Why Surface Colour Does Not Alone Determine Thermal Radiation at Every Wavelength

Two metal plates are heated to the same temperature. One is painted black, and the other is painted white. Which one emits more thermal radiation at a wavelength of \(10\,\mu\mathrm{m}\)?

You might choose the black plate. After all, black surfaces are usually described as good absorbers and emitters of radiation.

But the two plates could emit almost exactly the same amount at that wavelength. A polished metal plate at the same temperature might emit far less than either of them.

The reason is that visible colour tells us how a surface interacts with visible light, not how it behaves at every wavelength. Thermal emission depends on temperature and spectral emissivity: how efficiently a material emits radiation at a particular wavelength.

At ordinary temperatures, most thermal radiation is infrared, well outside the wavelengths your eyes detect. A surface can look white while behaving almost like an ideal black body in the infrared.

01Why physicists needed an experiment that controlled temperature

Suppose you compare a black metal object with a shiny metal object and find that the black one emits more radiation.

That sounds convincing. But what if the black object is slightly hotter? Or larger? Or positioned closer to your detector?

You haven’t isolated the effect of the surface.

This measurement problem motivated an experiment that remains useful today: Leslie’s cube.

Leslie’s cube: changing the surface without changing the temperature

In 1804, Scottish physicist John Leslie described experiments using hollow metal cubes with differently prepared surfaces.

Some surfaces were polished, others were covered with soot or paper, and different materials and finishes could be compared.

The crucial feature was that the surfaces belonged to the same container.

The cube could be filled with hot water through an opening at the top. A thermometer monitored the water temperature, while the metal walls were heated to approximately the same temperature.

Now the experimenter could compare surfaces without deliberately changing the temperature.

Leslie’s historical arrangement used a concave mirror and a differential air thermometer.

Radiation travelled from the selected face of the cube towards the mirror. The mirror concentrated it onto one of two air-filled thermometer bulbs.

If that bulb absorbed more energy, its temperature rose relative to the other bulb. The resulting pressure difference displaced liquid in the connecting tube.

The liquid displacement provided an indication of the radiation reaching the instrument.

Modern classroom versions often replace the mirror and differential thermometer with a thermopile or infrared radiation sensor connected to a meter.

The comparison still follows the same logic:

1. Heat the different surfaces to approximately the same temperature.
2. Position the detector at the same distance and angle from each face.
3. Measure the detector response for one face, then repeat for the others.
4. Keep the surroundings and measurement conditions as consistent as possible.

A good experiment also checks the actual surface temperatures rather than assuming that equal water temperature guarantees perfectly equal wall temperatures.

What should the experiment show?

Imagine two competing predictions.

If temperature alone determined thermal emission, every face at the same temperature would emit the same amount of radiation per square metre.

If the surface mattered, the detector could respond differently to equally hot faces.

Leslie’s investigations showed that surface condition affected radiation. Blackened surfaces generally produced a stronger thermal response than highly polished metal surfaces.

This was evidence that temperature alone was insufficient.

But there was a limitation.

The original instrument did not measure emission separately at every wavelength. Its response combined the radiation reaching it across a range of wavelengths, along with the effects of the detector’s own sensitivity.

It therefore could not establish that a visibly black surface emits more than every visibly white surface at every wavelength.

That requires a more precise measurement.

02Why visible colour is an unreliable guide to infrared emission

Your eyes detect light over roughly \(0.4\) to \(0.7\,\mu\mathrm{m}\).

An object appears white under ordinary illumination because it reflects much of the visible light reaching it. A black object absorbs much more of that light.

But a warm object may emit most strongly near \(10\,\mu\mathrm{m}\), more than ten times the wavelength of red light.

Those are different parts of the electromagnetic spectrum.

Think of the surface as having different responses to different wavelength bands.

Knowing its response to visible light tells you something about one band. It does not give you its response to the infrared band.

The underlying material, coating, thickness, oxidation, texture, and wavelength can all affect the result.

Spectral emissivity explains the difference

An ideal black body is a theoretical surface that absorbs all incident radiation and emits the maximum possible thermal radiation at each wavelength for its temperature.

Real surfaces usually emit less.

We describe their emission using spectral emissivity, \(\varepsilon_\lambda\).

