Temperature, Heat & Thermometers
JEE and NEET regularly ask: "Is this statement about heat or temperature correct?" Mixing these up in assertion-reason questions is the #1 source of lost marks here.
A measure of the average kinetic energy of molecules. Not total energy. SI unit: Kelvin (K).
Celsius to Kelvin conversion — used in ALL gas law and radiation formulas
Never use Celsius in Stefan's Law, Wien's Law, or Newton's Law of Cooling — unless the problem specifically asks for temperature difference (ΔT in °C = ΔT in K).
Energy transferred between objects due to temperature difference. Always flows from high T to low T. SI unit: Joule (J).
Old unit: calorie (1 cal = 4.186 J). Mechanical equivalent: J = 4.186 J/cal (by Joule's experiment).
Heat flow stops when thermal equilibrium is reached — temperatures become equal, NOT heat contents.
Temperature Scales Comparison
Universal temperature conversion — memorize this triple equality
The triple equality above directly gives you conversion without memorizing individual formulas. If C/100 = (F−32)/180, then: F = (9/5)C + 32. That's derived in 10 seconds.
Thermal Expansion
When temperature rises, interatomic distance increases → object expands. Three types based on geometry:
Key Relationship
Valid for isotropic solids. This is the most-asked relationship in JEE Main.
For anisotropic materials (different expansion in different directions), γ = α₁ + α₂ + α₃. JEE Advanced has tested this explicitly.
Derivation: Volume Expansion from Linear
When a rod is prevented from expanding:
Y = Young's modulus, α = linear expansion coefficient
Many students write thermal strain = αΔT correctly, then multiply by the wrong modulus. For a rod, it's Young's modulus (Y), not bulk modulus.
Water contracts from 0°C to 4°C (unlike all other substances). Maximum density at 4°C.
Why? Hydrogen bond reorganization during ice→liquid structure change.
This is why fish survive in frozen ponds — water at 4°C sinks, ice floats. NEET loves this application.
Expansion of Liquids: Real vs Apparent
Apparent expansion = what you observe (liquid expanding in a container). Real expansion = actual expansion of liquid. Real = Apparent + expansion of container vessel. Formula: γ_real = γ_apparent + γ_vessel
This is regularly tested in JEE Main numerical problems
Specific Heat & Calorimetry
Heat required per unit mass per unit temperature rise.
Units: J kg⁻¹ K⁻¹. Note: ΔT in K = ΔT in °C
Water: c = 4186 J/kg·K (highest among common substances = best coolant)
Heat Capacity (C) = total heat to raise temperature by 1K: C = mc
Molar Heat Capacity = C for one mole: C_m = Mc (M = molar mass)
Cp and Cv distinction is CRITICAL for JEE. For solids/liquids: Cp ≈ Cv. For gases: Cp - Cv = R (Mayer's relation). This links Thermal chapter with KTG.
Principle of Calorimetry
Heat lost by hot body = Heat gained by cold body (assuming no heat loss to surroundings). This is conservation of energy. The calorimeter equation: Σ(mc·ΔT) = 0 for all bodies in the system.
- Identify all bodies in the system and their initial temperatures.
- Determine final equilibrium temperature (T_f).
- For bodies losing heat: ΔT = T_initial − T_f (positive)
- For bodies gaining heat: ΔT = T_f − T_initial (positive)
- Write: Σ(m·c·ΔT)_hot = Σ(m·c·ΔT)_cold
- Check if any phase change occurs (use latent heat if so)
- Solve for T_f
If the equilibrium temperature comes out to be 0°C or 100°C after first pass, it means a phase change IS occurring. You need to check if all of the substance changes phase or only part of it.
Change of State & Latent Heat
When a substance changes phase (solid↔liquid↔gas), temperature does NOT change during the transition. Energy goes into breaking/forming molecular bonds.
L = Latent Heat; Units: J/kg. No ΔT term — temperature is constant during phase change.
Energy to convert solid → liquid at constant temperature.
For Water: L_f = 3.34 × 10⁵ J/kg = 80 cal/g
Process: Melting (solid→liquid), Freezing (liquid→solid)
Energy to convert liquid → vapor at constant temperature.
For Water: L_v = 22.6 × 10⁵ J/kg = 540 cal/g
L_v >> L_f because breaking gas bonds requires more energy than loosening solid bonds.
Heating Curve of Water (Visual)
On heating curve — the horizontal segments represent phase changes. Temperature is CONSTANT here. A question asking "what happens to temperature when 1 kg of water at 100°C is given 5×10⁵ J?" — the answer is: temperature stays at 100°C, and only 22.6×10⁵/5×10⁵ ≈ 22% of water vaporizes.
Triple Point of Water: 273.16 K, 0.006 atm. At this unique point, all three phases (solid, liquid, gas) coexist in equilibrium. CBSE boards ask this definitionally; JEE tests conceptual implications.
Modes of Heat Transfer
Transfer via molecular vibration. No bulk movement of matter. Dominant in solids.
H = heat flow rate (W), K = thermal conductivity
Transfer via bulk movement of fluid. Natural (buoyancy-driven) or forced. Dominant in liquids and gases.
Newton's Law: Q/t = hA(T_surface − T_fluid)
Transfer via electromagnetic waves. Needs no medium. Works in vacuum. All bodies above 0 K radiate.
For blackbody. σ = 5.67 × 10⁻⁸ W/m²K⁴
Thermal Resistance — The JEE Key Concept
Thermal resistance (R = L/KA) is ANALOGOUS to electrical resistance (R = ρL/A). This analogy is essential for JEE problems involving series/parallel slabs.
Same heat current through all slabs
Heat current = H = ΔT_total / R_total (same through each slab)
Same temperature difference across each slab
Equivalent K_eff = (K₁A₁ + K₂A₂)/(A₁ + A₂) for same L
Newton's Law of Cooling
Rate of cooling ∝ excess temperature over surroundings. Applies when temperature difference is small.
Graph of T vs t: Exponential decay curve. Graph of ln(T−T₀) vs t: Straight line with slope = −k. JEE often asks: "What is the nature of the graph?"
Newton's Cooling is a special case of Stefan's Law when temperature difference is small. For large ΔT, use Stefan's Law directly.
Radiation Laws
All bodies above absolute zero emit electromagnetic radiation. The nature of this radiation depends on the body's temperature and surface properties.
JEE Advanced has repeatedly tested: "If a body cools by radiation only, and T₀ is the surrounding temperature, show that the rate of cooling depends on (T⁴ − T₀⁴)." This requires starting from dQ/dt = eσA(T⁴ − T₀⁴) and using dQ = mcdT.
Blackbody Radiation — Key Facts
- Perfect blackbody: absorbs ALL incident radiation (a = 1, e = 1)
- Perfect blackbody is a theoretical concept — closest is a cavity with a small hole
- At thermal equilibrium: emission = absorption (energy balance)
- Higher temperature → peak wavelength shifts LEFT (Wien's Law) AND total emission increases (Stefan's Law T⁴)
- Sun's surface ≈ 5800 K → λ_max ≈ 500 nm (visible yellow-green) ✓
JEE twists radiation problems by making you compare two bodies or find temperature ratios. Key: if body A radiates 16× more than body B, and both have same area and emissivity, then T_A/T_B = (16)^(1/4) = 2. The T⁴ dependence is the exam trap.
Concepts clear? Now master the formulas.
Go to Formula & Dimensional Analysis — the searchable formula bank.