Emissivity: Why Dark Surfaces Radiate Better - Physics for Pilot Assessments

A matt black surface absorbs radiation better than a shiny pale one, and — this is the part that surprises people — it also emits better. By the end of this lesson you should be able to say what happens to radiation arriving at a surface, explain why a good absorber has to be a good emitter rather than a poor one, put an approximate number on how well a surface radiates, and predict which of two otherwise identical objects will cool faster.

What happens when radiation arrives at a surface

Radiation striking an object has three possible fates: it can be absorbed, becoming internal energy in the object and warming it; it can be reflected back off the surface; or it can be transmitted straight through, as light is through glass. For an opaque object nothing is transmitted, so everything that arrives is either absorbed or reflected, and those two shares have to add up to the whole.

Radiation arriving at an opaque surface

E_\text{incident} = E_\text{absorbed} + E_\text{reflected}

$E_\text{incident}$: the radiation energy arriving at the surface, in joules (J)
$E_\text{absorbed}$: the part taken in, which raises the object's internal energy, in joules (J)
$E_\text{reflected}$: the part bounced back off the surface, in joules (J)

Since the two shares are forced to add up to the total, a surface that reflects well must absorb badly and a surface that reflects badly must absorb well — they are not two independent properties but one property described from either end. A matt black surface reflects almost nothing, so it absorbs almost everything; a polished silver one reflects almost everything, so it absorbs almost nothing.

Two things are worth separating here, because the tests occasionally lean on the difference. Colour and finish are not the same property. Colour tells you which wavelengths of visible light a surface absorbs, and finish tells you whether what is reflected goes back in one direction, from a polished surface, or is scattered in all directions, from a matt one. For the infrared radiation that objects at everyday temperatures actually emit, visible colour matters much less than people assume: paint of any colour, including white, is a good infrared absorber and emitter, whereas bare polished metal is a poor one whether it looks bright or dull. So the honest version of the rule is that matt, non-metallic surfaces radiate well and polished metal surfaces radiate badly, and that in the visible band — sunlight, for instance — dark colours also beat pale ones. It is also, in every case, purely a property of the outermost surface. The same kettle, painted matt black on the outside or polished, behaves quite differently while the water inside it is identical.

A good absorber is also a good emitter

This is the step that closes the topic, and it deserves an argument rather than an assertion. Put two identical objects into a sealed room whose walls are held at a fixed temperature — one of them matt black, one of them polished — and leave them. Both will end up at the room's temperature and then stay there indefinitely. Now think about what that requires. Both objects are being bathed in radiation from the walls, and the matt black one is absorbing far more of it than the polished one, because it reflects hardly any. If the black object were not also losing energy faster, it would be taking in more than it gave out, and its temperature would climb — above the temperature of the walls, and then further, indefinitely, with no source of energy other than walls that are colder than it is. That is impossible: energy would be appearing from nowhere, and heat would be flowing from a cooler place to a hotter one unaided.

The only way the observed behaviour can hold is if each surface emits at exactly the rate it absorbs. A good absorber is therefore a good emitter, necessarily and by the same factor. That is the answer to the question the knowledge tests actually ask — why does a dark object radiate heat faster than a light one — and it is not because dark surfaces are somehow special. It is because absorbing and emitting are the same physical process running in opposite directions, so a surface cannot be selective about which direction it is good at.

Emissivity: putting a number on it

The measure of how good a surface is at this is called its emissivity, written with the Greek letter epsilon. It compares the surface against an ideal: a perfect absorber, which takes in every scrap of radiation that falls on it and reflects none, and which by the argument above must therefore also be the best possible emitter. Such a surface is called a black body, and its emissivity is defined as 1. Every real surface falls somewhere between that and 0.

Emissivity

\varepsilon = \frac{P_\text{surface}}{P_\text{black body}}

$\varepsilon$: emissivity — a bare ratio between 0 and 1, with no units
$P_\text{surface}$: the power the real surface radiates, in watts (W)
$P_\text{black body}$: the power a black body of the same area and temperature would radiate, in watts (W)

Because it is defined as a ratio of two powers measured at the same temperature and area, the same number also tells you what fraction of arriving radiation the surface absorbs — which is the previous section restated as arithmetic. Typical values: matt black paint about 0.95, ordinary paint of any colour and most brick, concrete and fabric between 0.85 and 0.95, oxidised or rough metal around 0.7 to 0.8, and polished aluminium about 0.05 with polished silver lower still at around 0.02. Notice how little of the range is actually occupied: almost everything non-metallic is up near 0.9, and it is only bright bare metal that drops to a few per cent. One further point without arithmetic: the power a surface radiates climbs very steeply as its temperature rises, much faster than in proportion, which is why a hot object cools fastest in the first minutes and why a fire feels so much stronger than a radiator.

Example 1 — what an emissivity means in watts

A surface is at a temperature and size at which a black body would radiate 400 watts. How much power does the surface actually radiate if it is coated in matt black paint, with an emissivity of 0.95? And if it is polished aluminium, with an emissivity of 0.05?

Emissivity is defined as the ratio of the surface's radiated power to the black body's, so multiplying the black-body figure by the emissivity gives the real output: power = emissivity × black-body power.
For the matt black paint: $P_\text{surface} = 0.95 \times 400 = 380\ \text{W}$ That is within a few per cent of the best any surface could possibly do.
For the polished aluminium: $P_\text{surface} = 0.05 \times 400$ Multiplying by 0.05 is taking a twentieth, and $\frac{400}{20} = 20$ watts.
Same area, same temperature, and one surface is radiating $\frac{380}{20} = 19$ times as fast as the other. Nothing differs but the outermost surface.
The same two numbers describe absorption. Bathe both surfaces in identical radiation and the painted one takes in 95 per cent of what arrives while the polished one takes in 5 per cent — one property, read in either direction.

