91Ó°ÊÓ

1. Radius of the sphere$r=0.500\text{m}$
2. Temperature of the sphere$T={27}^{\text{0}}\text{Cor}300.15\text{K}$
3. Emissivity,$∈=0.850\text{m}$
4. Temperature of the environment,$T_{env}={77}^{\text{0}}\text{Cor}350.15\text{K}$

Thermal radiation is the process of transferring heat through electromagnetic radiation that is produced by the thermal motion of matter particles. By using the concept of radiation, we can calculate the rate of emission and absorption via thermal radiation.

Formulae:

The rate at which the sphere emits thermal radiation, $P_{rad}=\mathrm{σ}\mathrm{Ï\mu }A{T}^4$ …(¾±)

where,$\mathrm{σ}$is the Stefan–Boltzmann constant and is equal to$(5.67×{10}^{−8}{\text{ W/³¾}}^{\text{2}}{\text{K}}^{\text{4}})$

The surface area of the sphere is given: $A=4\mathrm{Ï€}{r}^2$ …(¾±¾±)

The rate at which the sphere emits energy via thermal radiation using equation (ii) in equation (i) is given as:

$\begin{array}{c}{P}_{rad}=(5.67×{10}^{−8}\frac{\text{W}}{{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}})(0.850)\left(4\mathrm{π}\right){\left(0.500\text{m}\right)}^2{\left(300.15\text{K}\right)}^4\\ =1.23×{10}^3\text{W}\end{array}$

Hence, the rate at which the sphere emits is$1.23×{10}^3\text{W}$

The rate at which the sphere absorbs thermal radiation using equation (ii) in equation (i) is given as:

$\begin{array}{c}{P}_{abs}=(5.67×{10}^{−8}\frac{\text{W}}{{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}})\left(0.850\right)\left(4\mathrm{Ï€}\right){\left(0.500\text{″¾}\right)}^2{\left(350.15\text{‿é}\right)}^4\\ =2.28×{10}^3\text{W}\end{array}$

Hence, the rate of energy absorbed by the sphere is$2.28×{10}^3\text{W}$

Since an object both emits and absorbs thermal radiation, its net rate of energy exchange$P_{net}$is given by

$\begin{array}{c}{P}_{net}=P_{abs}−P_{rad}\\ =2.28×{10}^3\text{W}−1.23×{10}^3\text{W}\\ =1.05×{10}^3\text{W}\end{array}$

Hence, the net rate of energy exchange is $1.05×{10}^3\text{W}$