Thermal Radiation & the Black Body

Thermal-imaging cameras are equation (1) turned into a product. A body near $300\ \mathrm{K}$ emits most strongly at about $2.898\times10^{-3}/300 = 9.7\ \mathrm{\mu m}$, in the far infrared, so a sensor tuned to that band sees warm objects glowing in the dark with no illumination at all — firefighters finding people through smoke, electricians spotting an overheating joint, surveyors photographing the heat pouring from an un-lagged wall. The systematic error is always emissivity: a shiny pipe at $80\ \mathrm{^\circ C}$ looks cool because its $e$ is tiny, so operators tape a patch of matt black to it and read the patch. Radiation shields work the other end of the scale: the vacuum in a thermos kills conduction and convection but is transparent to radiation, so it is the silvering ($e \approx 0.02$) that stops the third mode — the same trick as the foil blanket, spacecraft multi-layer insulation, and the low-emissivity coating on modern double glazing.

Earth's energy balance is Kirchhoff's law at planetary scale, and the largest application of all. The planet absorbs about $1361/4 \times 0.70 = 238\ \mathrm{W\,m^{-2}}$ of sunlight, once the beam is averaged over the globe and the $30\%$ that is reflected is discarded — and in the long run it must radiate exactly as much back to space as infrared. Yet a surface at the observed mean temperature of $288\ \mathrm{K}$ radiates $\sigma T^4 = 390\ \mathrm{W\,m^{-2}}$, far more than the $238$ coming in. That gap is the greenhouse effect: the atmosphere is largely transparent at the visible wavelengths where sunlight arrives, but carbon dioxide, water vapour and methane absorb strongly in the infrared where Earth tries to emit — and, being good absorbers there, they are by Kirchhoff's law equally good emitters there, returning much of it straight back down. Add more of those gases and the outgoing window narrows, so the surface must warm until the books balance again. Every climate model on Earth is that sentence made quantitative. Infrared astronomy completes the circle: cold dust clouds and forming stars are far too cool to shine visibly but blaze in the infrared, so telescopes like JWST are chilled to a few tens of kelvin — otherwise their own thermal glow would drown the very sky they are built to read.

Heat a poker in a fire and, long before it glows, hold your hand beside it — not above it where the hot air rises, and not touching it. You still feel the heat, and it stops dead the instant you slide a card into the gap. What crosses that gap, in a straight line at the speed of light, is electromagnetic radiation: the same family of waves as light and radio, emitted because the charged particles inside every warm body are in ceaseless thermal agitation, and accelerating charges radiate. Warm the body and the agitation grows; the radiation grows with it, savagely. This is the mode of transfer that breaks the pattern of the other two. Conduction needs matter that stays put and hands energy along, atom to atom. Convection needs a fluid free to circulate and carry warm matter bodily from place to place. Radiation needs neither — it travels best through a vacuum, where there is nothing to scatter or absorb it. That is not a technicality; it is the reason life exists. Sunlight crosses $1.5\times10^{11}\ \mathrm{m}$ of empty space and arrives above our atmosphere carrying about $1361\ \mathrm{W}$ for every square metre facing the Sun. No other mechanism could have delivered a single joule. The second surprise is that radiation is not the preserve of hot things. Every object above absolute zero radiates, continuously, whatever its temperature. The ice cube radiates; the wall radiates; you radiate. Pierre Prévost saw this in 1791 and called it the theory of exchanges: a body is never simply "taking in heat" or "giving out heat" — it always does both at once, and what we call cooling is merely emitting more than one absorbs. A cup of tea on the table radiates hard, the room radiates back more feebly, and the cup loses the difference. When it finally reaches room temperature the exchange has not stopped; it has balanced. Why, then, do we only see the very hot things glow? Because temperature fixes not just how much is radiated but at what wavelength. At $300\ \mathrm{K}$ — you, the furniture, the room — the emission peaks around $10\ \mathrm{\mu m}$, deep in the infrared: invisible to the eye, blindingly obvious to a thermal camera. Only near $5800\ \mathrm{K}$, the Sun's surface, does the peak land in the visible band.

