Question

"A sphere of radius 0.50 m, temperature 27oC, and emissivity 0.85 is located in an environment of temperature 77oC. What is the net flow of energy transferred to the environment in 1 second?"

Answer 1

Explanation:

It is known that formula for area of a sphere is as follows.

                     A = $4 \pi r^{2}$

                        = $4 \times 3.14 \times (0.50 m)^{2}$

                        = 3.14 $m^{2}$

    $T_{a}$ = (27 + 273.15) K = 300.15 K

          T = (77 + 273.15) K = 350.15 K

Formula to calculate the net charge is as follows.

             Q = $esA(T^{4} - T^{4}_{a})$

where,    e = emissivity = 0.85

               s = stefan-boltzmann constant = $5.6703 \times 10^{-8} Wm^{-2} K^{-4}$

                A = surface area

Hence, putting the given values into the above formula as follows.

                 Q = $esA(T^{4} - T^{4}_{a})$

                     = $0.85 \times 5.6703 \times 10^{-8} Wm^{-2} K^{-4} \times 3.14 \times ((350.15)^{4} - (300.15)^{4})$

                     = 1046.63 W

Therefore, we can conclude that the net flow of energy transferred to the environment in 1 second is 1046.63 W.