Thermal Properties

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Key Formulas:

• Heat transfer: Q = mcΔT
• Latent heat: L = Q/m, so Q = mL
• Ideal gas: PV = nRT, PV = NkT
• Conduction: dQ/dt = -kA(dT/dx) (Fourier’s law)
• Stefan-Boltzmann: P = σAT⁴ (for blackbody)
• Newton’s law of cooling: dT/dt = -k(T - T_surr)
• Linear expansion: ΔL = αLΔT
• Area expansion: ΔA = 2αAΔT
• Volume expansion: ΔV = 3αVΔT

Exam Tips:

• Stefan-Boltzmann constant σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴
• Newton’s law of cooling only applies when temperature difference is small (< 30°C)
• Conduction requires a temperature gradient and a medium
• Radiation does not require a medium — works in vacuum
• Convection requires fluid motion
• Latent heat of fusion for ice = 336 kJ/kg; latent heat of vaporisation for water = 2260 kJ/kg

🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding of heat transfer and thermal properties.

Three Modes of Heat Transfer:

Conduction occurs through stationary matter via molecular collisions and electron movement (in metals). The rate of heat flow is given by dQ/dt = -kA(dT/dx), where k is the thermal conductivity, A is cross-sectional area, and dT/dx is the temperature gradient. Good conductors (metals) have high k; insulators have low k.

Convection involves bulk fluid movement driven by density differences caused by heating. Natural convection occurs due to buoyancy; forced convection uses fans or pumps. Convection is efficient for heating fluids and is responsible for weather patterns, ocean currents, and indoor heating.

Radiation is heat transfer via electromagnetic waves and requires no medium. All bodies radiate thermal energy according to P = εσAT⁴, where ε is the emissivity (0 ≤ ε ≤ 1). A perfect blackbody has ε = 1. Surfaces that are good emitters are also good absorbers (Kirchhoff’s law).

Thermal Conductivity Values (W/m·K):

Material	k (W/m·K)
Copper	~386
Aluminium	~205
Iron	~79
Steel	~45
Glass	~0.8
Water	~0.6
Brick	~0.6-1.0
Wood	~0.1-0.2
Air	~0.024
Wool	~0.04

Metals are good conductors due to free electrons; gases are poor conductors; liquids vary.

Stefan-Boltzmann Law Applications:

The law P = σAT⁴ applies to blackbody radiators. The intensity decreases with the fourth power of absolute temperature, explaining why small temperature changes cause large changes in radiative output. Applications include: stellar temperature estimation, infrared thermography, solar energy harvesting, and climate science. For a grey body (emissivity ε < 1): P = εσAT⁴.

Wien’s Displacement Law:

The peak wavelength of blackbody radiation λ_max is inversely proportional to absolute temperature: λ_max T = b = 2.898 × 10⁻³ m·K (Wien’s constant). As temperature increases, the peak shifts to shorter wavelengths (bluer). This explains why hot objects glow first red, then orange, then white (all colours present at high T). The Sun (T ≈ 5800 K) has λ_max ≈ 500 nm (green-yellow), which appears white against the dark sky.

Newton’s Law of Cooling:

The law states dT/dt = -k(T - T_surr), where k is a positive constant depending on the object’s properties. Integrating gives T(t) = T_surr + (T₀ - T_surr)e^(-kt). This exponential approach to ambient temperature is observed whenever the temperature difference is small (< 30°C). For larger differences, the cooling rate is non-linear. Applications include: estimating time of death in forensic science, cooling of food, and thermal comfort analysis.

Kirchhoff’s Law of Thermal Radiation:

At thermal equilibrium, the emissivity of a body equals its absorptivity: ε = α. A good emitter is a good absorber. This explains why a black-painted kettle cools faster than a shiny one (though shiny surfaces also radiate less, they absorb less too — for cooling, a dull black surface is preferable because it both absorbs and emits more radiation from the surroundings).

Heat Conduction Through Composite Walls:

For heat conduction through multiple layers in series, the total thermal resistance is the sum of individual resistances: R_total = Σ(d_i/k_i). The heat current is H = ΔT/R_total = ΔT/(Σ d_i/k_i). For parallel heat flow (rare but possible), the effective conductance is the sum of individual conductances. Composite walls in buildings use this principle: brick, insulation, plaster each contribute to overall thermal resistance.

