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JEE Mains Physics · Thermal Properties of Matter

Radiation and Newton's Law of Cooling

A body radiates power P = eσAT⁴ with T in kelvin, its spectrum peaks where λₘT = b, and a body only a little warmer than its surroundings cools at a rate proportional to its excess temperature.

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Why this matters

Eleven PYQs, three of them asking for a number, none from 2026. Three use Stefan's law to compare the power radiated by two bodies or to find an emissivity; eight are Newton's law of cooling, finding a time or a temperature for a second stage of cooling from the first. The average form of Newton's law turns each one into two short equations.

Concept 1 of 1: Stefan's law, Wien's law and Newton's law of cooling

Every body radiates, and the power rises steeply with temperature: as the fourth power of the kelvin temperature. That is Stefan's law. A hotter body also radiates its peak at a shorter wavelength, which is Wien's law. A body in a room also absorbs, so its net loss depends on how much hotter it is than the room. When that excess is small, the net loss is simply proportional to the excess: that is Newton's law of cooling. A body twice as far above room temperature cools twice as fast.

Definition

  • Stefan's law: P=eσAT4P = e\sigma AT^{4}, T in kelvin, σ=5.67×10−8 W m−2K−4\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\text{K}^{-4}. For a sphere A=4πr2A = 4\pi r^{2}, so P∝er2T4P \propto er^{2}T^{4}. For a wire A=πdLA = \pi dL.
  • Net loss to surroundings at T0T_0: Pnet=eσA(T4−T04)P_{net} = e\sigma A(T^{4} - T_0^{4}).
  • A good absorber is a good emitter; a perfect black body has e=1e = 1.
  • Wien's law: λmT=b\lambda_mT = b, with b≈2.9×10−3 m Kb \approx 2.9 \times 10^{-3}\ \text{m K}. Hotter means a shorter peak wavelength.
  • Newton's law of cooling (small excess): dTdt=−k(T−Ts)\dfrac{dT}{dt} = -k(T - T_s). Exactly, T−Ts=(T0−Ts)e−ktT - T_s = (T_0 - T_s)e^{-kt}.
  • Average form for exam questions: T1−T2t=k(T1+T22−Ts)\dfrac{T_1 - T_2}{t} = k\left(\dfrac{T_1 + T_2}{2} - T_s\right). Find k from the first stage, then use it for the second.
  • Over equal intervals the excess over the surroundings falls by the same factor each time, in the exact form and in the average form.

Radiation and cooling

P=eσAT4λmT=bT1−T2t=k(T1+T22−Ts)T−Ts=(T0−Ts)e−ktP = e\sigma AT^{4} \qquad \lambda_mT = b \qquad \frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_s\right) \qquad T - T_s = (T_0 - T_s)e^{-kt}

Worked example

A body cools from 70∘C70^{\circ}C to 50∘C50^{\circ}C in 6 minutes in a room at 30∘C30^{\circ}C. How long does it take to cool from 50∘C50^{\circ}C to 40∘C40^{\circ}C?
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The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q94Moderate

Example 1 · Thermal Properties of Matter · Radiation and Newton's Law of Cooling

A cup of coffee cools from 90∘C90^{\circ}C to 80∘C80^{\circ}C in t minutes when the room temperature is 20∘C20^{\circ}C. The time taken by the similar cup of coffee to cool from 80∘C80^{\circ}C to 60∘C60^{\circ}C at the same room temperature is :

Celsius in Stefan's law

P = eσAT⁴ needs the kelvin temperature. A body at 127°C against one at 27°C radiates (400/300)⁴ ≈ 3.2 times as much, not (127/27)⁴ ≈ 490 times.

Averaging the wrong quantity

In the average form, average the two TEMPERATURES and then subtract the surroundings. The rate on the left is the drop divided by the time, not the excess divided by the time.

Minutes and seconds

k comes out per minute if the times are in minutes. A time of 10/3 minutes is 200 s; check the unit the options use before choosing.

Doubling the excess does not quadruple the rate

Newton's law is linear in the excess temperature: double the excess, double the rate. The fourth power belongs to Stefan's law and the absolute temperature.
Drill 10 more on stefan's law, wien's law and newton's law of cooling

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

  • Stefan's law, Wien's law and Newton's law of cooling

    Radiation and cooling

    P=eσAT4λmT=bT1−T2t=k(T1+T22−Ts)T−Ts=(T0−Ts)e−ktP = e\sigma AT^{4} \qquad \lambda_mT = b \qquad \frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_s\right) \qquad T - T_s = (T_0 - T_s)e^{-kt}

Watch out for (4)

Test yourself on Thermal Properties of Matter

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