Take two identical blocks out of the same oven, one painted matt black and one polished to a mirror. Same $T$, same area — and the black one cools visibly faster. The property that separates them is the emissivity $e$: a pure number saying what fraction of the maximum possible radiation a surface actually emits. That maximum is set by an idealisation. A black body absorbs every wavelength that lands on it — nothing reflected, nothing transmitted — and is therefore, as we shall prove in a moment, also the best possible emitter at any temperature. It is the benchmark, $e = 1$; a perfect mirror sits at the other end, $e = 0$, absorbing nothing and radiating nothing. Every real surface lies between, and radiates the black-body value scaled down by $e$:

where: $P$ — the power radiated from the surface, in watts; $e$ — the emissivity, dimensionless, $0 \le e \le 1$; $\sigma$ — the Stefan-Boltzmann constant, $5.67\times10^{-8}\ \mathrm{W\,m^{-2}K^{-4}}$, a universal constant of nature; $A$ — the radiating surface area in $\mathrm{m^2}$; $T$ — the absolute temperature in kelvin, and never, under any circumstances, in degrees Celsius. Read it slowly, because it hides two very different kinds of statement. The $T^4$ is a law of nature — doubling the absolute temperature multiplies the radiated power by sixteen, which is why a fire feels so violently hotter than a radiator and why stars are so bright. That fourth power was found experimentally by Josef Stefan in 1879 and derived thermodynamically by Ludwig Boltzmann in 1884; its full treatment, with Wien's companion law for the colour, is the next article's job, and here we simply take $P = e\sigma A T^4$ as given and put it to work. The $e$, by contrast, is a statement about the surface: repaint the same block and $T$, $\sigma$, $A$ are untouched, yet $P$ can fall twentyfold.

That skin row is not a misprint: human skin has $e = 0.97$ whatever its visible colour, because "black" here means black in the infrared, at the $10\ \mathrm{\mu m}$ wavelengths a body near $310\ \mathrm{K}$ actually emits. Visible colour has nothing to do with thermal emission at body temperature.

The figure answers the question that troubles every student: if no real paint is perfectly black, how does anyone make a black body? You do not paint one — you build a cavity and drill a small hole. A ray entering strikes the inner wall, which absorbs most of it and reflects the remnant deeper in; that remnant strikes another wall, and so on. Suppose each bounce merely halves it: after ten bounces the survivor is $0.5^{10} = 0.00098$ of the original, and its chance of finding the tiny hole again is negligible. The hole therefore absorbs $1 - 0.00098 = 0.999$ of everything that enters it. And since absorbing perfectly means emitting perfectly, the radiation streaming back out of that hole from a hot cavity is the purest black-body radiation obtainable — exactly the apparatus Kirchhoff proposed in 1860, and the one Planck's spectrum was written to explain forty years later.

Equation (1) is only half the story — and the half examiners most enjoy watching students forget. A body never radiates into a void. It sits inside surroundings at some temperature $T_s$ which are radiating too: the walls, the furniture, the sky. That radiation lands on our body and, by Kirchhoff's law, is absorbed with exactly the same efficiency $e$ with which the surface emits. So the body gains $e\sigma A T_s^4$ while it loses $e\sigma A T^4$, and only the difference shows up as cooling:

where: $P_{\text{net}}$ — the net rate of energy loss by radiation, in watts (negative if the surroundings are hotter, in which case the body gains); $T$ — the body's absolute temperature; $T_s$ — that of the surroundings; $e$, $\sigma$, $A$ as before. Both temperatures go into kelvin before being raised to the fourth power: you may not subtract first and convert after, because $(T - T_s)^4$ is nothing whatever like $T^4 - T_s^4$. This is Prévost's theory of exchanges written as an equation. When $T > T_s$ the body wins the exchange and cools; when $T < T_s$ the sign flips and it is warmed by its surroundings, which is what a heat lamp or a patch of sunlight does for you. And when $T = T_s$ the terms cancel exactly, $P_{\text{net}} = 0$ — radiative equilibrium, in which the body has emphatically not stopped radiating. It pours out $e\sigma A T^4$ and takes back precisely as much. Equilibrium is a balance, not a shutdown. Now put yourself into equation (2), with skin area $1.8\ \mathrm{m^2}$ at $34\ \mathrm{^\circ C} = 307\ \mathrm{K}$, a room at $293\ \mathrm{K}$ and $e = 0.97$. You radiate a startling $879\ \mathrm{W}$ — more than a domestic kettle — and you would be dead within minutes if that were the whole story. But the room radiates back, and you absorb $730\ \mathrm{W}$ of it, so the net loss is only $879 - 730 = 150\ \mathrm{W}$, set against a resting adult's metabolic output of roughly $100\ \mathrm{W}$. That is why a merely mild room feels cold, and why you wear clothes; radiation, not conduction into the air, is the largest single channel by which an unclothed human loses heat indoors. The competitive worked example below builds all three numbers from scratch.

The most famous demonstration of Kirchhoff's law fits on a bench. A Leslie cube is a hollow metal cube filled with boiling water whose vertical faces carry different finishes — one blackened, one polished, one white, one dull. Every face has the same temperature, the same area and the same room in front of it; the only variable is the surface. Point an infrared detector at each in turn and the readings differ enormously.

Now run the experiment backwards. Shine a heater at the cube and put a thermometer inside: it is the blackened face that lets the heat in fastest. Best emitter, best absorber — one property, one number, exactly as the derivation demanded. This is why a black car is agony in July and a black kettle of tea goes cold soonest; why solar panels are matt black and satellites are wrapped in shiny foil.

Board level — the Leslie cube, face by face (CBSE/ICSE/IGCSE pattern)

A hollow cube of side $10\ \mathrm{cm}$ is filled with boiling water at $100\ \mathrm{^\circ C}$ and stands in a room whose walls are at $20\ \mathrm{^\circ C}$. One vertical face is coated matt black, $e = 0.95$; the opposite face is polished, $e = 0.05$. Take $\sigma = 5.67\times10^{-8}\ \mathrm{W\,m^{-2}K^{-4}}$. Find, for one face at a time: (a) the power radiated by the black face, (b) the power it absorbs from the room, (c) its net loss, (d) the net loss from the polished face, and (e) the ratio of the two.

Step 1. Kelvin and metres, before anything else. $T = 100 + 273 = 373\ \mathrm{K}$; $T_s = 20 + 273 = 293\ \mathrm{K}$. One face is $0.1\ \mathrm{m} \times 0.1\ \mathrm{m}$, so $A = 0.01\ \mathrm{m^2}$ — the question asks per face, not for the whole cube.

Step 2. Get the fourth powers once and reuse them: $373^4 = 1.936\times10^{10}$ and $293^4 = 7.370\times10^{9}$. Computing these twice is how students run out of time.

Step 3. (a) The black face emits, by equation (1), $P = e\sigma A T^4 = 0.95 \times 5.67\times10^{-8} \times 0.01 \times 1.936\times10^{10} = \boxed{10.4\ \mathrm{W}}$. (b) It also absorbs the room's radiation, with the same $e$ by Kirchhoff's law: $0.95 \times 5.67\times10^{-8} \times 0.01 \times 7.370\times10^{9} = \boxed{3.97\ \mathrm{W}}$.

Step 4. (c) The net loss is the difference, $10.4 - 3.97 = \boxed{6.46\ \mathrm{W}}$ — which is equation (2) in one line: $0.95 \times 5.67\times10^{-8} \times 0.01 \times (1.936\times10^{10} - 7.370\times10^{9}) = 6.46\ \mathrm{W}$.