For the same temperature and wavelength,

\[
M_\lambda
=
\varepsilon_\lambda M_{\lambda,\mathrm{bb}}
\]

where:

• \(M_\lambda\) is the surface’s emitted spectral radiant exitance.
• \(\varepsilon_\lambda\) is its spectral emissivity, a number between 0 and 1.
• \(M_{\lambda,\mathrm{bb}}\) is the spectral radiant exitance of an ideal black body at the same temperature.

Spectral radiant exitance means emitted power per square metre of surface per unit wavelength interval. Its units can be \(\mathrm{W\,m^{-2}\,\mu m^{-1}}\).

A spectral emissivity of \(0.90\) means the surface emits 90% as much as an ideal black body at that temperature and wavelength.

It does not mean the surface emits 90% of its entire thermal energy at that wavelength.

Compare three surfaces

Consider the following hypothetical coatings.

These are illustrative spectral emissivities, not measurements of particular commercial products.

At \(0.55\,\mu\mathrm{m}\), the black paint is a strong absorber. The white paint and polished aluminium absorb much less.

At \(10\,\mu\mathrm{m}\), however, the black and white paints have similar emissivities.

The polished aluminium remains a poor emitter.

There’s another subtle point. Even if black paint has high emissivity at \(0.55\,\mu\mathrm{m}\), it emits essentially no detectable visible thermal radiation at ordinary room temperatures.

Emissivity tells you the fraction of the black-body emission. Temperature determines how much black-body emission is available in the first place.

Why a good absorber can also be a good emitter

Kirchhoff’s law of thermal radiation connects absorption and emission.

At thermal equilibrium, a surface’s emissivity equals its absorptivity at the same wavelength, with corresponding directional conditions:

\[
\varepsilon_\lambda=\alpha_\lambda
\]

Here, \(\alpha_\lambda\) is the fraction of incident radiation absorbed at wavelength \(\lambda\).

For an opaque surface, where transmission is negligible,

\[
\alpha_\lambda+\rho_\lambda=1
\]

where \(\rho_\lambda\) is the reflected fraction.

These relationships explain why a surface that strongly absorbs infrared radiation can also emit it efficiently.

But the wavelength condition is essential.

A white coating may reflect visible sunlight efficiently while absorbing and emitting infrared radiation strongly.

This is useful in spacecraft thermal control. A white radiator coating can help limit absorbed solar radiation while still allowing internal energy to escape as thermal infrared radiation.

It would be a mistake to choose a spacecraft’s radiator coating using visible colour alone.

For the broader relationship between absorption, reflection, emission, and net energy transfer, see Thermal Radiation: Energy Transfer and Surface Effects.

03How a revised experiment separates the wavelengths

Return to Leslie’s cube.

Suppose the black and white painted faces produce nearly identical readings from a broadband infrared detector.

Does that prove their emissivities are identical everywhere?

No. The detector may be combining radiation from wavelengths where the two surfaces behave differently.

A more informative experiment measures separate wavelength bands.

A modern spectral comparison

Imagine an improved apparatus arranged as follows.

The heated cube sits inside a controlled enclosure. A temperature sensor measures its surface temperature.

Radiation from one face travels towards a fixed aperture that defines the region being observed.

It then passes through an infrared wavelength selector, such as a series of calibrated bandpass filters or a suitable infrared spectrometer.

Finally, the selected radiation reaches an infrared detector connected to a measuring instrument.

A black-body reference source provides calibration at known temperatures.

The investigator can now compare the same surface at, for example, \(8\,\mu\mathrm{m}\) and \(12\,\mu\mathrm{m}\).

To make those readings meaningful, several problems need controlling.

First, the detector may be more sensitive at one wavelength than another. Calibration must correct for this.

Second, filters do not necessarily transmit every selected wavelength equally.

Third, the surfaces may reflect radiation from the surroundings into the detector.

Finally, each face must occupy the same viewing geometry. Otherwise, a larger detector reading could simply result from collecting radiation from a larger apparent area.

The important improvement is not just a more sensitive sensor. It is the ability to distinguish emission at individual wavelengths from the combined detector response.

A common modern Leslie’s cube demonstration finds little difference between matte black and matte white painted faces in the thermal infrared, while the difference between painted and polished metal faces is much larger.

That is consistent with their infrared emissivities being similar, despite their very different visible appearances.

