$5.11.3.$ Determine the energy density of thermal radiation in the cavity of a body with temperature $T$. The speed of light is $c = 3 \cdot 10^8 \text{ m/s}$.
Solution
The energy density of thermal radiation (equilibrium black-body radiation) in a cavity is related to the radiant emittance by the formula: $$u = \frac{4\sigma}{c} T^4$$ where: * $\sigma = 5.67 \cdot 10^{-8} \text{ W/(m}^2 \cdot \text{K}^4\text{)}$ is the Stefan-Boltzmann constant, * $c = 3 \cdot 10^8 \text{ m/s}$ is the speed of light, * $T$ is the absolute temperature.
Let us substitute the numerical values of the constants to obtain the final calculation formula: $$u = \frac{4 \cdot 5.67 \cdot 10^{-8}}{3 \cdot 10^8} T^4 = 7.56 \cdot 10^{-16} T^4 \text{ J/m}^3$$ *(Note: the numerical coefficient $7.56 \cdot 10^{-16}$ itself has units of $\text{J}/(\text{m}^3 \cdot \text{K}^4)$).*