Thermal Radiation and Stefan's Law
On this page
Direct answer
A hot body radiates power P = σ e A T^4, where σ = 5.67 × 10^-8 W/m^2/K^4 is the Stefan-Boltzmann constant, e is emissivity (1 for an ideal black body), A the surface area and T the absolute temperature — the fourth power is what makes radiation explode with temperature: double the kelvin temperature and the emitted power rises sixteenfold. A body at T in surroundings at T_0 exchanges net power P_net = σ e A (T^4 − T_0^4). Two companion laws complete the NEET toolkit: Wien's displacement law λ_m T = 2.9 × 10^-3 m·K (hotter bodies radiate at shorter wavelengths, hence red-hot to white-hot), and Newton's law of cooling, the small-difference limit where dT/dt ∝ (T − T_0) and cooling becomes exponential in time.
What you must remember
- Stefan-Boltzmann law: P = σeAT^4 with σ = 5.67 × 10^-8 W/m^2/K^4; T must be kelvin — a Celsius slip is the single commonest numerical error here.
- Net exchange: P_net = σeA(T^4 − T_0^4); at T = T_0 the body is in radiative equilibrium, emitting and absorbing equally.
- Ratio shortcut: for the same body, P_2/P_1 = (T_2/T_1)^4 — most NEET questions are ratio questions and need no value of σ at all.
- Wien's law: λ_m T = 2.9 × 10^-3 m K; the Sun's λ_m ≈ 500 nm (visible green) puts its surface near 5800 K, an NCERT-quoted example.
- Newton's law of cooling: dT/dt = −k(T − T_0), valid for small excess (about 30 K or less); integrated, T − T_0 falls exponentially with time constant 1/k.
- Emissivity and absorptivity: a good absorber is a good emitter (Kirchhoff's law); a black body has e = a = 1, polished surfaces have low e and radiate poorly.
- Cooling-time symmetry: the time to cool from T_1 to T_2 uses the mean temperature in the rate expression, T_1 + T_2 over 2 — the arithmetic convention Indian textbooks use for average-rate cooling problems.
A worked ratio and a cooling estimate
A body cools from 80 °C to 60 °C in 5 minutes in surroundings at 20 °C; estimate the time to cool from 60 °C to 40 °C. Using Newton's law with the mean-temperature method: in the first interval the average excess is ((80 − 20) + (60 − 20))/2 = 50 K, and the drop achieved is 20 K in 5 min, so 20 = k × 50 × 5, giving k = 0.08 per minute. In the second interval the average excess is ((60 − 20) + (40 − 20))/2 = 30 K, so the required time is t = 20/(0.08 × 30) = 8.3 minutes. The second interval takes longer because the driving excess shrank — the qualitative half of the answer ("later coolings are slower") is what NEET tests alongside the number. Contrast the Stefan side: raising a furnace from 500 K to 1000 K multiplies its radiation 2^4 = 16 times, which is why the same poker blazes dull red and then searing orange within a small temperature range.
Where students slip
Celsius-versus-kelvin kills more Stefan numericals than any conceptual gap: 100 °C is 373 K, and (373/300)^4 ≈ 2.4, not (100/27)^4 — options are printed for both. In Newton's law questions, students quote the law as linear in time and predict the body cools to exactly T_0 and below; the law's rate vanishes at T_0, so the approach is asymptotic — a graph question in disguise. Wien-confusion is subtler: λ_m is the wavelength of peak emission, not the only wavelength emitted; a body whose peak lies in infrared still emits some visible light when hot enough, which is why metal glows before λ_m enters the visible band. Finally, in net-exchange problems remember the surroundings radiate back; setting emission alone as "net loss" overestimates cooling, and the (T^4 − T_0^4) form is the expected answer.
Frequently asked questions
What does Stefan-Boltzmann law state and what is sigma?
A body's total radiant power is P = σeAT^4, with σ = 5.67 × 10^-8 W/m^2/K^4; temperature must be absolute (kelvin) and the fourth power dominates all other factors.
How does radiation change when the absolute temperature doubles?
It increases 2^4 = 16 times, since P ∝ T^4 for the same body — the ratio form P_2/P_1 = (T_2/T_1)^4 needs no constants.
When is Newton's law of cooling valid and what is its form?
For small temperature excess over surroundings (roughly under 30 K), dT/dt = −k(T − T_0); it is the linearised limit of Stefan's net-exchange equation and gives exponential cooling.
What does Wien's displacement law tell about a hot body?
The wavelength of maximum emission obeys λ_m T = 2.9 × 10^-3 m K, so hotter bodies peak at shorter wavelengths — the physics behind red-hot turning white-hot.
What is a black body in radiation terminology?
An ideal emitter and absorber with emissivity e = 1, which radiates σAT^4 and absorbs all incident radiation; real surfaces fall short with e < 1.