Thermal Expansion in Bimetallic Strips:

Different materials expand by different amounts when heated. A bimetallic strip made of two different metals bonded together will bend when heated because one metal expands more than the other. This is used in thermostats: the bending activates an electrical switch when temperature changes. Linear expansion coefficients: brass ≈ 19 × 10⁻⁶ /°C, iron ≈ 12 × 10⁻⁶ /°C.

Worked NEET Example: A copper rod (α = 17 × 10⁻⁶ /°C) is 2 m long at 20°C. Find its length at 100°C.

• ΔT = 80°C
• ΔL = αLΔT = 17×10⁻⁶ × 2 × 80 = 2.72 × 10⁻³ m = 2.72 mm
• Final length = 2.00272 m

Previous Year NEET Numerical Examples:

• Stefan-Boltzmann calculations: comparing power radiated by two bodies at different temperatures
• Newton’s law of cooling: finding time of death or cooling time
• Composite wall heat conduction: calculating heat flow and temperature difference across each layer
• Linear expansion: calculating change in length for bridges, railway tracks, metal pipes with temperature change
• Latent heat: ice converting to water and then heating

🔴 Extended — Deep Study (3mo+)

Comprehensive derivation-based theory for complete mastery of thermal physics.

Stefan-Boltzmann Derivation from Blackbody Radiation:

The Stefan-Boltzmann law P = σAT⁴ can be derived from thermodynamics and electromagnetic theory. For a blackbody cavity at temperature T, the energy density inside is u = αT⁴ (proportional to T⁴). The radiation pressure p = u/3. The energy flux leaving through a small hole is J = pc = (α/3)cT⁴. Setting this equal to σT⁴ identifies σ = (α/3)c, where α is a constant from cavity radiation theory. Modern value: σ = 5.670374 × 10⁻⁸ W m⁻² K⁻⁴.

Planck’s Law Connection:

Planck’s law for spectral energy density u_λ = (8πhc/λ⁵) × 1/(e^(hc/λkT) - 1) reduces to the Rayleigh-Jeans law at long wavelengths and correctly predicts the fall-off at short wavelengths (ultraviolet catastrophe solved). Integrating Planck’s law over all wavelengths gives the total energy density u ∝ T⁴, confirming Stefan-Boltzmann. Wien’s displacement law λ_max T = b follows from differentiating Planck’s law.

Wien’s Displacement Law Derivation:

Setting derivative of Planck’s spectral distribution to zero gives the transcendental equation 5e^x = 5 + x, where x = hc/(λ_max kT). Solving gives x ≈ 4.965, giving b = hc/(xk) = 2.898 × 10⁻³ m·K. This constant is Wien’s displacement constant and applies to any blackbody.

Newton’s Law of Cooling Derivation:

Consider a body at temperature T in surroundings at T_s. The rate of heat loss is proportional to the temperature difference: dQ/dt = -hA(T - T_s), where h is the heat transfer coefficient (W/m²·K). Since dQ = mc dT (for small temperature changes where specific heat is constant): mc(dT/dt) = -hA(T - T_s). Rearranging: dT/dt = -(hA/mc)(T - T_s) = -k(T - T_s), where k = hA/mc. Integrating: ∫[T₀ to T] dT/(T - T_s) = -∫[0 to t] k dt, giving ln[(T - T_s)/(T₀ - T_s)] = -kt. Hence T(t) = T_s + (T₀ - T_s)e^(-kt). The time constant τ = 1/k = mc/(hA) is the characteristic cooling time.

Heat Conduction in Layered Slabs (Series Combination):

Consider two slabs of thicknesses d₁, d₂ and thermal conductivities k₁, k₂ with temperature difference ΔT_total across them. The steady-state heat current H must be the same through both (continuity). Therefore H = ΔT₁/(d₁/k₁) = ΔT₂/(d₂/k₂). Total ΔT = ΔT₁ + ΔT₂. Hence H = ΔT_total/(d₁/k₁ + d₂/k₂) and the total thermal resistance is R_total = d₁/k₁ + d₂/k₂. Generalising to n layers: R_total = Σ(d_i/k_i). The temperature drop across layer i is ΔT_i = H × (d_i/k_i).