Step 5. (d) For the polished face nothing changes but $e$, so scale rather than recompute: $6.46 \times (0.05/0.95) = \boxed{0.34\ \mathrm{W}}$. (e) The ratio is therefore just the ratio of emissivities, $0.95/0.05 = \boxed{19}$. Same cube, same temperature, same room — and nineteen times the heat leaving one face rather than the other. That number is Kirchhoff's law made visible.

Competitive level — the 150-watt human (JEE-Main/NEET/AP-2 pattern)

A person has exposed skin area $A = 1.8\ \mathrm{m^2}$ at a surface temperature of $34\ \mathrm{^\circ C}$, with emissivity $e = 0.97$. The room's walls are at $20\ \mathrm{^\circ C}$. Find (a) the power radiated, (b) the power absorbed from the walls, (c) the net radiative loss. (d) The walls now cool to $15\ \mathrm{^\circ C}$ — what is the new net loss? (e) The person is wrapped in a metallised emergency blanket of emissivity $0.05$ at the same skin temperature. What is the net loss now?

Step 1. Convert. $T = 34 + 273 = 307\ \mathrm{K}$, $T_s = 20 + 273 = 293\ \mathrm{K}$ (and later $288\ \mathrm{K}$). Note the skin is at $34\ \mathrm{^\circ C}$, not the $37\ \mathrm{^\circ C}$ core — examiners give you the surface temperature because the surface is what radiates.

Step 2. Fourth powers, done once: $307^4 = 8.883\times10^{9}$, $293^4 = 7.370\times10^{9}$, so $T^4 - T_s^4 = 1.513\times10^{9}\ \mathrm{K^4}$. Notice how modest the difference is against either term — that is the whole reason the net is small.

Step 3. (a) Emitted: $P = 0.97 \times 5.67\times10^{-8} \times 1.8 \times 8.883\times10^{9} = \boxed{879\ \mathrm{W}}$. (b) Absorbed: the same prefactor against $293^4$ gives $\boxed{730\ \mathrm{W}}$.

Step 4. (c) Net: $879 - 730 = \boxed{150\ \mathrm{W}}$ — or in one shot, $0.97 \times 5.67\times10^{-8} \times 1.8 \times 1.513\times10^{9} = 150\ \mathrm{W}$. Against a resting adult's $\approx 100\ \mathrm{W}$ of metabolic heat, an unclothed person in a $20\ \mathrm{^\circ C}$ room really is losing ground — which is why you feel cold, and why you wear clothes.

Step 5. (d) Cool the walls by just $5\ \mathrm{K}$: $288^4 = 6.880\times10^{9}$, so $T^4 - T_s^4$ jumps to $2.003\times10^{9}$ and the loss becomes $0.97 \times 5.67\times10^{-8} \times 1.8 \times 2.003\times10^{9} = \boxed{198\ \mathrm{W}}$ — a third more heat for a five-degree change in surfaces you never touch. This is why a stone church feels bitter even when the air is mild.

Step 6. (e) Only $e$ changes, so scale again: $P = 150 \times (0.05/0.97) = \boxed{7.7\ \mathrm{W}}$. A few grams of aluminised plastic have cut the radiative loss $0.97/0.05 = 19.4$-fold. That is the whole physics of the foil blanket handed to marathon finishers: it works purely by lowering $e$, and has almost no insulating thickness at all. Timing note: four minutes for all five parts, and it hinges on one habit. Compute $307^4$, $293^4$ and $288^4$ once, write them down, and never touch them again; then (a) to (c) are one multiplication each, and (d) and (e) are pure ratios. Candidates who rebuild the prefactor $e\sigma A$ five times lose two minutes and, usually, a power of ten. In an MCQ, part (e) falls to ratio alone in five seconds — no calculator.

Putting Celsius into the fourth power. The single most punished slip in the chapter. $T$ in $P = e\sigma A T^4$ is absolute: $27\ \mathrm{^\circ C}$ means $300\ \mathrm{K}$. Radiation is not conduction, where a temperature difference lets you keep Celsius — here $T$ itself is raised to a power and the offset never cancels. CBSE, NEET and JEE all set distractor options built from the un-converted number.