04Spectral intensity, total power and peak wavelength are different measurements

A radiation spectrum answers three different questions. They must not be confused.

For an actual surface, its spectral emission can be thought of as the ideal black-body spectrum modified by its wavelength-dependent emissivity.

The peak wavelength is not the only emitted wavelength

For an ideal black body, Wien’s displacement law gives

\[
\lambda_{\max}T=b
\]

where:

• \(\lambda_{\max}\) is the peak wavelength in metres.
• \(T\) is absolute temperature in kelvin.
• \(b=2.898\times10^{-3}\,\mathrm{m\,K}\) is Wien’s displacement constant.

For a black body at \(320\,\mathrm{K}\),

\[
\begin{aligned}
\lambda_{\max}
&=\frac{2.898\times10^{-3}}{320}\\
&=9.06\times10^{-6}\,\mathrm{m}\\
&=9.06\,\mu\mathrm{m}
\end{aligned}
\]

That peak lies in the infrared.

However, Wien’s law gives the peak of an ideal black-body wavelength spectrum.

If a real surface has approximately constant emissivity across the relevant wavelengths, its spectrum has the same shape and peak position as the black-body spectrum at the same temperature.

If its emissivity changes strongly with wavelength, the real surface’s emission peak can shift away from the ideal black-body value.

So a measured peak from a wavelength-selective surface does not necessarily give its temperature directly through Wien’s law.

05Why an infrared detector can make a shiny surface look hotter

There is another experimental complication.

An infrared detector does not necessarily receive only radiation emitted by the surface it is viewing.

Some radiation may have come from elsewhere and been reflected by that surface.

Imagine pointing an infrared camera at a polished metal panel near a warm heater.

The panel can reflect infrared radiation from the heater towards the camera, rather like an ordinary mirror reflects visible light.

The camera may therefore receive a large signal even though the metal panel is a poor thermal emitter.

For an opaque surface in a simplified diffuse-radiation model, the outgoing signal at a particular wavelength can be written as

\[
S_\lambda
=
\varepsilon_\lambda B_s
+
(1-\varepsilon_\lambda)B_b
\]

Here:

• \(S_\lambda\) is the calibrated outgoing spectral signal.
• \(B_s\) is the black-body spectral emission at the surface temperature.
• \(B_b\) is the effective black-body spectral emission of the surroundings being reflected towards the detector.
• \(\varepsilon_\lambda\) is the surface’s spectral emissivity.

All three spectral quantities have the same units, such as \(\mathrm{W\,m^{-2}\,\mu m^{-1}}\).

The first term describes emitted radiation. The second describes reflected radiation.

This model assumes the surface is opaque and diffuse, the surrounding radiation field is uniform, and the instrument’s geometric and spectral responses have been corrected. Real mirror-like surfaces can require a directional reflection model.

Extension: What about total power?

For an ideal black body, the Stefan-Boltzmann law gives the total emitted power across all wavelengths.

For an approximately grey surface, whose emissivity can be treated as constant over the important thermal wavelengths,

\[
P_{\mathrm{emit}}
\approx\varepsilon\sigma AT^4
\]

where \(P_{\mathrm{emit}}\) is emitted power in watts, \(\varepsilon\) is the total hemispherical emissivity, \(A\) is emitting area in square metres, \(T\) is temperature in kelvin, and

\[
\sigma
=5.67\times10^{-8}
\ \mathrm{W\,m^{-2}\,K^{-4}}
\]

is the Stefan-Boltzmann constant.

This total-power calculation is an extension beyond the central Wien’s-law analysis in the NSW Physics Stage 6 (2017) course.

It is useful here because it makes an important limitation explicit: you cannot automatically substitute the emissivity measured at \(10\,\mu\mathrm{m}\) into a total-power equation.

For a non-grey material, total emissivity depends on how its emissivity varies across the complete thermal spectrum.

For an opaque grey surface surrounded by a large, uniform black-body enclosure at temperature \(T_b\), the net radiative power leaving the surface can be approximated by

\[
P_{\mathrm{net,out}}
\approx
\varepsilon\sigma A(T^4-T_b^4)
\]

This assumes the enclosure fills the surface’s radiative surroundings, and ignores conduction and convection.

Notice the distinction between total emitted power and net energy loss. Even a hot surface absorbs radiation from its surroundings while emitting its own.