Cylindrical Shell Heat Conduction:

For a hollow cylinder of inner radius r₁, outer radius r₂, length L, with T₁ > T₂ (inner surface hotter), the heat flow rate is H = 2πkL(T₁ - T₂)/ln(r₂/r₁). This follows from integrating dQ/dt = -kA(dT/dr) with A = 2πrL varying with radius. The thermal resistance of a cylindrical shell is R = ln(r₂/r₁)/(2πkL).

Critical Radius of Insulation for Cylinders:

Adding insulation to a pipe initially increases heat loss (insulation has lower k but increases surface area). The critical radius r_c = k_insulation/h, where h is the convective heat transfer coefficient. If the insulation radius exceeds r_c, further insulation reduces heat loss. For bare pipes in still air, r_c ≈ 10-20 mm, so typical pipe insulation is thicker than this threshold.

Latent Heat Calculations:

During phase change at constant temperature, Q = mL, where L is the specific latent heat. For ice at 0°C melting to water at 0°C: Q = m × 336 kJ/kg. For water at 100°C vaporising to steam at 100°C: Q = m × 2260 kJ/kg. The enthalpy of vaporisation is much larger than enthalpy of fusion because breaking intermolecular bonds completely (vaporisation) requires more energy than merely loosening molecular arrangement (fusion).

Triple Point Phase Diagrams:

At the triple point, all three phases (solid, liquid, gas) coexist in equilibrium. For water, the triple point occurs at 0.01°C (273.16 K) and 611.73 Pa (0.006 atm). This defines the Kelvin temperature scale. Phase boundaries: fusion curve (solid-liquid), vaporisation curve (liquid-gas), sublimation curve (solid-gas). The Clapeyron equation dP/dT = L/(TΔV) gives the slope of phase boundaries.

Entropy Change in Phase Transitions:

During a reversible isothermal phase change, ΔS = Q_rev/T = mL/T. For melting ice at 273 K: ΔS = (m × 336×10³)/273 ≈ 1230 J/kg·K. For vaporising water at 373 K: ΔS = (m × 2260×10³)/373 ≈ 6060 J/kg·K. Total entropy change of universe is positive for irreversible processes, consistent with the second law.

Carnot Refrigerator and Heat Pump:

A refrigerator absorbs heat Q_c from a cold reservoir at T_c and rejects heat Q_h to a hot reservoir at T_h, requiring work W = Q_h - Q_c. The coefficient of performance (COP) = Q_c/W = T_c/(T_h - T_c). A heat pump works similarly but the purpose is to heat the hot reservoir: COP_hp = Q_h/W = T_h/(T_h - T_c). Both COP values are greater than 1 and increase as the temperature difference decreases.

Heat Pumps:

Practical heat pumps use vapour-compression cycles with refrigerants. They can provide 2-5 kW of heating for every 1 kW of electrical input, making them 200-500% efficient (compared to resistive heating at 100%). They are most effective when temperature difference between reservoirs is small — ideal for underfloor heating in moderate climates.

Detailed NEET Problem Strategies:

1. Read the question twice — identify what’s being asked and what’s given
2. For conduction problems, draw the thermal circuit, calculate total resistance, find heat current, then find temperature drops
3. For Newton’s law of cooling, check if the temperature difference is small enough for the linear law to apply
4. For expansion problems, remember linear expansion applies to length, but calculate final length, not just the change
5. For radiation problems, convert all temperatures to Kelvin before using Stefan-Boltzmann law
6. Latent heat problems often involve two stages: melting/boiling (phase change) plus temperature change

Previous Year NEET Question Patterns:

• Composite wall conduction appears regularly, with students needing to find temperature at the interface
• Newton’s law of cooling frequently tested with coffee/tea cooling examples
• Stefan-Boltzmann comparisons of power radiated by bodies at different temperatures
• Linear expansion of railway tracks and bimetallic strips in thermostat applications
• Heat current H = ΔT/R_total for series conduction