Absolute temperature of the blackbody — the only quantity that sets the color.
While playing, the temperature sweeps automatically between 1000K and 12000K. Moving the slider pauses it.
$$B_\lambda(\lambda,T)=\frac{2hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda k T}-1}$$
Planck’s law: the spectral radiance of a blackbody at wavelength $\lambda$ and temperature $T$. $h$ Planck constant, $c$ speed of light, $k$ Boltzmann constant.
$$X=\int B_\lambda\,\bar x(\lambda)\,d\lambda,\quad Y=\int B_\lambda\,\bar y\,d\lambda,\quad Z=\int B_\lambda\,\bar z\,d\lambda$$
Integrate the spectrum against the CIE color-matching functions $\bar x,\bar y,\bar z$ to obtain the XYZ tristimulus values. Chromaticity is $x=X/(X{+}Y{+}Z)$, $y=Y/(X{+}Y{+}Z)$. An XYZ→sRGB transform yields the on-screen color.
$$\lambda_{max}=\frac{b}{T},\quad b=2.898\times10^{-3}\,\mathrm{m\cdot K}$$
Wien’s displacement law: the higher the temperature, the shorter (bluer) the peak wavelength. Joining these points over temperature traces the Planckian locus.
