Question

The radiant energy emitted by a perfect blackbody at $400K$per unit area per second is:
$\left( \sigma =5.7\times {{10}^{-8}}W{{m}^{-2}}{{K}^{-4}} \right)$
A) $5700J$
B) $1459J$
C) $256J$
D) $1000J$

Answer 1

Hint:
A blackbody is the perfect absorber which absorbs all electromagnetic radiations incident on its surface.
The radiant energy emitted by a perfect blackbody depends on the temperature of the surface of the body.
The radiant energy emitted by a perfect blackbody depends on the surface area of the body.
To find the radiant energy per unit area per second, we use Stefan-Boltzmann’s law.

Step-by-step solution:
A black body is an ideal object that absorbs all the electromagnetic energy it receives, without reflecting or transmitting it. Thus when it is illuminated with electromagnetic radiations it totally absorbs light and should appear black, hence it is called blackbody.
The radiant energy emitted by a perfect blackbody per unit area per second is calculated using Stefan-Boltzmann’s law. According to Stefan-Boltzmann’s law the energy radiated per unit area per second from a perfect blackbody is given as,
$\phi =\sigma {{T}^{4}}$
Where,
$\sigma =$Stefan-Boltzmann’s constant
$T=$The temperature of the surface of the blackbody
It is given that the temperature of the surface of the blackbody is$400K$
Putting the values in the expression for radiant energy per unit area per second, we get
$ \phi =\left( 5.7\times {{10}^{-8}} \right){{\left( 400 \right)}^{4}}W{{m}^{-2}} $
$ =1459.2W{{m}^{-2}} $
As $W$ is the unit of the power, which represents energy per unit second
The unit$W{{m}^{-2}}$can be rewritten as radiant energy per unit area per second.
Therefore, the radiant energy per unit area per second is approximately equal to $1459J$
Hence, option B is the correct option.

Note:
-For the values used in the calculation, we should use only SI unit
-The radiant energy per unit area per second is also known as the radiant energy flux.