Answer to Question #129792 in Physics for Dr. Horus

Question

Calculate the rate at which a blackbody at temperature 300°C emits radiant energy. The blackbody is a sphere of diameter 20cm.

Answer 1

According to the Stefan–Boltzmann law, the total energy radiated per unit surface area of a black body across all wavelengths per unit time is:

$$j = \sigma T^4$$

where $\sigma = 5.67\times 10^{-8} Wm^{-2}K^{-4}$ is the Stefan–Boltzmann constant and $T = 300\degree C = 573K$ is the a blackbody temperature.

In order to obtain the radiation rate one should multiply $j$ by the area of blackbody surface:

$$P = A\cdot j = \pi d^2j$$

where $d = 20cm = 0.2m$ is the diameter of the sphere. Thus, obtain:

$$P = \pi d^2j = \pi d^2\sigma T^4 = \pi\cdot 0.2^2\cdot 5.67\times 10^{-8}\cdot 573^4 \approx 768.1\space W$$

Answer. 768.1 W.