How to Compare Black-Body Curves at Different Temperatures

A black body is heated from 2400 K to 4800 K. Its radiation curve moves towards shorter wavelengths, but does that mean it now emits less infrared radiation at the longer wavelengths it has moved away from?

No. For two ideal black bodies plotted on the same calibrated wavelength and spectral-intensity axes, the hotter curve is higher at every wavelength. Its peak moves left, towards shorter wavelengths, and the total area under the curve increases.

That’s the method for comparing black-body curves at different temperatures:

• Peak position: the hotter body has a shorter peak wavelength.
• Height at a fixed wavelength: the hotter body emits more radiation per unit surface area in that wavelength interval.
• Area under the curve: the hotter body emits more total power per unit surface area.

These are three different measurements. Confusing them leads to some surprisingly convincing wrong answers.

01How did physicists measure black-body curves?

In the late nineteenth century, physicists needed to answer a measurement problem: could one mathematical law describe all the radiation emitted by a hot object?

A glowing metal surface was not an ideal starting point. Different materials emit different amounts of radiation, and their surfaces can change as they heat up.

A better approach was to use a heated cavity with a small opening.

The apparatus: from a heated cavity to a detector

Imagine a hollow furnace with an electrically heated wall and a small aperture.

The apparatus works in this order:

1. The furnace walls are brought to a nearly uniform temperature, monitored using a temperature sensor.
2. Radiation inside the cavity undergoes repeated absorption and emission by the walls. At thermal equilibrium, radiation escaping the small opening closely approximates black-body radiation.
3. Radiation passes from the opening through an optical slit into a spectrometer.
4. A prism or diffraction grating separates the radiation by wavelength.
5. A sensitive thermal detector, such as a bolometer, measures the radiation reaching it. A bolometer responds to heating through a change in electrical resistance.
6. By selecting different wavelengths and recording the detector signal, the experimenter constructs a spectrum. The cavity temperature is then changed, and the measurement is repeated.

The optical alignment, aperture, detector response, and background radiation must be controlled. Otherwise, a change in detector reading might be caused by the apparatus rather than the source.

What did the early measurements reveal?

One early proposed description, Wien’s radiation distribution law, agreed well with measurements in parts of the shorter-wavelength spectrum.

If it were universally correct, measurements at longer infrared wavelengths should also follow its predicted intensities.

They did not.

Measurements by Otto Lummer and Ernst Pringsheim around 1899-1900 revealed increasing disagreement at longer wavelengths and higher temperatures.

Heinrich Rubens and Ferdinand Kurlbaum extended the investigation into the far infrared in 1900. Their method used repeated reflections from suitable crystals to isolate narrow bands of long-wavelength radiation. The selected radiation reached a thermal detector, allowing its intensity to be compared at different cavity temperatures.

This was an important revision. Rather than relying on measurements concentrated in a wavelength region where the earlier formula worked reasonably well, the researchers could test its predictions where the disagreement was greatest.

The observed long-wavelength radiation was stronger than the earlier distribution formula predicted. Max Planck developed a new radiation law in 1900 that described the measured spectrum much more successfully.

There are two different laws associated with Wien, and they’re easy to mix up:

• Wien’s earlier radiation distribution formula attempted to describe the entire spectrum and failed in some regions.
• Wien’s displacement law describes the position of the spectrum’s peak and remains valid for an ideal black body.

The experimentally established change in peak position is what makes temperature comparisons possible.

02Read the axes before comparing the curves

A black-body spectrum is usually drawn with wavelength, \(\lambda\), on the horizontal axis and spectral intensity on the vertical axis.

For a precise physical comparison, we’ll use spectral radiant exitance, \(M_\lambda\). This means emitted power per square metre of surface per unit wavelength interval.

For example, its units might be \(\mathrm{W\,m^{-2}\,\mu m^{-1}}\).

The distinction between a curve’s height, peak position, and area is essential.

1. Compare peak wavelengths using Wien’s displacement law

Think of a hot object emitting a mixture of electromagnetic wavelengths.

At lower temperatures, the strongest emission occurs at relatively long wavelengths. Heating the object changes the balance of the mixture, making shorter-wavelength emission more prominent.