06Questions and solutions

Question 1

A black phone case and a white phone case are both at \(300\,\mathrm{K}\).

(a) Calculate the peak wavelength of an ideal black-body spectrum at this temperature using \(b=2.898\times10^{-3}\,\mathrm{m\,K}\).

(b) State whether the colour of the cases is enough to determine which emits more thermal radiation at \(10\,\mu\mathrm{m}\).

Solution 1

The ideal black-body peak is approximately \(9.66\,\mu\mathrm{m}\), and visible colour alone cannot determine which case emits more radiation at \(10\,\mu\mathrm{m}\).

(a) Using Wien’s displacement law,

\[
\lambda_{\max}=\frac{b}{T}
\]

Substitute the given temperature:

\[
\begin{aligned}
\lambda_{\max}
&=\frac{2.898\times10^{-3}}{300}\\
&=9.66\times10^{-6}\,\mathrm{m}\\
&=9.66\,\mu\mathrm{m}
\end{aligned}
\]

The ideal black-body emission peaks in the infrared, not the visible spectrum.

(b) Black and white appearances depend on the cases’ interactions with visible light.

Their thermal emission at \(10\,\mu\mathrm{m}\) depends on their infrared emissivities, which have not been provided.

The tempting conclusion is that the black case must radiate more because it absorbs more visible light. That assumes visible absorptivity determines infrared emissivity.

It does not.

At the same temperature and wavelength, the case with the greater infrared emissivity emits more radiation per square metre.

Question 2

Two experimental coatings, A and B, are applied to equal-area plates held at \(330\,\mathrm{K}\).

A spectrometer measures their emitted spectral radiant exitance after correcting for reflected background radiation.

The following illustrative values are averages over two selected wavelength bands. Assume each value is approximately constant within its stated band.

All spectral values are in \(\mathrm{W\,m^{-2}\,\mu m^{-1}}\).

(a) Determine the spectral emissivity of each coating in each band.

(b) Explain whether either coating can be treated as a greybody with constant emissivity across the two bands.

(c) Calculate which coating emits more power per square metre across the two measured bands combined.

(d) Can the experiment establish which coating emits more total power across all wavelengths, or which has the longer actual peak wavelength?

Solution 2

Coating B emits more across the measured bands: \(79\,\mathrm{W\,m^{-2}}\), compared with \(64\,\mathrm{W\,m^{-2}}\) for A. Neither coating has constant emissivity across these bands, and the incomplete data cannot establish their total emitted powers or actual peak wavelengths.

(a) Use the ratio of measured emission to black-body emission:

\[
\varepsilon_\lambda
=
\frac{M_\lambda}{M_{\lambda,\mathrm{bb}}}
\]

For coating A,

\[
\begin{aligned}
\varepsilon_{A,8-9}&=\frac{40}{50}=0.80\\
\varepsilon_{A,11-13}&=\frac{12}{40}=0.30
\end{aligned}
\]

For coating B,

\[
\begin{aligned}
\varepsilon_{B,8-9}&=\frac{15}{50}=0.30\\
\varepsilon_{B,11-13}&=\frac{32}{40}=0.80
\end{aligned}
\]

These emissivities are dimensionless.

Coating A is the stronger emitter in the shorter-wavelength band. Coating B is stronger in the longer-wavelength band.

(b) Neither coating has constant emissivity across both bands.

A changes from \(0.80\) to \(0.30\), while B changes from \(0.30\) to \(0.80\).

They are wavelength-selective emitters, so multiplying an entire ideal black-body curve by one constant emissivity would not reproduce their measured behaviour.

(c) The emitted power per square metre in each band is approximately the spectral radiant exitance multiplied by the band’s width.

For A,

\[
\begin{aligned}
M_{A,\mathrm{bands}}
&=40(1.0)+12(2.0)\\
&=64\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

For B,

\[
\begin{aligned}
M_{B,\mathrm{bands}}
&=15(1.0)+32(2.0)\\
&=79\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

Coating B emits \(15\,\mathrm{W\,m^{-2}}\) more across the measured bands.

This is easy to miss if you compare only the largest spectral value, \(40\), with the other readings.

The longer-wavelength band is twice as wide. Its contribution to the integrated power must be weighted accordingly.

(d) Neither total emitted power nor the true peak wavelength can be established.

