Question

A 1.00 cm2 blackbody surface emits 201 watts of radiant power. What is the temperature of the surface? Give to the nearest whole degree.

Answer 1

Explanation:

Formula for black body radiation is as follows.

                 $\frac{P}{A} = \sigma \times T^{4} J/m^{2}s$

where,       P = power emitted

                 A = surface area of black body

          $\sigma$ = Stephen's constant = $5.6703 \times 10^{-8} watt/m^{2}.K^{-4}$

As area is given as 1.0 $cm^{2}$. Converting it into meters as follows.

           $1.00 cm^{2} \times \frac{(10^{-2})^{2} m^{2}}{1 cm^{2}}$      (as 1 m = 100 cm)

            = $1 \times 10^{-4} m^{2}$

It is given that P = 201 watts. Hence,  

       $\frac{201 watts}{1 \times 10^{-4} m^{2}}$ = $5.6703 \times 10^{-8} watt/m^{2}.K^{-4} \times T^{4}$

                $T^{4}$ = $35.45 \times 10^{12}$

                       T = $(35.45 \times 10^{12})^{1/4}$

                          = 8862.5 K

Thus, we can conclude that the temperature of the surface is 8862.5 K.