The entire spectrum doesn’t simply slide left unchanged. Its shape also grows taller, and the total emitted power increases.

The relationship between peak wavelength and temperature is Wien’s displacement law:

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

where:

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

Rearranging gives:

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

This means peak wavelength is inversely proportional to temperature.

If temperature doubles, peak wavelength halves.

For two black bodies, A and B, an efficient comparison is:

\[ \frac{T_B}{T_A} = \frac{\lambda_{\max,A}} {\lambda_{\max,B}} \]

Notice that the wavelength ratio is reversed.

03Spectral intensity is not total emitted power

Suppose a detector records a spectral intensity of \(200\ \mathrm{W\,m^{-2}\,\mu m^{-1}}\) at a particular wavelength.

That does not mean the surface emits only \(200\ \mathrm{W\,m^{-2}}\) in total.

The reading describes power distributed across wavelength.

If the intensity stayed approximately constant across a narrow interval of \(0.20\,\mu\mathrm{m}\), the power per square metre in that interval would be approximately:

\[ \begin{aligned} M_{\text{band}} &\approx M_\lambda\Delta\lambda\\ &=(200)(0.20)\\ &=40\ \mathrm{W\,m^{-2}} \end{aligned} \]

Here, \(\Delta\lambda\) is the wavelength interval’s width in micrometres.

For a changing curve, a better estimate comes from dividing the spectrum into narrow intervals, multiplying the average spectral emission in each interval by its width, and adding the results.

For the entire spectrum, this becomes the area under the calibrated wavelength curve:

\[ M=\int_0^\infty M_\lambda\,d\lambda \]

Here, \(M\) is the total emitted power per square metre in \(\mathrm{W\,m^{-2}}\). This integral is the mathematical description of adding all the wavelength contributions.

You don’t need to perform this integration analytically to compare areas in an HSC-style graph.

04Three situations where a curve comparison can mislead you

The straightforward rule – hotter means a higher curve at every wavelength – assumes ideal black bodies, the same physical spectral quantity, and a shared calibrated scale.

Real graphs can violate those conditions without violating black-body physics.

Case 1: Each curve has been normalised

Suppose software divides every spectral reading by that curve’s own maximum.

Both curves now have a peak height of 1, even though their original peak intensities were different.

At some wavelengths, the hotter curve may appear below the cooler curve on this normalised graph.

That does not mean the hotter surface emits less radiation there.

It only means that wavelength contributes a smaller fraction of the hotter curve’s own maximum.

Normalisation preserves each curve’s peak wavelength, but it removes the original absolute intensity scale.

Case 2: The emitting areas are different

A black-body curve plotted per square metre describes emission from the surface, not necessarily the total power from an entire object.

A small, extremely hot surface can emit less total power than a much larger, cooler surface.

For a graph of spectral power from the complete objects, the emitting areas affect the vertical scale.

This can even allow total spectral-power curves from differently sized ideal black bodies to cross.

Case 3: The detector changes the spectrum

A spectrometer might be more sensitive to some wavelengths than others.

A filter or window could also transmit a different fraction of radiation at each wavelength.

In that case, the detector’s raw signal is not the same as the source’s emitted spectrum.

A grey filter, which transmits the same fraction at every wavelength, reduces the whole curve’s height without changing its peak wavelength.

A wavelength-dependent filter can change the apparent peak position.

Neither effect should be mistaken for a change in the source’s actual temperature.

Extension: Stefan-Boltzmann law and total emitted power

The Stefan-Boltzmann law gives the total radiation emitted by an ideal black body across all wavelengths:

\[ P=A\sigma T^4 \]

where:

• \(P\) is total emitted power in watts (W),
• \(A\) is emitting surface area in square metres (\(\mathrm{m^2}\)),
• \(T\) is absolute temperature in kelvin (K),
• \(\sigma=5.67\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}\) is the Stefan-Boltzmann constant.

This is an extension beyond the central Wien’s-law comparison in the Physics Stage 6 (2017) course, but it is useful for understanding the area under black-body curves.

