Question

The power of a black body at temperature 200K is 544 watt. Its surface area is $(\sigma = 5.67 \times {10^{ - 8}}w{m^2}{K^{ - 4}})$
(A) $6 \times {10^{ - 2}}{m^2}$
(B) $6{m^2}$
(C) $6 \times {10^{ - 6}}{m^2}$
(D) $6 \times {10^2}{m^2}$

Answer 1

Hint
We have to calculate the surface area of the black body, for this we will use Stefan-Boltzmann law formula.
Then we will calculate the surface area, for this we will equate the Stefan-Boltzmann law equation. After that we will put the values of emissivity, absolute temperature, Stefan–Boltzmann constant and the power.
After solving the equation by putting these values we will get the solution.

Complete step by step answer
We know when there comes power, we use Stefan-Boltzmann law which states that the total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature.
So according to Stefan-Boltzmann law the power radiated from a black body can be expressed as
$\Rightarrow P = \varepsilon \sigma A{T^4}$ where P is the power, $\varepsilon$ is the emissivity,
$\Rightarrow \sigma$ is the Stefan–Boltzmann constant, A area of radiating surface and T is absolute temperature of the radiating body.
Now we have to calculate the surface area, for this we will use to determine the area as
$\Rightarrow A = \dfrac{E}{{\sigma {T^4}}}$ . Now we put the values of emissivity, Stefan–Boltzmann constant and absolute temperature of the radiating body to find the area of the radiating surface.
So, $\Rightarrow A = \dfrac{{544}}{{5.67 \times {{10}^{ - 8}} \times ({{200}^4})}}$ .
After calculating the above equation, we get $A = 5.99{\kern 1pt} = 6mete{r^2}$ .
Option (B) is correct.

Note
Note we can take approx. values if no option is matching with the solution we got. Also remember Stefan-Boltzmann law applies only to blackbodies, surfaces that absorb all incident heat radiation.