🎯 Strategy Tip
Don't memorize formulas in isolation. Understand the dimensional structure — it tells you what each term represents. In an exam, if you forget a formula, dimensional analysis can reconstruct it.
🌡️ Temperature & Thermal Expansion
Temperature Scale Conversions
C/100 = (F − 32)/180 = (K − 273.15)/100 C = Celsius, F = Fahrenheit, K = Kelvin
Also: K = C + 273.15 | F = (9/5)C + 32
Also: K = C + 273.15 | F = (9/5)C + 32
Linear Thermal Expansion
ΔL = α · L₀ · ΔT → L = L₀(1 + αΔT) α = coefficient of linear expansion (K⁻¹), L₀ = original length, ΔT = temperature change
Dimensional formula of α: [K⁻¹] or [θ⁻¹]
Superficial (Area) Expansion
ΔA = β · A₀ · ΔT where β = 2α β = coefficient of superficial expansion, A₀ = original area
Key relation: β = 2α (for isotropic solids)
Volumetric (Cubic) Expansion
ΔV = γ · V₀ · ΔT where γ = 3α γ = coefficient of cubic expansion, V₀ = original volume
α : β : γ = 1 : 2 : 3 (most asked relationship)
Thermal Stress in Constrained Rod
Thermal Stress = Y · α · ΔT Y = Young's Modulus (Pa), α = linear expansion coeff., ΔT = temp change
Thermal Strain = α · ΔT | Force = Y · A · α · ΔT
Thermal Strain = α · ΔT | Force = Y · A · α · ΔT
⚠ Use Young's Modulus (not Bulk) for rods. Common error source.
Real vs Apparent Expansion of Liquid
γ_real = γ_apparent + γ_vessel (γ_vessel = 3α_vessel) γ_real = actual expansion of liquid, γ_apparent = observed expansion
For glass vessel: γ_vessel = 3α_glass ≈ 3 × 9×10⁻⁶ K⁻¹
⚗️ Heat Capacity & Calorimetry
Heat Transfer (No Phase Change)
Q = m · c · ΔT Q = heat (J), m = mass (kg), c = specific heat (J/kg·K), ΔT = temperature change (K or °C)
Dimensional formula: [ML²T⁻²] for Q; [L²T⁻²θ⁻¹] for c
Latent Heat (Phase Change)
Q = m · L L = latent heat (J/kg). No ΔT term — temperature is CONSTANT during phase change.
L_fusion (ice) = 3.34×10⁵ J/kg | L_vaporization (water) = 22.6×10⁵ J/kg
L_fusion (ice) = 3.34×10⁵ J/kg | L_vaporization (water) = 22.6×10⁵ J/kg
⚠ Biggest mistake: using Q = mcΔT for phase change. T is constant during phase transition.
Principle of Calorimetry
Heat lost = Heat gained Σ(mcΔT) = 0
Assuming no heat exchange with surroundings (adiabatic mixing)
♨️ Heat Transfer — Conduction, Convection, Radiation
Fourier's Law of Conduction
H = dQ/dt = K · A · (T₁ − T₂) / L H = heat current (W), K = thermal conductivity (W/m·K), A = cross-section area (m²), L = length (m)
Dimensional formula of K: [MLT⁻³θ⁻¹]
Thermal Resistance
R_th = L / (K · A)
Series: R_total = R₁ + R₂ | Parallel: 1/R_total = 1/R₁ + 1/R₂
Heat current H = ΔT / R_th (analogous to V = IR)
Heat current H = ΔT / R_th (analogous to V = IR)
This electrical analogy is the key to solving all multi-slab conduction problems in JEE.
Stefan-Boltzmann Law
P = e · σ · A · T⁴ P_net = e · σ · A · (T⁴ − T₀⁴) σ = 5.67×10⁻⁸ W/m²·K⁴ (Stefan constant), e = emissivity (0 to 1)
For blackbody: e = 1. T must be in Kelvin.
For blackbody: e = 1. T must be in Kelvin.