For equal emitting areas:

\[ \frac{P_B}{P_A} = \left(\frac{T_B}{T_A}\right)^4 \]

Doubling the temperature multiplies the total emitted power by 16.

For different areas:

\[ \frac{P_B}{P_A} = \frac{A_B}{A_A} \left(\frac{T_B}{T_A}\right)^4 \]

This equation describes total emitted radiation, not necessarily net heat transfer to cooler surroundings. Net radiative heat transfer requires accounting for absorbed radiation as well.

It also assumes ideal black-body emission. Real surfaces can have emissivity below 1, and that emissivity may change with wavelength.

Here’s the distinction that matters in an exam: if the graph’s vertical axis is calibrated spectral emission per square metre, its complete area gives total emitted power per square metre. To find total power from the object, you also need its emitting area.

If the graph is labelled only “relative intensity”, the numerical area generally cannot be turned directly into watts without further calibration.

05Questions and solutions

Question 1

Two ideal black bodies have peak wavelengths of \(2.90\,\mu\mathrm{m}\) and \(1.45\,\mu\mathrm{m}\).

The curves are plotted using the same calibrated spectral emission scale per square metre.

(a) Calculate the temperature of each black body.

(b) State which curve has the greater spectral emission at \(2.0\,\mu\mathrm{m}\).

(c) Explain which curve has the greater total area underneath it.

Use \(b=2.898\times10^{-3}\ \mathrm{m\,K}\).

Solution 1

Black body A is approximately 1000 K, black body B is approximately 2000 K, and B has both greater spectral emission at \(2.0\,\mu\mathrm{m}\) and a greater area under its calibrated curve.

(a) Calculate the temperatures.

Using Wien’s displacement law:

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

For A:

\[ \begin{aligned} T_A &=\frac{2.898\times10^{-3}} {2.90\times10^{-6}}\\ &\approx999\ \mathrm{K} \end{aligned} \]

For B:

\[ \begin{aligned} T_B &=\frac{2.898\times10^{-3}} {1.45\times10^{-6}}\\ &\approx1999\ \mathrm{K} \end{aligned} \]

To suitable precision, these are 1000 K and 2000 K.

(b) Compare the spectral emission.

B has the greater spectral emission at \(2.0\,\mu\mathrm{m}\).

The reason is not simply that its peak is closer to that wavelength. For ideal black bodies, increasing temperature increases spectral emission at every fixed wavelength.

(c) Compare the areas.

B has the greater area because it emits more total radiation per square metre.

The tempting mistake is to compare only the peaks and imagine that the hotter body’s curve has moved left without changing its total area.

Heating changes the peak position, the spectral heights, and the integrated emission.

Question 2

A spectrometer measures two ideal black bodies at 2400 K and 3600 K.

The data are then normalised so that each curve’s own maximum has a value of 1.00.

At \(1.21\,\mu\mathrm{m}\), the normalised readings are:

An independent calibration establishes that the actual spectral peak height at 3600 K is approximately 7.6 times the peak height at 2400 K.

A student concludes that heating the source reduced its radiation at \(1.21\,\mu\mathrm{m}\) by 29%.

(a) Determine whether this conclusion is correct and calculate the actual spectral-intensity ratio at that wavelength.

(b) Using the Stefan-Boltzmann law as an extension, calculate the ratio of the total emitted powers per square metre.

(c) Calculate the ratio of the areas under the two normalised wavelength curves. Explain why this result can differ from the ratio in part (b).

Assume the complete, calibrated wavelength spectra are ideal black-body spectra.

Solution 2

The student’s conclusion is incorrect. At \(1.21\,\mu\mathrm{m}\), the hotter body actually emits approximately 5.4 times as much radiation per square metre per unit wavelength. It also emits about 5.06 times as much total power per square metre, even though the area under its normalised curve is smaller.

(a) Recover the actual spectral-intensity ratio.

Normalisation divides a spectral reading by its own curve’s peak.

To reverse that operation, multiply the normalised value by the original peak height.

