The surface area of a black body is $5 imes 10^{- 4}\ m^{2}$ and its temperature is 727$℃$ . the energy radiated by it per minute is ($\sigma = 5.67 imes 10^{- 8}\ Jm^{- 2} - s^{- 1} - K^{- 4})$

Question

The surface area of a black body is $5 	imes 10^{- 4}\ m^{2}$ and its temperature is 727$℃$ . the energy radiated by it per minute is ($\sigma = 5.67 	imes 10^{- 8}\ Jm^{- 2} - s^{- 1} - K^{- 4})$

Dr. Abubacker · Accepted Answer

(a) From Stefan’s law
$E = \sigma T^{4}A$
Given, $T = 727℃ = (727 + 273) = 1000$ K
$\ \ \ \ \ \ \ \ \ A = 5 	imes 10^{- 4}\ m^{2}$
$	herefore\ \ \ \ Energy = \left( 5.67 	imes 10^{- 8} ight)(1000)^{4}\left( 5 	imes 10^{- 4} ight)60$
$	herefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = 1.7 	imes 10^{3}$ J

Q. The surface area of a black body is $5 \times 10^{- 4}\ m^{2}$ and its temperature is 727$℃$ . the energy radiated by it per minute is ($\sigma = 5.67 \times 10^{- 8}\ Jm^{- 2} - s^{- 1} - K^{- 4})$

Explanation

Reviewed by Dr. Abubacker

Want more practice?