The instrument has measured only two regions of the spectrum. There could be substantial additional emission between or outside those regions.

Even though coating B emits more across the tested bands, coating A could emit more over the unmeasured wavelengths and reverse the overall ranking.

Similarly, the greatest measured band-average value is not proof of the location of the true spectral maximum.

The exact constraint is incomplete wavelength coverage, not insufficient arithmetic.

Question 3

Two visibly white panels, P and Q, are held at exactly the same temperature by independent heaters.

Panel P has a matte coating. Panel Q has a glossy coating.

They are viewed by a calibrated infrared detector at one narrow wavelength band. Both panels are opaque, and assume a diffuse-radiation model with the same viewing geometry.

When the surrounding enclosure is hotter than the panels, Q produces the larger outgoing infrared signal.

The enclosure is then cooled to below the panels’ temperature without changing either panel’s temperature. Now P produces the larger signal.

(a) Explain why the original detector reading does not prove Q emits more radiation.

(b) Under the stated assumptions, which panel has the higher emissivity at the measured wavelength?

(c) Explain why the ranking reverses when the enclosure temperature changes.

(d) Identify an experimental control that would help separate surface emission from reflection.

Solution 3

Panel P has the higher spectral emissivity, while panel Q is more reflective. Q’s larger signal in the warm enclosure comes from reflected background radiation, not necessarily from stronger emission.

(a) The infrared detector records outgoing radiation consisting of both emitted and reflected components.

Since the panel temperatures are independently confirmed to be equal, temperature differences cannot explain the reversal.

The original result could be caused by Q reflecting more of the hot enclosure’s infrared radiation towards the detector.

(b) P must have the higher emissivity in the simplified model.

For an opaque surface,

\[
\varepsilon_\lambda+\rho_\lambda=1
\]

The less reflective panel is therefore the more emissive one at the same wavelength.

P’s stronger signal when the enclosure is colder than the panels is consistent with its own emission making up a larger fraction of the radiation seen by the detector.

(c) The two contributions compete differently depending on the enclosure temperature.

When the surroundings are hotter, their radiation is stronger than the black-body radiation corresponding to the panel temperature.

Q reflects more of that strong radiation, so it produces the larger outgoing signal.

When the surroundings become cooler, the reflected contribution weakens.

P, with higher emissivity, then produces the larger signal because its own thermal emission is more important.

Mathematically, the outgoing signal is

\[
S_\lambda
=
\varepsilon_\lambda B_s+
(1-\varepsilon_\lambda)B_b
\]

The sign of \(B_s-B_b\) determines whether greater emissivity increases or decreases the outgoing signal.

This is the central constraint that reverses the ranking.

(d) A useful control is to repeat the comparison inside a large enclosure with a known, uniform infrared background temperature, while independently measuring the panel temperatures.

Measurements using at least two controlled background temperatures help distinguish reflection from intrinsic emission.

A narrow-band measurement still establishes behaviour only in that band. Ranking total thermal emission requires broader spectral information.

Question 4

Two equally sized panels are used in an idealised thermal-control experiment.

One has a black coating, and the other has a white coating.

Each has an exposed area of \(1.00\,\mathrm{m^2}\) and is maintained at \(320\,\mathrm{K}\).

A filtered lamp delivers \(700\,\mathrm{W\,m^{-2}}\) of visible radiation to each panel. The panels also exchange infrared radiation with a large, uniform black-body enclosure at \(300\,\mathrm{K}\).

The coatings have the following properties:

Treat the coatings as grey emitters over the relevant thermal infrared wavelengths. Assume their infrared absorptivities equal their emissivities, the enclosure fills their radiative surroundings, and conduction and convection can be ignored.

For this extension question, use

\[
P_{\mathrm{net,in}}
=
\alpha_v FA
–
\varepsilon\sigma A(T_s^4-T_b^4)
\]

where \(\alpha_v\) is visible absorptivity, \(F\) is incident visible radiant power per square metre, \(A\) is exposed area, \(T_s\) is panel temperature, and \(T_b\) is enclosure temperature. Take positive net power to mean energy entering the panel.

Use \(\sigma=5.67\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}\).

(a) Calculate the visible radiant power absorbed by each panel.

(b) Calculate each panel’s net infrared power loss to the enclosure.