The ratio is:

\[ \begin{aligned} \frac{M_{\lambda,3600}} {M_{\lambda,2400}} &=\frac{0.71(7.6)}{1.00}\\ &\approx5.40 \end{aligned} \]

The ratio is dimensionless because the spectral-intensity units cancel.

So the 3600 K source emits about 5.4 times as strongly at the specified wavelength, not 29% less.

The misleading comparison was between fractions of two different maximum values.

(b) Compare total emitted power per square metre.

Using Stefan-Boltzmann scaling:

\[ \begin{aligned} \frac{M_{3600}}{M_{2400}} &=\left(\frac{3600}{2400}\right)^4\\ &=1.5^4\\ &=5.0625 \end{aligned} \]

Therefore, total emitted power per square metre increases by a factor of approximately 5.06.

(c) Compare the normalised areas.

The area under a normalised spectrum equals the area under the original spectral curve divided by that curve’s peak height.

Therefore, the ratio of normalised areas is approximately:

\[ \begin{aligned} R_{\text{norm}} &=\frac{5.0625}{7.6}\\ &\approx0.666 \end{aligned} \]

The hotter curve’s normalised area is only about two-thirds of the cooler curve’s normalised area.

This result looks strange until you recognise what normalisation removes.

The hotter body’s total emission has increased by a factor of about 5.06, but its peak spectral height has increased by a larger factor, about 7.6.

Dividing by that larger peak produces a narrower-looking, lower-area normalised curve.

The conclusion is that normalised graph area and actual emitted power are different quantities. You cannot infer one from the other without the original scale.

Question 3

Two isolated ideal black-body radiators have different temperatures and emitting areas.

A calibrated system measures the total emitted spectral power from each radiator, with geometry and collection losses already corrected.

Independent spectral calculations provide the following ratios of emission per square metre:

(a) Calculate each radiator’s peak wavelength.

(b) Using the Stefan-Boltzmann law, determine which complete radiator emits more power across all wavelengths.

(c) Determine which radiator produces greater total spectral power at each wavelength shown in the table.

(d) Explain why the total spectral-power curves can cross even though the per-square-metre black-body curves cannot.

Assume both radiators emit as ideal black bodies over their stated areas.

Solution 3

The cooler, larger radiator emits 25% more total power than the hotter one, even though radiator H emits more per square metre at every wavelength. H dominates the complete-object spectral power at \(0.50\,\mu\mathrm{m}\), while C dominates at \(5.0\,\mu\mathrm{m}\).

(a) Calculate the peak wavelengths.

For H:

\[ \begin{aligned} \lambda_{\max,H} &=\frac{2.898\times10^{-3}}{4000}\\ &=7.245\times10^{-7}\ \mathrm{m}\\ &\approx0.725\,\mu\mathrm{m} \end{aligned} \]

For C:

\[ \begin{aligned} \lambda_{\max,C} &=\frac{2.898\times10^{-3}}{2000}\\ &=1.449\times10^{-6}\ \mathrm{m}\\ &\approx1.45\,\mu\mathrm{m} \end{aligned} \]

(b) Compare total powers.

First convert the areas:

\[ \begin{aligned} A_H&=1.0\times10^{-5}\ \mathrm{m^2}\\ A_C&=2.0\times10^{-4}\ \mathrm{m^2} \end{aligned} \]

Using Stefan-Boltzmann:

\[ \frac{P_H}{P_C} = \frac{A_H}{A_C} \left(\frac{T_H}{T_C}\right)^4 \]

Substituting:

\[ \begin{aligned} \frac{P_H}{P_C} &=\frac{0.10}{2.0} \left(\frac{4000}{2000}\right)^4\\ &=0.05(16)\\ &=0.80 \end{aligned} \]

The hotter radiator emits only 80% of the cooler radiator’s total power.

For an absolute check:

\[ \begin{aligned} P_H&=A_H\sigma T_H^4\\ &\approx145\ \mathrm{W}\\ P_C&=A_C\sigma T_C^4\\ &\approx181\ \mathrm{W} \end{aligned} \]

The cooler radiator therefore emits \(181/145\approx1.25\) times as much total power.

(c) Compare spectral power at each wavelength.

