Heat transfer by radiation

Stand three metres from a bonfire on a cold night and one side of your face goes warm within seconds — long before any hot air current could have drifted across to touch you, and with nothing physically bridging the gap between the flames and your skin. That warmth arrived as radiation: electromagnetic waves streaming outward from every hot surface, needing no medium at all to carry heat energy from one place to another. It is the same mechanism, scaled up almost unimaginably, that lets the Sun warm the Earth across 150 million kilometres of near-perfect vacuum, and the same mechanism that quietly drains heat from a mug of tea even while conduction and convection are also doing their share of the work. Radiation is the third and strangest of the three heat-transfer mechanisms this unit builds — after conduction (heat creeping through a solid, particle jostling particle) and convection (heat riding along with a moving fluid), radiation needs neither particles nor a moving medium, only line of sight, and it works exactly as well through empty space as through air. Every object warmer than absolute zero is, right now, glowing with thermal radiation, even if that glow is entirely invisible infrared. A hand held near a switched-off but still-warm oven, a black car baking in a parking lot, a thermos flask's silvered walls, a satellite drifting through orbit with no air to conduct or convect against it at all — all of these are radiation problems, and this article builds the vocabulary and the one central law, Stefan's law, that lets you reason about them quantitatively rather than just qualitatively. We will pin down what makes a surface a good emitter versus a good absorber, connect that directly to Stefan's law $E = \sigma T^4$ , derive the net rate at which any hot object loses heat to cooler surroundings, and then work two examples — one board-level, one competitive-level — that turn the everyday black-can-versus-shiny-can intuition into real, checkable numbers.

Key takeaways

Radiation transfers heat via electromagnetic waves and needs no medium — unlike conduction and convection, it works through a perfect vacuum, which is how the Sun heats the Earth.
Every object above absolute zero continuously emits thermal radiation, mostly invisible infrared at everyday temperatures; only very hot objects glow visibly.
A dark, matte surface is a good absorber and a good emitter; a light, polished surface is a poor absorber and a poor emitter — good reflectors are poor radiators, not good ones.
Stefan's law, $E = \sigma T^4$ , gives the power radiated per unit area by a body, with $\sigma = 5.67\times10^{-8}\ \text{W/m}^2\text{K}^4$ ; a real surface radiates $\varepsilon$ times this, where $\varepsilon$ (emissivity) measures how close it is to an ideal blackbody.
Net radiative heat loss is a two-way balance: a hot body both emits and absorbs radiation simultaneously, and the net outflow is $\varepsilon\sigma A(T^4 - T_0^4)$ , which vanishes exactly at thermal equilibrium.
A perfect blackbody absorbs all incident radiation and is also the best possible emitter at its temperature — this absorb/emit link is the seed of the fuller blackbody-radiation treatment (Wien's law, spectral peaks) in the next article.
Key idea: every object above absolute zero radiates thermal energy through electromagnetic waves needing no medium, at a rate set by Stefan's law E = σT⁴ scaled by surface emissivity, so the net radiative exchange εσA(T⁴ − T₀⁴) always flows from hotter to cooler and vanishes only at thermal equilibrium.

The core insight

Every body above absolute zero radiates electromagnetic energy at a rate given by Stefan's law $E=\sigma T^4$ scaled by its surface emissivity, and because it also absorbs radiation from its surroundings at the same rate law, the net radiative heat flow $\varepsilon\sigma A(T^4-T_0^4)$ is always from the hotter body to the cooler one, needing no medium at all to travel.

Where it fits in the syllabus

Radiation is the third heat-transfer mechanism in the Thermal Properties of Matter chapter, taught right after conduction and convection, and it is the mechanism that leads directly into the more detailed blackbody radiation article that follows. The 2025-26 CBSE Class 11 prescription expects you to know that radiation transfers heat via electromagnetic waves without needing a medium, that all bodies emit and absorb thermal radiation depending on temperature and surface, and to be able to use Stefan's law $E = \sigma T^4$ for the power radiated by a body. The fuller quantitative picture — Wien's displacement law, spectral distributions, and how blackbody curves shift with temperature — is developed in the next node; Newton's law of cooling as a derived exponential decay and the greenhouse effect are explicitly kept out of this year's board syllabus and are only pointed at here, never derived.