(c) Determine the net radiative power gain or loss for each panel.

(d) Explain why the result does not contradict the principle that good absorbers are good emitters.

Solution 4

The black panel gains approximately \(503\,\mathrm{W}\) by radiation, while the white panel loses approximately \(19.5\,\mathrm{W}\). The large difference comes mainly from visible-light absorption, not infrared emission.

(a) The absorbed visible power is

\[
P_{\mathrm{abs}}=\alpha_v FA
\]

For the black panel,

\[
P_{\mathrm{abs,black}}
=0.90(700)(1.00)
=630\ \mathrm{W}
\]

For the white panel,

\[
P_{\mathrm{abs,white}}
=0.15(700)(1.00)
=105\ \mathrm{W}
\]

The black panel absorbs \(525\,\mathrm{W}\) more from the visible lamp.

(b) Calculate the net infrared radiation exchange.

First, the temperature-dependent term is

\[
\begin{aligned}
\sigma(T_s^4-T_b^4)
&=5.67\times10^{-8}\\
&\quad\times(320^4-300^4)\\
&\approx135.3\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

For the black panel,

\[
\begin{aligned}
P_{\mathrm{IR,black}}
&=0.94(135.3)(1.00)\\
&\approx127.2\ \mathrm{W}
\end{aligned}
\]

For the white panel,

\[
\begin{aligned}
P_{\mathrm{IR,white}}
&=0.92(135.3)(1.00)\\
&\approx124.5\ \mathrm{W}
\end{aligned}
\]

These are positive outward infrared power losses.

The white panel’s infrared loss is only slightly smaller.

(c) Take absorbed visible power as positive and net outward infrared power as negative.

For the black panel,

\[
\begin{aligned}
P_{\mathrm{net,in}}
&=630-127.2\\
&=+502.8\ \mathrm{W}
\end{aligned}
\]

For the white panel,

\[
\begin{aligned}
P_{\mathrm{net,in}}
&=105-124.5\\
&=-19.5\ \mathrm{W}
\end{aligned}
\]

The black panel has a net radiative energy gain of approximately \(503\,\mathrm{W}\).

The white panel has a net radiative energy loss of approximately \(19.5\,\mathrm{W}\).

Because their temperatures are being held fixed, an external cooling system would need to remove energy from the black panel. An external heater would need to replace the white panel’s net loss, under the stated assumptions.

(d) There is no contradiction because the visible illumination and thermal infrared emission occupy different wavelength regions.

Kirchhoff’s law connects absorption and emission at corresponding wavelengths.

It does not require a surface with low visible absorptivity to have low infrared emissivity.

The white panel reflects most of the incoming visible radiation while still emitting infrared radiation efficiently.

The tempting approach is to classify the black panel as a universally better emitter and absorber. The missing condition is the wavelength at which each process occurs.

Question 5

Two experimental surfaces, A and B, have identical visible appearances.

Both are opaque, diffuse, and maintained at \(340\,\mathrm{K}\) by independent heaters.

They sit inside a large enclosure whose walls behave as ideal black bodies. A calibrated infrared instrument measures their outgoing spectral radiation in two separate bands.

Each band is \(1.0\,\mu\mathrm{m}\) wide, centred at \(8\,\mu\mathrm{m}\) or \(12\,\mu\mathrm{m}\). Assume spectral emission and emissivity are approximately constant within each band.

The enclosure completely surrounds the surfaces, and all geometric and detector corrections have been made.

First, the enclosure is also held at \(340\,\mathrm{K}\).

The detector records identical signals from A and B at both wavelengths:

The enclosure is then cooled to \(280\,\mathrm{K}\), while both surfaces remain at \(340\,\mathrm{K}\).

The new detector readings are:

All spectral readings are in \(\mathrm{W\,m^{-2}\,\mu m^{-1}}\).

Independent black-body calibration provides the following illustrative reference values:

Use the model

\[
S_\lambda
=
\varepsilon_\lambda B_s+
(1-\varepsilon_\lambda)B_b
\]

where \(B_s\) and \(B_b\) are the black-body spectral values at the surface and enclosure temperatures, respectively.

(a) A student argues that the identical detector readings in the \(340\,\mathrm{K}\) enclosure prove the surfaces have identical emissivities. Determine whether this is correct.

