Calorimetry & Heat Transfer: JEE Main Guide

Thermal physics in JEE Main spans three chapters — calorimetry, thermodynamics, and kinetic theory — and together they contribute five to seven questions per paper. This guide focuses on calorimetry and heat transfer, the most formula-driven part of thermal physics, where every mark is predictable and recoverable with thorough preparation.

Calorimetry: Specific Heat and Latent Heat

Heat absorbed or released: Q = mcΔT, where m is mass, c is specific heat, and ΔT is temperature change. Units: c in J/(kg·K) or cal/(g·°C). Water's specific heat = 4200 J/(kg·K) = 1 cal/(g·°C) — the benchmark value. During a phase change, temperature remains constant: Q = mL, where L is the latent heat (L_fusion for melting, L_vaporisation for boiling). JEE Main mixes both types in a single calorimetry problem: a hot solid dropped into cold water that partially freezes or partially boils.

Principle of calorimetry: heat lost by hot body = heat gained by cold body (for an isolated system). Set up: m₁c₁(T₁ − T_f) = m₂c₂(T_f − T₂), and solve for T_f. Include phase change terms (mL) if the final temperature crosses a phase transition. A common JEE Main trap: add enough ice to just completely melt without raising temperature, then check whether remaining heat raises the water temperature. For the thermodynamics chapter that follows calorimetry, see our Thermodynamics Physics Guide.

Newton's Law of Cooling

Newton's law of cooling: dT/dt = −k(T − T_s), where T_s is the surrounding temperature and k is a positive constant. This gives exponential decay: T(t) = T_s + (T₀ − T_s)e^(−kt). For small temperature differences, the cooling rate is proportional to the excess temperature. JEE Main tests: (1) given cooling from T₁ to T₂ in time t₁, find time to cool from T₂ to T₃; (2) graph-reading — identify exponential cooling curve vs. linear; (3) average temperature approximation: ΔT/Δt = k[(T₁+T₂)/2 − T_s] for finite intervals.

Average rate method: if a body cools from 60°C to 50°C in 5 min with surroundings at 20°C, then (10/5) = k × [(60+50)/2 − 20] = k × 35, giving k = 2/35. Use this k to find time for next cooling interval. Take a free mock test with calorimetry and Newton's cooling questions to practise the calculation chain.

Modes of Heat Transfer

Conduction: heat flows through a solid from high to low temperature. Rate of heat flow: dQ/dt = kA(T₁−T₂)/L, where k is thermal conductivity, A is area, and L is thickness. Units of k: W/(m·K). Good conductors have high k (metals); insulators have low k (wood, air). For slabs in series: R_total = R₁ + R₂ + ... where R = L/(kA). For slabs in parallel: 1/R_total = 1/R₁ + 1/R₂ + ... JEE Main tests thermal resistance combinations exactly like electrical resistance combinations.

Convection: heat transfer through bulk fluid motion. Natural convection is driven by density differences (warm air rises). Forced convection uses a fan or pump. JEE Main tests convection conceptually — qualitative questions about which mode is fastest or which is relevant to a specific situation.

Radiation: heat transfer by electromagnetic waves. Stefan-Boltzmann law: power radiated P = εσAT⁴, where ε is emissivity (0 ≤ ε ≤ 1), σ = 5.67 × 10⁻⁸ W/(m²·K⁴), A is surface area, and T is absolute temperature in Kelvin. Net power radiated by a body at temperature T in surroundings at T₀: P_net = εσA(T⁴ − T₀⁴). Wien's displacement law: λ_max·T = b = 2.898 × 10⁻³ m·K. JEE Main uses Wien's law to find the peak wavelength of radiation given temperature, or to compare temperatures of two stars from their peak wavelengths.

Kirchhoff's Law and Black Body

Kirchhoff's law: at thermal equilibrium, absorptivity = emissivity for a given surface. A perfect black body has ε = 1 and absorbs all incident radiation. A black body is both the perfect absorber and the perfect emitter. Greenhouse effect: glass is transparent to visible sunlight (short wavelength) but opaque to infrared radiation (long wavelength) re-emitted by the Earth — trapping heat. JEE Main tests Kirchhoff's law in the form: a surface that absorbs 60% of incident radiation at a given wavelength emits with emissivity 0.6 at that wavelength.

Revision Summary

Calorimetry and heat transfer questions are some of the most systematically solvable in JEE Main. Focus on: (1) phase change calorimetry calculation chain; (2) Newton's cooling average-rate method; (3) thermal resistance analogy; (4) Stefan-Boltzmann and Wien's law. Two full topic-tests on this chapter should make it reliable exam day. For kinetic theory which completes thermal physics, see our Thermodynamics Guide. Upgrade for ₹149/month for chapter-level practice with detailed solutions.