Final answer:
The rate of energy radiation of a black body at 273K can be calculated using the Stefan-Boltzmann law, I = σeT^4, where σ is the Stefan-Boltzmann constant and e is the emissivity. For a black body e=1, and at 273K the intensity is found using I=5.67 x 10⁻¸ J/s · m² · K⁴ × (273K)^4.
Step-by-step explanation:
The rate of energy radiation per unit area of a black body at a temperature of 273K is given by the Stefan-Boltzmann law, which states that this rate, also known as intensity, I, is proportional to the fourth power of the absolute temperature (T) of the object.
The equation for this law is I = σeT^4, where σ is the Stefan-Boltzmann constant (σ = 5.67 × 10⁻¸ J/s · m² · K⁴) and e is the emissivity of the body. For a black body, e = 1 which means it is a perfect radiator.
Thus the calculation for the intensity at 273K would be: I = (5.67 × 10⁻¸ J/s · m² · K⁴) × (273K)^4.