(b) Calculate the emissivity of each surface at \(8\,\mu\mathrm{m}\) and \(12\,\mu\mathrm{m}\).

(c) Find the emitted spectral radiant exitance of each surface at both wavelengths. Explain why these values differ from the detector readings in the cooled enclosure.

(d) Determine which surface has the greater net radiative energy loss per square metre across the two measured wavelength bands when the enclosure is at \(280\,\mathrm{K}\).

(e) Can you determine which surface emits more total power across all wavelengths, or which has the greater total net radiative loss? Explain what additional evidence is needed.

Solution 5

The student’s claim is incorrect. The surfaces can have very different emissivities while producing identical detector readings when the enclosure and surfaces are at the same temperature.

In the cooled enclosure, A has higher emissivity at \(8\,\mu\mathrm{m}\), while B has higher emissivity at \(12\,\mu\mathrm{m}\). A loses more energy across the two measured bands combined, but the full-spectrum ranking remains undetermined.

(a) Why identical signals do not prove identical emission

When the enclosure and surface temperatures are equal,

\[
B_s=B_b
\]

Substituting into the detector model,

\[
\begin{aligned}
S_\lambda
&=\varepsilon_\lambda B_s\\
&\quad+(1-\varepsilon_\lambda)B_s\\
&=B_s
\end{aligned}
\]

The emissivity disappears from the final result.

A highly emissive surface produces much of the outgoing radiation itself.

A highly reflective surface produces less radiation of its own but reflects more radiation from the equally warm enclosure.

Both can produce the same outgoing signal.

This is the crucial result: equal detected radiation in a uniform enclosure at thermal equilibrium does not establish equal intrinsic emissivity.

It also explains why the experimental change in enclosure temperature is useful.

(b) Calculate the emissivities

Rearrange the detector equation:

\[
\varepsilon_\lambda
=
\frac{S_\lambda-B_b}{B_s-B_b}
\]

At \(8\,\mu\mathrm{m}\), the black-body values are 58 and 19.

For A,

\[
\begin{aligned}
\varepsilon_{A,8}
&=\frac{50.2-19}{58-19}\\
&=\frac{31.2}{39}\\
&=0.80
\end{aligned}
\]

For B,

\[
\begin{aligned}
\varepsilon_{B,8}
&=\frac{26.8-19}{58-19}\\
&=\frac{7.8}{39}\\
&=0.20
\end{aligned}
\]

At \(12\,\mu\mathrm{m}\), the black-body values are 45 and 21.

For A,

\[
\begin{aligned}
\varepsilon_{A,12}
&=\frac{25.8-21}{45-21}\\
&=\frac{4.8}{24}\\
&=0.20
\end{aligned}
\]

For B,

\[
\begin{aligned}
\varepsilon_{B,12}
&=\frac{40.2-21}{45-21}\\
&=\frac{19.2}{24}\\
&=0.80
\end{aligned}
\]

Therefore,

Their spectral emissivity rankings reverse with wavelength.

Neither is accurately described by one constant emissivity across these two bands.

(c) Calculate actual emitted radiation

The emitted spectral radiant exitance is

\[
M_\lambda=\varepsilon_\lambda B_s
\]

At \(8\,\mu\mathrm{m}\),

\[
\begin{aligned}
M_{A,8}&=0.80(58)=46.4\\
M_{B,8}&=0.20(58)=11.6
\end{aligned}
\]

At \(12\,\mu\mathrm{m}\),

\[
\begin{aligned}
M_{A,12}&=0.20(45)=9.0\\
M_{B,12}&=0.80(45)=36.0
\end{aligned}
\]

All values are in \(\mathrm{W\,m^{-2}\,\mu m^{-1}}\).

The actual emitted spectral values are:

These differ from the cooled-enclosure detector readings because the detector also receives reflected radiation.

For example, at \(8\,\mu\mathrm{m}\), A emits \(46.4\,\mathrm{W\,m^{-2}\,\mu m^{-1}}\) but also reflects

\[
(1-0.80)(19)
=3.8\ \mathrm{W\,m^{-2}\,\mu m^{-1}}
\]

The total outgoing signal is therefore

\[
46.4+3.8
=50.2\ \mathrm{W\,m^{-2}\,\mu m^{-1}}
\]

which matches the measurement.

Emission and outgoing detector signal are not interchangeable quantities.