The physical picture

Unlike conduction, which needs a continuous solid to pass vibrations along, or convection, which needs a fluid free to circulate, radiation is the transfer of heat as electromagnetic waves — the same family of waves as visible light, just at wavelengths our eyes usually cannot see. Because electromagnetic waves propagate perfectly well through a vacuum, radiation is the only heat-transfer mechanism available in empty space, and it is why sunlight can cross 150 million kilometres of nothing and still warm your skin. At room temperature, an object's thermal radiation peaks in the infrared, invisible to us; only when an object gets hot enough — an electric heater's coil, molten metal, the Sun's 5778 K surface — does a noticeable fraction shift into the visible band and the object visibly glows. Every real surface differs in how efficiently it emits and absorbs this radiation, and the two abilities are locked together. A surface that is dark and rough tends to absorb incident radiation well and, for the same underlying reason, emits it well too. A surface that is light-coloured and polished tends to reflect radiation, which means it neither absorbs nor emits efficiently. This single fact explains a long list of everyday observations: a black car parked in the sun gets hotter than an identical white one because the black paint absorbs more of the Sun's radiant energy; a thermos flask's inner wall is silvered because a poor emitter loses little heat by radiation, keeping the contents hot (or cold) far longer; a matte-black radiator panel is chosen specifically to maximise the radiative heat it throws off into a room. An idealised surface that absorbs every wavelength of incident radiation completely, reflecting none of it, is called a perfect blackbody. Because absorbing ability and emitting ability go hand in hand, a blackbody is also the theoretical best possible emitter at any given temperature — no real surface can radiate more power per unit area than a blackbody at the same temperature. That upper limit is set precisely by Stefan's law:

E = \sigma T^4 \qquad (1)

where: $E$ is the power radiated per unit area of a blackbody surface (in $\text{W/m}^2$ ), $T$ is the absolute temperature of the surface in kelvin, and $\sigma = 5.67\times10^{-8}\ \text{W/m}^2\text{K}^4$ is the Stefan-Boltzmann constant, a fixed number of nature. The physical reading is stark: radiated power does not grow linearly or even quadratically with temperature, it grows with the fourth power — double the absolute temperature of an object and it radiates sixteen times as much power per unit area. This is why a filament at 3000 K glows brilliantly while the same filament at 300 K (room temperature) radiates a completely imperceptible, purely infrared trickle of energy, even though the temperature only dropped by a factor of ten. Real surfaces are not perfect blackbodies. A real surface at temperature $T$ radiates a fraction $\varepsilon$ of the ideal blackbody rate, where $\varepsilon$ (the emissivity, a pure number between 0 and 1) measures how close that particular surface comes to ideal behaviour — matte black surfaces have $\varepsilon$ close to 1, and polished metals can have $\varepsilon$ as low as 0.02 to 0.1. The total power radiated by a surface of area $A$ is then $P = \varepsilon\sigma A T^4$ , and because the same surface is simultaneously absorbing radiation arriving from its surroundings, the two effects combine into a single net exchange, which the derivation below builds up from scratch.

Figure

Two identical cans side by side in a room, the matte black can shown with many long outward arrows representing strong radiative emission and the shiny polished can shown with only a few short arrows representing weak radiative emission, both cans also receiving thin inward arrows from the room representing absorbed radiation

Stefan's law and net radiative heat loss

Derivation ANet rate of radiative heat loss from a hot object

The goal here is to combine emission and absorption into one working formula that tells you the net rate at which a hot object loses heat to its surroundings by radiation alone, and to show that this net rate always points from the hotter body to the cooler one, collapsing to zero exactly at thermal equilibrium — the same general direction rule that governs conduction and convection, now derived for radiation using nothing beyond Stefan's law and simple bookkeeping of power in versus power out.

Step 1. Write the power the object emits. A surface of area $A$ at absolute temperature $T$ and emissivity $\varepsilon$ radiates power $P_{\text{emit}} = \varepsilon\sigma A T^4$ , a direct scaling of Stefan's law $E=\sigma T^4$ by the surface area and by $\varepsilon$ , which accounts for the surface being a less-than-perfect blackbody.

Step 2. Write the power the object absorbs. The object sits inside surroundings at a (typically lower) temperature $T_0$ , which are themselves radiating at that temperature and bathing the object in radiant energy from every direction. Because a good emitter is equally good at absorbing — the same surface property $\varepsilon$ governs both directions — the object absorbs power $P_{\text{absorb}} = \varepsilon\sigma A T_0^4$ from this ambient radiation field.

Step 3. Subtract to get the net rate. The net power lost by the object is simply what it emits minus what it absorbs: $P_{\text{net}} = \varepsilon\sigma A\left(T^4 - T_0^4\right)$ . When $T > T_0$ this is positive — a net outflow of heat, exactly matching the everyday fact that a hot object cools by radiating into a cooler room. When $T < T_0$ the expression turns negative, meaning the object is a net gainer of heat by radiation, which is exactly what happens to a cold object placed in a warm room.