Let \(S_\lambda\) represent spectral power from the whole radiator, in watts per unit wavelength.

Since \(S_\lambda=A M_\lambda\), the ratio of complete-object spectral powers is the per-area spectral ratio multiplied by the area ratio.

At \(0.50\,\mu\mathrm{m}\):

\[ \begin{aligned} \frac{S_{\lambda,H}}{S_{\lambda,C}} &=0.05(1.3\times10^3)\\ &=65 \end{aligned} \]

H produces approximately 65 times as much total spectral power at that wavelength.

At \(5.0\,\mu\mathrm{m}\):

\[ \begin{aligned} \frac{S_{\lambda,H}}{S_{\lambda,C}} &=0.05(3.1)\\ &=0.155 \end{aligned} \]

H produces only 15.5% as much total spectral power as C at this longer wavelength.

(d) Explain the crossing.

Per square metre, H emits more at every wavelength because its temperature is higher.

But for the complete objects, H’s much smaller emitting area reduces its total spectral power.

The temperature advantage is extremely large at short wavelengths. At longer wavelengths, that advantage is smaller, so C’s much larger area becomes decisive.

The complete-object spectral curves therefore reverse their ordering between the two specified wavelengths and must cross somewhere between them, assuming the spectra vary continuously.

The tempting argument is that black-body curves never cross, so this result must be impossible.

That rule only applies to the same spectral emission quantity per unit area on a common scale. Multiplying by very different emitting areas changes the comparison.

Question 4

An ideal black-body cavity remains at a constant temperature of 3000 K, independently confirmed by a temperature sensor.

A spectrometer records its radiation using four narrow wavelength bands. A wavelength-selective window is then inserted between the cavity and the spectrometer.

The following readings are illustrative. The unfiltered values are on one consistent relative spectral-signal scale.

Assume the window transmits the stated fraction of the radiation at each sampled wavelength and does not otherwise affect the source.

(a) Calculate the filtered detector signal at each wavelength.

(b) Identify the wavelength with the greatest sampled reading before and after the window is inserted.

(c) A student uses the largest filtered reading to calculate a temperature from Wien’s law. Calculate that apparent temperature and explain why it is unreliable.

(d) Calculate the cavity’s actual ideal black-body peak wavelength. Explain why even the largest unfiltered sampled reading need not occur at the exact peak.

(e) Explain why multiplying every filtered reading by one constant factor cannot generally restore the unfiltered spectrum.

Solution 4

The cavity remains at 3000 K. The window changes the measured spectral shape, making \(0.70\,\mu\mathrm{m}\) the strongest sampled band even though the actual black-body peak remains near \(0.966\,\mu\mathrm{m}\). Applying Wien’s law directly to the filtered reading would give a false estimate of about 4140 K.

(a) Calculate the filtered readings.

The measured signal is the incoming signal multiplied by the window transmission:

\[ S_{\text{filtered}}=\tau S_{\text{original}} \]

Here, \(\tau\) is the dimensionless transmitted fraction.

The results are:

(b) Compare the largest sampled readings.

Before the window, the strongest sampled signal is 99 at \(0.90\,\mu\mathrm{m}\).

After the window, the strongest sampled signal is 75 at \(0.70\,\mu\mathrm{m}\).

The apparent peak among these four samples has moved to a shorter wavelength without any change in the cavity’s temperature.

(c) Calculate the misleading apparent temperature.

If a student incorrectly assumes that \(0.70\,\mu\mathrm{m}\) is the black-body peak:

\[ \begin{aligned} T_{\text{apparent}} &=\frac{2.898\times10^{-3}} {0.70\times10^{-6}}\\ &\approx4140\ \mathrm{K} \end{aligned} \]

This disagrees with the independently measured temperature of 3000 K.

The problem is that the filtered detector signal is not an undistorted black-body spectrum.

Wien’s displacement law applies to the actual black-body spectral distribution, not automatically to the maximum of an instrument’s wavelength-dependent response.

(d) Determine the actual peak wavelength.

