Stefan Boltzmann Law - Practice Questions & MCQ

• According to Stefan Boltzmann law, the radiant energy emitted by a perfectly black body per unit area per sec is directly proportional to the fourth power of its absolute temperature,

       or emissive power of the black body is directly proportional to the fourth power of its absolute temperature ($\theta$).

i.e $E \propto \theta^4$

$
\Rightarrow E=\sigma \theta^4
$

where
$\sigma=$ Stefan's constant
and its value is

$
\sigma=5.67 \times 10^{-8} W / m^2 K^4
$

- For ordinary body
1. Emissive power is given by $e=\epsilon E$

So according to Stefan Boltzmann law

$
e=\epsilon E=\epsilon \sigma \theta^4
$

where $\epsilon=$ represents emissivity of the material
2. Radiant energy-

If Q is the total energy radiated by the ordinary body then

$
e=\frac{Q}{A \times t}=\epsilon \sigma \theta^4 \Rightarrow Q=A \epsilon \sigma \theta^4 t
$

3. Radiant power ( P ): It is defined as the energy radiated per unit area.

$
P=\frac{Q}{t}=A \epsilon \sigma \theta^4
$

4. If an ordinary body at temperature $\theta$ is surrounded by a body at a temperature $\theta_0$

                    Then according to Stefan Boltzmann law

                                $e=\epsilon \sigma\left(\theta^4-\theta_0^4\right)$