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MixedJEE Physics · Original learning card15 original chapter questions

Newton Cooling and Radiative Cooling Dynamics

For a lumped body of heat capacity C, C dT/dt = -P_loss(T); Newton cooling gives P_loss=hA(T-T_s), while pure radiation gives P_loss=e sigma A(T^4-T_s^4).

Why this shows up in the exam

Comparing cooling rates of bodies · Solar-heating steady temperature · Time ratios in radiative cooling

Learn the idea

Cooling rate equals net heat-loss power divided by the body's heat capacity. Temperature does not fall because of temperature alone: the body loses power to its environment, and its thermal inertia converts that power into a rate of temperature change.

🧠 Memory hook: Loss power sets the fall; heat capacity sets the delay.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • C dT/dt = -P_loss(T) — lumped energy balance for cooling
  • T-T_s = (T_i-T_s)e^(-t/tau) — Newton-cooling solution for constant coefficients
  • tau = C/(hA) — Newton-cooling time constant
  • C dT/dt = -e sigma A(T⁴-T_s⁴) — radiative cooling equation

How to approach it

  1. 1Write the body's total heat capacity
  2. 2Write the correct net loss law
  3. 3Form or integrate dT/dt and check the sign and limiting behavior

Common slip-ups that cost marks

  • •Comparing power without dividing by heat capacity
  • •Linearizing radiation over a large temperature range
  • •Ignoring the surroundings temperature

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 15

A wire 1 m long and cross-sectional area 2 mm^2 extends by 1 mm under a 200 N load. Find Young modulus.

Take a timed JEE Physics sectional mock

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