Using the independently known source temperature:

\[ \begin{aligned} \lambda_{\max} &=\frac{2.898\times10^{-3}}{3000}\\ &=9.66\times10^{-7}\ \mathrm{m}\\ &=0.966\,\mu\mathrm{m} \end{aligned} \]

The true peak lies between the sampled wavelengths of \(0.90\) and \(1.10\,\mu\mathrm{m}\).

The highest sampled reading does not have to coincide exactly with the continuous curve’s maximum. Four wavelength samples are too widely spaced to determine that maximum precisely.

A finely sampled, properly calibrated spectrum would give a better experimental estimate.

(e) Explain the required correction.

A single multiplier cannot correct the whole spectrum because the window transmits different fractions at different wavelengths.

The correction must be made separately at each wavelength:

\[ S_{\text{original}} = \frac{S_{\text{filtered}}}{\tau} \]

Dividing the four filtered readings by their respective transmission fractions recovers 75, 99, 96, and 67.

If the window were truly grey, meaning \(\tau\) was constant across all wavelengths, it would reduce the curve uniformly without moving the true spectral peak.

The key experimental constraint is wavelength-dependent transmission, not merely reduced detector brightness.

Question 5

Two optically isolated ideal black-body emitters are measured separately. Their complete, calibrated spectral powers are then added to produce a combined wavelength spectrum.

The emitters have these properties:

Assume both are ideal black bodies, their temperatures are uniform, their emissions do not affect each other, and all spectra are expressed as total emitted power per unit wavelength, not power per square metre.

For this extension challenge, you are given two scaling relationships for ideal black bodies on a wavelength-based spectrum:

• The peak spectral emission height per square metre is proportional to \(T^5\).
• The total emitted power per square metre is proportional to \(T^4\).

You may also use Wien’s displacement law.

A finely sampled combined spectrum contains two distinct humps, one at shorter wavelengths and one at longer wavelengths.

(a) Calculate each emitter’s peak wavelength in isolation.

(b) Determine which emitter has the taller individual peak after its emitting area is included, and by what factor.

(c) Determine which emitter contributes more total power integrated across all wavelengths.

(d) A student notices that the combined spectrum’s tallest hump is near the hot emitter’s peak and concludes that almost all the emitted power must come from H. Explain why this conclusion fails.

(e) The maxima of the two humps in the combined spectrum are not exactly at the isolated emitters’ peak wavelengths. Without using Planck’s full equation, predict which way each hump’s maximum shifts. Use the rising and falling portions of the component curves to justify your answer.

(f) If H is completely blocked, what happens to the spectral power at every wavelength, and which way does the remaining long-wavelength hump’s maximum move?

Solution 5

The hot emitter has a spectral peak twice as tall as the cool emitter’s individual peak, but it contributes only half as much total power. The combined spectrum is not itself the spectrum of one uniform-temperature black body, and its two local maxima shift slightly towards each other.

(a) Find the individual peak wavelengths.

For C:

\[ \begin{aligned} \lambda_{\max,C} &=\frac{2.898\times10^{-3}}{1000}\\ &=2.898\,\mu\mathrm{m} \end{aligned} \]

For H:

\[ \begin{aligned} \lambda_{\max,H} &=\frac{2.898\times10^{-3}}{4000}\\ &=0.7245\,\mu\mathrm{m} \end{aligned} \]

The hotter emitter peaks at one-quarter of the cooler emitter’s wavelength.

(b) Compare the individual peak heights after allowing for area.

The total spectral power from each surface is its spectral emission per square metre multiplied by its emitting area.

Using the supplied \(T^5\) scaling for peak spectral emission:

\[ \frac{S_{\max,H}}{S_{\max,C}} = \frac{A_H}{A_C} \left(\frac{T_H}{T_C}\right)^5 \]

Substituting:

\[ \begin{aligned} \frac{S_{\max,H}}{S_{\max,C}} &=\frac{1}{512}(4)^5\\ &=\frac{1024}{512}\\ &=2 \end{aligned} \]

H has twice the peak spectral-power height of C, despite having only \(1/512\) of its emitting area.

(c) Compare total integrated power.

