Thermal Properties of Matter

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

Overview Notes NEET Questions Interactive Learning

7 interactive concept widgets for Thermal Properties of Matter. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

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Temperature scale converter

Thermal expansion (linear / areal / volumetric)

Heat exchange (calorimetry)

Heating curve of water (with phase changes)

Fourier's law and composite slabs

Stefan-Boltzmann + Wien

Newton's law of cooling

Temperature and expansion

Convert between temperature scales, and watch how solids expand on heating.

Temperature

Temperature scale converter

Convert between Celsius, Fahrenheit and Kelvin. Pick a preset or type any value.

Temperature in °C

Quick presets (Celsius)

Celsius

25.00 °C

Fahrenheit

77.00 °F

Kelvin

298.15 K

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0 K = −273.15 °C is absolute zero. Nothing can be colder.
−40 °C = −40 °F is the only point where Celsius and Fahrenheit scales coincide.
Body temperature 37 °C = 98.6 °F = 310 K.

Thermal expansion

Thermal expansion (linear / areal / volumetric)

Toggle between linear, areal, and volumetric expansion. β = 2α and γ = 3α relate them.

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

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For an isotropic solid, β = 2α and γ = 3α. The 1 : 2 : 3 ratio is a NEET classic.
A 1 m steel rod expands by ~1 mm for a 100 K rise. Small but matters in railway tracks and bridges.
Bimetallic strips work because two metals have different α — the strip bends on heating.

Heat exchange and phase changes

Calorimetry between two bodies, plus the heating curve of water with both latent-heat plateaus.

Calorimetry

Heat exchange (calorimetry)

Two bodies at different temperatures placed in thermal contact reach a common equilibrium temperature determined by their masses and specific heats.

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

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Equal masses of water at 80 °C and 20 °C → mid-point 50 °C (water has same c on both sides).
A small hot iron (low c, low m) hardly raises the temperature of a large body of water (high c, high m).
Heat lost by hot = heat gained by cold: the principle of calorimetry.

Latent heat

Heating curve of water (with phase changes)

Heat water through phase changes. The widget shows energy needed for each segment, including the latent-heat plateaus at 0 °C and 100 °C.

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

Try this

From −20 °C to +120 °C: ice heats up, melts at 0 °C (plateau), water heats, vaporises at 100 °C (huge plateau), steam heats. Two big spikes from latent heat.
L_v / L_f ≈ 6.8 — vaporising water needs ~7× more energy than melting. That's why steam burns are far worse than ice burns.
Doubling the mass doubles every term — the curve's shape is unchanged.

Heat transfer and radiation

Conduction (single + composite slabs), Stefan-Boltzmann + Wien together, and Newton's law of cooling.

Conduction

Fourier's law and composite slabs

Heat current through a single slab, or through series / parallel composite slabs. Conductivity adds in parallel; resistance adds in series.

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

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Series: same heat flows through both slabs. Total thermal resistance adds.
Parallel: same temperature difference across each slab. Effective conductivity is the area-weighted average.
Doubling A doubles the heat current. Doubling L (single slab) halves it.

Radiation

Stefan-Boltzmann + Wien

Power radiated per m² scales as T⁴. The peak wavelength shifts as 1/T. Together they describe black-body radiation.

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}

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Sun (5800 K) peaks at ~500 nm — green-yellow visible light. We see white light by combining all visible.
Body at 310 K → peak ~10 μm in deep infrared. That is what thermal cameras see.
Doubling T multiplies radiated power by 16 (T⁴ scaling). A red-hot iron at 1000 K emits 16× as much per m² as one at 707 K.

Newton cooling

Newton's law of cooling

The temperature of a hot body decays exponentially toward the ambient temperature, with a rate proportional to the temperature excess.

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}

Try this

The temperature excess (above ambient) decays exponentially with time constant 1/k.
Half-life: time for the excess temperature to drop to half. ln 2 / k.
After 5 half-lives (~3.5 / k), the body is within ~3% of ambient.

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