Temperature alone sets the whole curve

$$\lambda_{\text{max}} T = 2.90\times10^{-3}\ \text{m}\cdot\text{K}.$$

The product is a fixed constant, so peak wavelength and temperature move in opposite directions. Doubling $T$ halves $\lambda_{\text{max}}$: a $3000$ K star peaks near $970$ nm and a $6000$ K star near $480$ nm.

Reading it as a direct proportion predicts a heated metal running from white toward red, and slides the peak of a sketched curve the wrong way when the temperature is raised.

The check is one everybody already owns: a stove element goes red, then orange, then white as it heats. Colour moves toward the short-wavelength end as the temperature climbs, which is exactly what an inverse relation gives.

Working method: write the product as a fixed constant and solve for whichever quantity is unknown.

$$P = \sigma e A T^4.$$

So the radiated power is extraordinarily sensitive to temperature. Tripling the absolute temperature multiplies the power by $3^4 = 81$.

Three failure modes, all mechanical:

1. Scaling linearly: a filament at $3000$ K does not radiate twice what it did at $1500$ K; it radiates $2^4 = 16$ times as much.
2. Using Celsius: a fourth power of a Celsius reading is meaningless. The scale has to be one where zero means zero.
3. Forgetting the area. If the size changes too, fold that in separately, remembering a sphere's area goes as $r^2$ while the temperature goes as $T^4$.

Comparison questions are where this surfaces: two stars of equal size differing by a factor of two in temperature differ by sixteen in power. Take the ratio with the exponent intact rather than computing two absolute values.

A blackbody emits a continuous spectrum whose shape is fixed by temperature alone. Copper and iron at $1200$ K glow the same colour, because the curve does not ask what the sample is.

The name causes the other half of the confusion. "Black" describes what the object absorbs, not what it emits: a blackbody absorbs everything landing on it. And a good absorber is necessarily a good emitter, since at a steady temperature it has to return the energy it takes in. That is precisely why a perfect absorber is the model for thermal emission.

One boundary worth keeping sharp: this is thermal emission, a continuum. A hot low-density gas emits at discrete wavelengths set by its energy levels, which is the previous topic. Same object can do both under different conditions, and the two spectra look nothing alike.

It is worth knowing what this topic is doing in a modern physics unit. Classical physics predicted that a blackbody's emission should rise without limit toward short wavelengths, which is plainly false: a stove element does not emit lethal ultraviolet.

The measured curve instead rises to a peak and falls away. Planck reproduced it by assuming energy is exchanged in discrete quanta of size $hf$, which is where $h$ enters physics for the first time.

So the shape of this curve is the original evidence for quantization, and everything else in this unit follows from the same idea. The two laws above are the measurable summaries of a curve whose very shape required a new physics to explain.