Using $T^4$ alone for cooling. If a body sits in surroundings at $T_s$, it is absorbing as well as emitting, and the rate it cools is $e\sigma A(T^4 - T_s^4)$. Quote $e\sigma A T^4$ and you overstate our person's loss by a factor of six. Only when $T_s \ll T$ — a red-hot filament, a star — is the $T_s^4$ term safely dropped.

Writing $(T - T_s)^4$. Fourth powers first, subtract second — the two differ by five orders of magnitude for our person, and boards plant this as a deliberate distractor.

Believing that only hot things radiate. Every body above $0\ \mathrm{K}$ radiates continuously. NEET and ICSE assertion-reason items are built on the false claim that a cold body only absorbs. Prévost settled it in 1791: a cold body radiates too — it simply absorbs more than it emits.

Divorcing emissivity from absorptivity. A surface with $e = 0.05$ is also a poor absorber, $a = 0.05$. Any answer implying a surface that drinks sunlight greedily but refuses to radiate is proposing a perpetual-motion machine — Kirchhoff's law forbids it, and examiners set this exact trap.

Reading emissivity off the visible colour. "Black in the infrared" is not "black to the eye": white and black paint both sit near $e \approx 0.9$ at room temperature, and skin is $0.97$ regardless of complexion. It is shiny metal, not dark colour, that means low $e$. Nor does a vacuum flask's gap stop radiation — vacuum is perfectly transparent to it, which is exactly why the walls must be silvered.

Getting the area wrong. $A$ is the radiating surface: a sphere gives $4\pi r^2$; a cube of side $a$ gives $6a^2$ in total but $a^2$ per face.

Texts. NCERT Physics Class 11, Ch. 11 (Thermal Properties of Matter) — radiation, black body, Kirchhoff's law · Cambridge IGCSE Physics 0625, Thermal Physics (emission and absorption by black and shiny surfaces; the Leslie-cube experiment) · Halliday, Resnick & Walker, "Heat Transfer Mechanisms" · College Board AP Physics 2 CED, Topic 15.4 Blackbody Radiation.

Provenance. Standard results; every number on this page is recomputed in the verifier. The anchors: the fourth powers $307^4 = 8.883\times10^{9}$, $293^4 = 7.370\times10^{9}$, $288^4 = 6.880\times10^{9}$, $373^4 = 1.936\times10^{10}$; the person's $879\ \mathrm{W}$ emitted, $730\ \mathrm{W}$ absorbed, $150\ \mathrm{W}$ net (rising to $198\ \mathrm{W}$ for walls at $288\ \mathrm{K}$, falling to $7.7\ \mathrm{W}$ at $e = 0.05$, and $154\ \mathrm{W}$ for an ideal black body); the Leslie cube's $6.46\ \mathrm{W}$ against $0.34\ \mathrm{W}$, ratio $19$; the cavity's $0.5^{10} = 0.00098$; the radiator's $377\ \mathrm{W}$ radiative against $600\ \mathrm{W}$ convective; and Earth's $238\ \mathrm{W\,m^{-2}}$ absorbed against $\sigma(288)^4 = 390\ \mathrm{W\,m^{-2}}$ emitted.

Prerequisites. Heat vs Temperature. Leads to. Stefan-Boltzmann & Wien's Laws (the quantitative $T^4$ and $\lambda_{\max}T = b$ laws that this article states and uses but does not derive) · Newton's Law of Cooling · ARCHETYPE: Cooling & Radiation Numericals. Go deeper (undergraduate). "Photon Gas and Planck Radiation Law" treats the radiation inside Kirchhoff's cavity as a gas of photons and derives the entire spectrum from statistical mechanics — recovering Stefan's $\sigma$ as an integral over it and Wien's $b$ as its peak, and answering the question this page can only pose: why is the cavity's radiation what it is?