⚠ Always use Kelvin for T in Stefan's Law. Using Celsius will give wrong answer.
Wien's Displacement Law
λ_max · T = b = 2.898 × 10⁻³ m·K λ_max = wavelength at peak intensity, T = absolute temperature (K)
As T increases, λ_max decreases (peak shifts to UV/visible)
As T increases, λ_max decreases (peak shifts to UV/visible)
Kirchhoff's Law of Radiation
e / a = E_blackbody → e = a (at same T) e = emissive power (emissivity), a = absorptivity
For any body at thermal equilibrium: emissivity = absorptivity
For any body at thermal equilibrium: emissivity = absorptivity
Newton's Law of Cooling
dT/dt = −k(T − T₀) → T(t) = T₀ + (T_i − T₀)e^(−kt) T₀ = surrounding temperature, T_i = initial temperature of body
Valid when temperature excess is small (ΔT ≪ T₀)
Valid when temperature excess is small (ΔT ≪ T₀)
Graph: T vs t → exponential decay; ln(T−T₀) vs t → straight line (slope = −k)
📏 Dimensional Analysis — Complete Table
🔬 Exam Insight
JEE Advanced occasionally asks for dimensional formula of quantities like thermal conductivity or emissive power. Know the derivation, not just the answer.
| Quantity | Symbol | Formula | Dimensional Formula | SI Unit |
|---|---|---|---|---|
| Temperature | T | — | [θ] or [K] | Kelvin (K) |
| Heat Energy | Q | Q = mcΔT | [ML²T⁻²] | Joule (J) |
| Specific Heat Capacity | c | c = Q/(mΔT) | [L²T⁻²θ⁻¹] | J kg⁻¹ K⁻¹ |
| Latent Heat | L | L = Q/m | [L²T⁻²] | J kg⁻¹ |
| Linear Expansion Coeff. | α | α = ΔL/(L₀ΔT) | [θ⁻¹] or [K⁻¹] | K⁻¹ |
| Thermal Conductivity | K | K = HL/(AΔT) | [MLT⁻³θ⁻¹] | W m⁻¹ K⁻¹ |
| Stefan Constant | σ | σ = P/(AT⁴) | [MT⁻³θ⁻⁴] | W m⁻² K⁻⁴ |
| Wien's Constant | b | b = λ_max·T | [Lθ] | m·K |
| Emissive Power | E | E = P/A | [MT⁻³] | W m⁻² |
| Thermal Resistance | R_th | R = L/(KA) | [M⁻¹L⁻²T³θ] | K/W |
🔢 Important Constants & Values
Stefan-Boltzmann Constant
σ = 5.67 × 10⁻⁸ W/m²·K⁴
Wien's Constant
b = 2.898 × 10⁻³ m·K
Specific Heat of Water
c_w = 4186 J/kg·K
= 1 cal/g·°C
Latent Heat of Fusion (Ice)
L_f = 3.34 × 10⁵ J/kg
= 80 cal/g
Latent Heat of Vaporization
L_v = 22.6 × 10⁵ J/kg
= 540 cal/g
Thermal Conductivity (Copper)
K_Cu = 385 W/m·K
Thermal Conductivity (Iron)
K_Fe = 50 W/m·K
α for Steel
α = 12 × 10⁻⁶ K⁻¹
Triple Point of Water
T = 273.16 K
P = 611.6 Pa
Mechanical Equivalent of Heat
J = 4.186 J/cal
❌ Top 5 Formula Mistakes in Exams
- Using Celsius instead of Kelvin in Stefan's Law (P = eσAT⁴)
- Using Bulk modulus instead of Young's modulus for thermal stress in rods
- Not checking if a phase change occurs before applying Q = mcΔT
- Confusing γ_real and γ_apparent for liquid expansion
- Forgetting that α:β:γ = 1:2:3 requires isotropic solid assumption
Next: Thermal Precision & Estimation
Order-of-magnitude thinking and typical values — critical for JEE.