(d) Compare net radiation losses in the measured bands

For a surface exchanging radiation with the uniform black-body enclosure, the net outward spectral exchange is

\[
M_{\mathrm{net},\lambda}
=
\varepsilon_\lambda(B_s-B_b)
\]

At \(8\,\mu\mathrm{m}\), the black-body difference is

\[
58-19=39
\]

So,

\[
\begin{aligned}
M_{\mathrm{net},A,8}&=0.80(39)=31.2\\
M_{\mathrm{net},B,8}&=0.20(39)=7.8
\end{aligned}
\]

At \(12\,\mu\mathrm{m}\),

\[
45-21=24
\]

Therefore,

\[
\begin{aligned}
M_{\mathrm{net},A,12}&=0.20(24)=4.8\\
M_{\mathrm{net},B,12}&=0.80(24)=19.2
\end{aligned}
\]

These intermediate results are spectral power densities in \(\mathrm{W\,m^{-2}\,\mu m^{-1}}\).

Each measured band is \(1.0\,\mu\mathrm{m}\) wide.

Multiplying by the band widths gives the net power lost per square metre across the two measured bands.

For A,

\[
\begin{aligned}
M_{\mathrm{net},A}
&=31.2(1.0)+4.8(1.0)\\
&=36.0\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

For B,

\[
\begin{aligned}
M_{\mathrm{net},B}
&=7.8(1.0)+19.2(1.0)\\
&=27.0\ \mathrm{W\,m^{-2}}
\end{aligned}
\]

A loses \(9.0\,\mathrm{W\,m^{-2}}\) more than B across these two bands.

But look at the individual results: A dominates the shorter-wavelength band, and B dominates the longer-wavelength band.

Their relative performance changes with wavelength.

(e) Why the total-power ranking remains unknown

Neither total emitted power nor total net radiative loss across all wavelengths can be ranked from the supplied information.

The measurements cover only two wavelength bands.

Surface B could have substantially higher emissivity over a broad unmeasured region, allowing it to emit more total power despite losing less energy across the two tested bands.

Alternatively, A could remain the stronger emitter over most of the remaining thermal spectrum.

Both possibilities are consistent with the measurements.

The correct next experiment is to measure the calibrated emission spectrum across the significant thermal wavelengths, accounting for reflection, and compare the complete areas.

With the enclosure at \(280\,\mathrm{K}\), the net radiative exchange across the full spectrum must also include the radiation absorbed from that enclosure.

A final subtlety is that neither surface’s actual peak wavelength can be established from these two measurements.

Even though A emits more at \(8\,\mu\mathrm{m}\) and B emits more at \(12\,\mu\mathrm{m}\), an unmeasured wavelength could contain the maximum for either surface.

Wien’s law gives the ideal black-body peak at \(340\,\mathrm{K}\), but a strongly wavelength-selective coating need not have its actual emission maximum at that position.

The tempting approach is to identify the surface with the highest detector reading and call it the better radiator.

The experiment shows why that fails: detector readings combine emitted and reflected radiation, emissivity depends on wavelength, and total power depends on the entire spectrum.

07What this changes when interpreting radiation experiments

The lesson from Leslie’s cube is not simply that black surfaces emit more than shiny ones.

It is that the surface changes radiation even when temperature is controlled. The more precise lesson from spectral measurements is that this surface effect depends on wavelength.

When comparing thermal radiation, identify the quantity being measured:

• For radiation at one wavelength, compare temperatures and spectral emissivities at that wavelength.
• For emitted power across a wavelength range, account for the area under the spectral emission curve across that range.
• For peak wavelength, use Wien’s law directly only when an ideal black-body model, or an appropriate approximation, is justified.
• For an infrared detector reading, establish whether reflected environmental radiation contributes to the signal.

These distinctions become especially useful when interpreting an apparently strange radiation curve.

Two objects may look completely different while emitting almost identical infrared spectra. Two surfaces at the same temperature may have different spectral peaks. And a reflective surface can produce a stronger infrared detector signal while emitting less radiation of its own.

The next useful step is to separate changes caused by the surface from changes caused by temperature. In How to Compare Black-Body Curves at Different Temperatures, the surface is idealised so that temperature can be isolated. Once that reference behaviour is understood, deviations caused by real materials become much easier to recognise.