Step 4. Read the special case $T = T_0$ . The net rate becomes exactly zero, even though both the object and its surroundings are still radiating and absorbing continuously and at full strength — emission has not switched off, it has simply been balanced by an equal absorption. This is the resolution to the most common misconception about radiation: an object in thermal equilibrium with its surroundings is not radiation-silent, it is radiation-balanced. (A closely related result — how fast an object's temperature falls with time under this net loss — is Newton's law of cooling; deriving the resulting exponential decay curve is university-adjacent and outside this year's board syllabus, so it is only named here, not developed.) Consistency check: at $T=T_0=300\ \text{K}$ , $T^4-T_0^4=0$ exactly, so $P_{\text{net}}=0$ regardless of $\varepsilon$ or $A$ — the formula correctly predicts no net heat flow at thermal equilibrium, as physically required.

Worked examples

Board level - comparing the black can and the shiny can quantitatively

A matte black can ( $\varepsilon = 0.9$ ) and an identical, polished shiny can ( $\varepsilon = 0.05$ ) each hold hot water at $T = 350\ \text{K}$ ( $77^\circ\text{C}$ ) in a room at $T_0 = 300\ \text{K}$ ( $27^\circ\text{C}$ ). Each can has a radiating surface area $A = 0.02\ \text{m}^2$ . Find the net radiative power lost by each can, and compare.

Step 1. Compute the temperature term common to both cans. $T^4 = 350^4 = 1.5006\times10^{10}\ \text{K}^4$ and $T_0^4 = 300^4 = 8.1\times10^{9}\ \text{K}^4$ , so $T^4 - T_0^4 = 6.906\times10^{9}\ \text{K}^4$ . This factor is identical for both cans since both start at the same water temperature in the same room.

Step 2. Apply the net-loss formula to the black can. $P_{\text{black}} = \varepsilon\sigma A\left(T^4-T_0^4\right) = 0.9\times(5.67\times10^{-8})\times0.02\times(6.906\times10^{9})$ . Multiplying step by step: $0.9\times5.67\times10^{-8}\times0.02 = 1.021\times10^{-9}$ , and $1.021\times10^{-9}\times6.906\times10^{9} \approx 7.05\ \text{W}$ .

Step 3. Apply the same formula to the shiny can, changing only $\varepsilon$ . $P_{\text{shiny}} = 0.05\times(5.67\times10^{-8})\times0.02\times(6.906\times10^{9}) \approx 0.39\ \text{W}$ .

Step 4. Compare. $P_{\text{black}}/P_{\text{shiny}} = 0.9/0.05 = 18$ , so the black can radiates heat away roughly eighteen times faster than the shiny can under identical conditions. The matte black can loses heat by radiation at about $7.0\ \text{W}$ , versus only about $0.39\ \text{W}$ for the shiny can — confirming quantitatively that the black can cools faster, matching the qualitative reasoning that a good absorber is also a good emitter.

Competitive level - equilibrium temperature of a satellite in deep space

A spherical satellite of radius $r = 1\ \text{m}$ orbits far from Earth, in surroundings so cold ( $T_0 \approx 0\ \text{K}$ ) that absorbed radiation can be neglected. Its onboard electronics dissipate a steady $P = 1000\ \text{W}$ of heat, which must leave entirely by radiation since there is no air for conduction or convection. Treating the surface as an ideal blackbody ( $\varepsilon = 1$ ), find its steady-state surface temperature.

Step 1. Set up the energy balance. In steady state, the satellite's temperature is no longer changing, so the power radiated away must exactly equal the power generated internally: $\sigma A T^4 = P$ .

Step 2. Compute the radiating area. A sphere of radius $r=1\ \text{m}$ has surface area $A = 4\pi r^2 = 4\pi(1)^2 \approx 12.57\ \text{m}^2$ .

Step 3. Solve for $T^4$ . $T^4 = \dfrac{P}{\sigma A} = \dfrac{1000}{(5.67\times10^{-8})(12.57)} = \dfrac{1000}{7.126\times10^{-7}} \approx 1.403\times10^{9}\ \text{K}^4$ .

Step 4. Take the fourth root by taking a square root twice. $\sqrt{1.403\times10^{9}} \approx 3.746\times10^{4}$ , and $\sqrt{3.746\times10^4} \approx 193.5$ . So the satellite settles at a steady-state surface temperature of about $T \approx 193\ \text{K}$ , roughly $-80^\circ\text{C}$ — cold by everyday standards, but a stable equilibrium reached entirely through radiation, with no medium available for any other heat-loss route. Timing note: this is a two-mark-worthy setup dressed as a four-mark problem — the only real work is one division and two square roots; don't reach for a calculator's fourth-root button if you don't have one, just square-root twice, and budget under 90 seconds once the balance equation $\sigma A T^4 = P$ is written down.