Two cans, one experiment

The standard demonstration is worth describing because items are written around it. Take two identical metal cans, one painted matt black on the outside and one left polished. Fill each with the same mass of water at the same starting temperature, put a thermometer in each, stand them side by side in the same room, and record both temperatures every minute. Everything that could differ has been made identical except the surface, which is the point of the design. The black can cools measurably faster, every time, and the gap between the two readings widens for as long as the experiment runs.

Same water, same starting temperature, same room. The only difference is the outside surface, and it is enough to change the cooling rate by a factor of two or more. The matt can is drawn with a rough hatch and the polished one smooth with highlight streaks, rather than one dark block beside a pale one: these figures are drawn in the page's own ink, so a block that is dark on a white page is the pale one on a dark page, which would stand the comparison on its head.

It is worth being clear that radiation is not the only thing cooling those cans — conduction into the bench and convection into the surrounding air are happening as well, and they are the same for both. That is exactly why the comparison works: the mechanisms the surface does not affect are held identical, so the difference between the two curves is attributable to radiation alone.

Example 2 — turning a temperature drop into an energy loss

Each can holds 0.2 kilograms of water starting at 80 °C. After 10 minutes the matt black can reads 60 °C and the polished one reads 72 °C. How much energy has each lost, and what average power does that represent? Take water's specific heat capacity as 4200 joules per kilogram per degree.

Work out each temperature drop: the black can has fallen 80 − 60 = 20 °C, and the polished one 80 − 72 = 8 °C.
Convert each drop into joules with heat = mass × specific heat capacity × temperature change. For the black can: $E = 0.2 \times 4200 \times 20$ Take it in order — 0.2 × 4200 = 840, and 840 × 20 = 16 800 joules.
For the polished can: $E = 840 \times 8 = 6720\ \text{J}$
Now the average power, which from the energy module is energy divided by time. Ten minutes is 10 × 60 = 600 seconds.
Black can: $P = \frac{16\,800}{600} = 28\ \text{W}$ Polished can: $P = \frac{6720}{600} = 11.2\ \text{W}$
So the matt black surface is shedding energy at two and a half times the rate, from the same water at the same starting temperature in the same room.

Average power

P = \frac{E}{t}

$P$: average power, in watts (W) — one watt is one joule per second
$E$: the energy transferred, in joules (J)
$t$: the time it took, in seconds (s)

That relationship runs backwards just as usefully, which the next example does.

Example 3 — running the power calculation the other way

A matt black container of 0.2 kilograms of water is losing energy at an average of 28 watts. How long does it take to cool by 20 °C? And if the container were polished instead, losing only 11.2 watts, how long would the same 20 °C drop take?

First find the energy involved: $E = 0.2 \times 4200 \times 20 = 16\,800\ \text{J}$ That is exactly as before.
Rearrange $P = \frac{E}{t}$ for the time by multiplying both sides by the time and then dividing by the power, which gives $t = \frac{E}{P}$ .
For the matt black container: $t = \frac{16\,800}{28} = 600\ \text{s}$ That is 10 minutes.
For the polished one: $t = \frac{16\,800}{11.2}$ Multiply both numbers by 10 to clear the decimal, giving $\frac{168\,000}{112} = 1500$ seconds, which is 25 minutes.
So the same drop takes two and a half times as long behind a shiny surface — the same factor as before, which it has to be, since the energy was identical and only the rate changed.

Where it is used

Once the rule is in place the design decisions read off it. Anything meant to shed heat gets a matt dark non-metallic surface: the fins on the back of a refrigerator, the cooling fins of an air-cooled engine, the black-painted absorber plate of a solar water heater, which wants to be a good absorber and is unavoidably a good emitter as well. Anything meant to keep heat gets a polished metal one: the silvered walls of a vacuum flask, the shiny foil of a survival blanket, which works in both directions by reflecting body radiation back inwards and sunlight outwards, and the polished skin of a fuel tank or a storage vessel standing in strong sun. There is one honest caveat, and it is the same one from the previous lesson: a household radiator loses most of its heat by convection rather than radiation, so painting it a different colour changes very little. The surface only dominates where radiation is the main route out.

Why it matters in aviation

A surface radiates according to what it is and how warm it is, not according to what the air around it is doing, and that distinction has a direct consequence on the ground. An upward-facing surface radiates towards whatever is above it, and on a clear night that is the sky, which behaves as something extremely cold. An aircraft parked in the open therefore loses energy upwards faster than the air can conduct it back, and the skin — the thin upper surface of a wing especially — settles below the temperature of the air around it.

That is the mechanism by which frost forms on an airframe on a night when the reported air temperature never reached freezing. Cloud cover changes it, because a cloud base radiates back downwards at something close to the air temperature and the surface then stays near it. Two things follow. The state of the surface has to be established by inspecting it rather than inferred from a thermometer, and the finish is part of the story: a painted, matt skin radiates far more freely than bare polished metal, so two surfaces under the same sky do not cool by the same amount.

What the test asks

The distractor here is unusually seductive, and it is the one this lesson exists to close: an option stating that a good absorber must be a poor emitter, sometimes dressed up as a dark object holding on to heat because it soaked so much up. It has the shape of a trade-off and there is no trade-off available — absorbing and emitting are the same process in opposite directions, so a surface is either good at both or bad at both. Expect the two-cans item in one of its two forms, cooling or warming, and note that the answer is the black can either way: fastest to cool when hot, and fastest to warm when radiation is falling on it. A third distractor attributes the effect to what is inside rather than to the surface, offering an explanation in terms of the water rather than the paint. And watch for an option claiming that a shiny object does not emit at all. Every object emits; a polished one simply emits a few per cent of what a black body would.