A perfect blackbody, maintained at 27 °C, radiates energy at the rate

1.	A perfect blackbody, maintained at 27 °C, radiates energy at the rate of 551.1 W. Find the surface area of the body.
Answer» Data : \(\frac{dQ}{dt}\) = 551.1 W, T = 27 °C = (27 + 273) K = 300 K, σ = 5.67 × 10^-8 W/m² .K⁴ Let A be the surface area of the body. Energy radiated per unit time, \(\frac{dQ}{dt}\) = AσT⁴ \(\therefore\) A = \(\frac{\frac{dQ}{dt}}{σT^4}\) = \(\frac{551.1}{(5.67\times10^{-8})(300)^4}\) = 1.2 m²

A perfect blackbody, maintained at 27 °C, radiates energy at the rate of 551.1 W. Find the surface area of the body.

Answer»

Data : \(\frac{dQ}{dt}\) = 551.1 W,

T = 27 °C = (27 + 273) K = 300 K,

σ = 5.67 × 10^-8 W/m² .K⁴

Let A be the surface area of the body. Energy radiated per unit time,

\(\frac{dQ}{dt}\) = AσT⁴

\(\therefore\) A = \(\frac{\frac{dQ}{dt}}{σT^4}\) = \(\frac{551.1}{(5.67\times10^{-8})(300)^4}\)

= 1.2 m²