Common traps

Thinking radiation needs a medium the way conduction and convection do — it is the one mechanism that works through a vacuum, which is precisely why it dominates heat exchange in space.
Assuming only very hot, visibly glowing objects emit radiation — every object above absolute zero radiates continuously; most of it at room temperature is invisible infrared, not visible light.
Believing a good reflector is also a good emitter — it is the reverse: good reflectors are poor absorbers and, by the same surface property, poor emitters, while dark matte surfaces are good at both.
Treating an object at thermal equilibrium as if it has "stopped radiating" — it is still emitting and absorbing at full strength, just with the two exactly balanced, so the net flow is zero, not the individual emission.
Applying $E = \sigma T^4$ with a Celsius temperature instead of kelvin — the fourth power makes this error catastrophic, since it silently changes the answer by orders of magnitude; always convert to kelvin first.
Forgetting the emissivity factor $\varepsilon$ for real (non-blackbody) surfaces and using $\sigma T^4$ directly — that formula gives only the theoretical upper limit for an ideal blackbody, not the actual radiated power of a real can, wall, or satellite skin.

You should be able to…

Explain why radiation, unlike conduction and convection, can transfer heat through a vacuum.
Connect surface colour and texture (dark/matte versus light/polished) to how strongly a surface absorbs and emits thermal radiation.
State and apply Stefan's law $E=\sigma T^4$ , including using emissivity $\varepsilon$ for real, non-blackbody surfaces.
Compute the net radiative power exchanged between an object and its surroundings using $P_{\text{net}} = \varepsilon\sigma A(T^4-T_0^4)$ , always converting temperatures to kelvin first.
Explain why an object at thermal equilibrium with its surroundings still emits and absorbs radiation, even though the net flow is zero.
Give real-life examples (thermos flasks, radiator panels, car colours, spacesuits) where surface choice is used deliberately to control radiative heat transfer.

Ask the tutor: curious how the same Stefan's law leads into Wien's displacement law and why hotter stars look bluer — ask the AI tutor to bridge this article into blackbody radiation.

Why does a thermos flask have a shiny silvered inner wall instead of a black one

Because a shiny, polished surface is a poor emitter of thermal radiation, and the whole point of a thermos is to minimise heat loss (or gain) from the liquid inside. If the inner wall were matte black, it would radiate heat away from hot contents very efficiently — exactly the opposite of what you want in a flask. The silvering reflects most incident radiation instead of absorbing and re-emitting it, and by the same emissivity logic used in the worked example above, a low-emissivity wall loses many times less power by radiation than a high-emissivity one would, for the same temperature difference.

Does an ice cube radiate heat too, since it is so cold

Yes — any object above absolute zero radiates, including an ice cube at around 273 K. It radiates at a much lower rate than something hot, because Stefan's law scales with the fourth power of absolute temperature, but the rate is never exactly zero as long as the temperature is above 0 K. In fact the ice cube is simultaneously absorbing radiation from the warmer room around it, and because the room's $T_0^4$ term is larger than the ice cube's own $T^4$ term, the net radiative flow is actually into the ice cube, which is one reason (along with conduction and convection) that it eventually melts.

Is Stefan's law only for perfect blackbodies, or can I use it for ordinary objects like a kettle

Stefan's law in its bare form, $E=\sigma T^4$ , gives the theoretical maximum power per unit area for an ideal blackbody — no real surface can beat that limit at a given temperature. For an ordinary object like a kettle, you scale it down by the surface's emissivity $\varepsilon$ , giving $P=\varepsilon\sigma A T^4$ for the power it actually radiates. A shiny stainless-steel kettle has a low emissivity (often under 0.1), which is precisely why kettles are usually shiny rather than matte black — it keeps the water hotter for longer by cutting radiative loss.

Why doesn't this article just derive Wien's displacement law and finish blackbody radiation here

Because CBSE splits the topic deliberately: this article covers radiation as a heat-transfer mechanism — how heat moves via electromagnetic waves, what makes surfaces good or poor emitters, and Stefan's law for the total power radiated. The next article, on blackbody radiation, builds the fuller picture — how that radiated power is spread across different wavelengths, why hotter objects glow at shorter wavelengths (Wien's displacement law), and why the Sun looks white-hot while a heating coil glows dull red. Keeping the two separate mirrors exactly how your NCERT textbook sequences the chapter, and how board papers tend to split the marks.