law of cooling

For small temperature differences (and convective surroundings), the rate of cooling is proportional to the difference between the body's temperature and the surroundings:

\frac{d T}{d t} = - k (T - T_{s}) .

Solution: $T (t) - T_{s} = (T_{0} - T_{s}) e^{- k t}$ . The temperature difference decays exponentially with time constant $1/ k$ .

Initial T₀: 80 °C

Surroundings T_s: 20 °C

Cooling constant k: 0.050 /min

Time t: 10.0 min

Temperature at t = 10.0 min

56.39 °C

Half-life of excess temperature

τ½ = ln 2 / k = 13.86 min

T (t) - T_{s} = (T_{0} - T_{s}) e^{- k t}

Worked NEET problems

NEET-style problem · Calorimetry

Question

A piece of iron of mass

200 g

100 ° C

is dropped into

500 g

of water at

20 ° C

in a calorimeter. Take

c_{iron} = 460 J/(kg\cdotK)

c_{w} = 4186 J/(kg\cdotK)

. Find the equilibrium temperature.

Solution

T_{f} = \frac{m _{1} c _{1} T _{1} + m _{2} c _{2} T _{2}}{m _{1} c _{1} + m _{2} c _{2}} .

Numerator: $(0.2) (460) (100) + (0.5) (4186) (20) = 9200 + 41860 = 51060$ .

Denominator: $(0.2) (460) + (0.5) (4186) = 92 + 2093 = 2185$ .

$T_{f} = 51060/2185 \approx 23.4 ° C$ .

NEET-style problem · Latent heat

Question

How much heat is needed to convert

100 g

of ice at

0 ° C

into water at

50 ° C

? Take

L_{f} = 334 kJ/kg

c_{w} = 4.186 kJ/(kg\cdotK)

Solution

Two stages:

Q_{1} = m L_{f} = 0.1 \times 334 = 33.4 kJ .

Q_{2} = m c_{w} Δ T = 0.1 \times 4.186 \times 50 = 20.93 kJ .

Q_{total} = 33.4 + 20.93 \approx 54.3 kJ .

NEET-style problem · Conduction

Question

A copper rod of length

1 m

and area

1 0^{- 4} m^{2}

conducts heat between

100 ° C

and

0 ° C

. With

k = 400 W/(m\cdotK)

, the steady-state heat current is:

Solution

\frac{d Q}{d t} = \frac{k A Δ T}{L} = \frac{400 \times 1 0 ^{- 4} \times 100}{1} = 4 W .

NEET-style problem · Stefan-Boltzmann

Question

A black body radiates at

500

K. By what factor does the radiated power per unit area change if the temperature is doubled?

Solution

$P / A \propto T^{4}$ . Doubling $T$ multiplies $P / A$ by $2^{4} = 16$ .

NEET-style problem · Thermal expansion

Question

A steel rod has length

1 m

20 ° C

. Its length at

120 ° C

is approximately (

α = 1.1 \times 1 0^{- 5} K^{- 1}

Solution

Δ L = L_{0} α Δ T = 1 \times 1.1 \times 1 0^{- 5} \times 100 = 1.1 \times 1 0^{- 3} m = 1.1 mm .

New length ≈ 1.0011 m.

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Summary cheat sheet

Temperature: $T_{K} = T_{C} + 273.15$ , $T_{F} = \frac{9}{5} T_{C} + 32$ .
Expansion: $β = 2 α$ , $γ = 3 α$ .
Heat for ΔT: $Q = m c Δ T$ .
Calorimetry: $T_{f} = \frac{\sum m _{i} c _{i} T _{i}}{\sum m _{i} c _{i}}$ .
Latent heat: $Q = m L$ . Water L_f = 334 kJ/kg, L_v = 2260 kJ/kg.
Conduction: $d Q / d t = k A Δ T / L$ . Series: R adds. Parallel: 1/R adds.
Stefan-Boltzmann: $P / A = σ T^{4}$ , $σ = 5.67 \times 1 0^{- 8} W/(m^{2} \cdotK^{4})$ .
Wien: $λ_{m a x} T = 2.898 \times 1 0^{- 3} m\cdotK$ .
Newton's cooling: $d T / d t = - k (T - T_{s})$ .

Next: try the interactive widgets for thermal expansion, calorimetry and Stefan-Boltzmann radiation, or work through the 30+ NEET PYQs with full solutions. To time yourself, take the free 10-question mock test.

Frequently asked questions

How many questions come from Thermal Properties of Matter in NEET 2027?

You can expect 1 to 2 questions from this chapter in NEET 2027. The chapter has medium PYQ frequency. Calorimetry, latent heat, conduction (especially composite slabs), Stefan-Boltzmann's law and Newton's law of cooling are the favourite topics.

What is the relation between Celsius, Fahrenheit and Kelvin?

T_K equals T_C plus 273.15. T_F equals (9 over 5) times T_C plus 32. So 0 °C is 273.15 K and 32 °F. The size of one Celsius degree equals the size of one Kelvin (no division). One Fahrenheit degree is 5/9 of a Celsius degree.

What is the relation between linear, areal and volumetric expansion coefficients?

For an isotropic solid, the area expansion coefficient beta equals 2 alpha (twice the linear coefficient), and the volume expansion coefficient gamma equals 3 alpha. So alpha : beta : gamma equals 1 : 2 : 3. NEET expects you to know this ratio.

What is specific heat capacity?

Specific heat capacity c is the heat needed to raise the temperature of unit mass of a substance by one Kelvin. Q equals m c delta T. Water has the highest specific heat among common substances, about 4186 J per kg per K, which is why oceans moderate climate.

What is latent heat?

Latent heat L is the heat absorbed (or released) per unit mass during a phase change without a temperature change. Q equals m L. Water has L_fusion equals 334 kJ per kg (ice to water at 0 °C) and L_vaporization equals 2260 kJ per kg (water to steam at 100 °C).

What is Fourier's law of conduction?

Heat flow rate dQ over dt equals minus k A dT over dx. Here k is thermal conductivity, A is cross-section, and dT over dx is the temperature gradient. Steady-state through a slab gives dQ over dt equals k A (T_hot minus T_cold) over L.

What does Stefan-Boltzmann's law say?

Power radiated per unit area by a black body is sigma T to the fourth, where sigma is the Stefan-Boltzmann constant equals 5.67 times 10 to the minus 8 W per m squared per K to the fourth. For a real body multiply by emissivity e. Doubling temperature multiplies radiated power by 16.

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Thermal Properties of MatterNEET Physics · Class 11 · NCERT Chapter 10

Introduction

Temperature and temperature scales

Thermal expansion

Anomalous expansion of water

Specific heat capacity

Molar specific heat

Calorimetry — heat exchange

Latent heat and phase changes

Conduction (Fourier's law)

Composite slabs in series

Composite slabs in parallel

Convection

Thermal radiation

Stefan-Boltzmann's law

Wien's displacement law

Newton's law of cooling

Worked NEET problems

Summary cheat sheet

Frequently asked questions

How many questions come from Thermal Properties of Matter in NEET 2027?

What is the relation between Celsius, Fahrenheit and Kelvin?

What is the relation between linear, areal and volumetric expansion coefficients?

What is specific heat capacity?

What is latent heat?

What is Fourier's law of conduction?

What does Stefan-Boltzmann's law say?

Continue with the next chapter notes

Track Your NEET Score Across All 90 Chapters