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

Step-by-step explanation:

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

A =
`4 \pi r^(2)`

=
`4 * 3.14 * (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 * 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 * 5.6703 * 10^(-8) Wm^(-2) K^(-4) * 3.14 * ((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.