Now use the different temperature dependence for total emitted power:

\[ \frac{P_H}{P_C} = \frac{A_H}{A_C} \left(\frac{T_H}{T_C}\right)^4 \]

Therefore:

\[ \begin{aligned} \frac{P_H}{P_C} &=\frac{1}{512}(4)^4\\ &=\frac{256}{512}\\ &=0.50 \end{aligned} \]

The hotter emitter contributes only half as much total power as the cooler emitter.

For an absolute check, convert the areas:

\[ \begin{aligned} A_H&=1.00\times10^{-4}\ \mathrm{m^2}\\ A_C&=5.12\times10^{-2}\ \mathrm{m^2} \end{aligned} \]

Using \(P=A\sigma T^4\):

\[ \begin{aligned} P_H&\approx1.45\times10^3\ \mathrm{W}\\ P_C&\approx2.90\times10^3\ \mathrm{W} \end{aligned} \]

So C contributes about twice as much power as H.

(d) Explain the apparently contradictory results.

The student’s conclusion fails because the tallest point of a spectrum is not necessarily associated with the greatest integrated power.

Imagine comparing two hills. One is narrow and steep, while the other is broad and lower. The tallest point belongs to the steep hill, but the broad hill can cover more ground.

Here, hill height represents spectral power per unit wavelength, and hill area represents total emitted power. The analogy breaks because a radiation spectrum is not a physical hill; the mathematical relationship is between a density and its integral.

The hot emitter’s radiation is concentrated into a much shorter wavelength scale. Its spectral peak is therefore exceptionally high.

The cooler emitter’s spectrum is spread over a broader wavelength range and comes from a much larger surface. Its integrated power is greater.

The combined spectrum also cannot be assigned one genuine black-body temperature using its tallest peak alone.

Wien’s law connects peak wavelength to temperature for a single ideal black-body spectrum. A sum of spectra from two different temperatures does not generally have that single-temperature shape.

(e) Predict the shifts of the combined humps.

This is the more subtle part.

The combined spectral power is:

\[ S_\lambda = A_C M_{\lambda,C} + A_H M_{\lambda,H} \]

Each term is positive, and each component has a smoothly rising side before its peak and a falling side after its peak.

Consider the hot emitter’s isolated peak at \(0.7245\,\mu\mathrm{m}\).

At that wavelength:

• H is exactly at its own maximum, so its curve is locally flat.
• C is still on the rising side of its spectrum, because C peaks much later, at \(2.898\,\mu\mathrm{m}\).

Adding C’s still-rising contribution means that the combined curve is still increasing at H’s isolated peak.

Therefore, the shorter-wavelength hump’s maximum occurs slightly to the right of H’s isolated peak, at a longer wavelength.

Now consider C’s isolated peak at \(2.898\,\mu\mathrm{m}\).

At that wavelength:

• C is locally flat at its maximum.
• H is already on the falling side of its spectrum.

Adding H’s decreasing contribution means that the combined curve is already falling at C’s isolated peak.

Therefore, the longer-wavelength hump’s maximum occurs to the left of C’s isolated peak, at a shorter wavelength.

The two maxima move towards each other.

The surprising point is that adding a second source doesn’t merely raise the height of the existing peaks. Because the added spectrum is rising or falling at those positions, it can also move the maxima.

(f) Predict the effect of blocking H.

Blocking H decreases the measured spectral power at every wavelength, because a positive contribution has been removed.

The remaining spectrum is simply C’s original black-body spectrum.

Its peak is therefore at \(2.898\,\mu\mathrm{m}\), so the long-wavelength maximum moves right, back towards the cooler emitter’s isolated peak.

Removing radiation can shift a spectral maximum towards a longer wavelength without increasing the spectral power anywhere.

The exact constraint is that a peak’s position depends on the shape of the complete spectrum, not just the temperature of whichever source contributes most strongly near that peak.

This is why an experimentalist must establish that a measured spectrum represents one thermal source before using its peak to infer a temperature. Establishing that the radiation really follows a black-body distribution leads to the next question: why does that distribution have its particular shape? That is the problem developed in Black-body radiation and the ultraviolet catastrophe explained, where Planck’s quantum model provides the deeper explanation.