Question

A sphere of radius $\mathbf{0}\mathbf{.500}\mathbf{}\mathbf{\text{m}}$, temperature$\mathbf{27}\mathbf{.0}\mathbf{\text{°C}}$, and emissivity$\mathbf{0}\mathbf{.850}$is located in an environment of temperature$\mathbf{77}\mathbf{.0}\mathbf{\text{°C}}$.

(a) At what rate does the sphere emit and

(b) At what rate does the sphere absorb thermal radiation?

(c)What is the sphere’s net rate of energy exchange?

Short answer

1. The sphere emits at the rate of$1.23\times {10}^3\text{W}$
2. The sphere absorbs thermal radiation at the rate of$2.28\times {10}^3\text{W}$
3. The sphere’s net rate of energy exchange is$1.05\times {10}^3\text{W}$

Step-by-step solution

The given data

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,$\in =0.850\text{m}$
4. Temperature of the environment,$T_{env}={77}^{\text{0}}\text{Cor}350.15\text{K}$

Understanding the concept of thermal radiation

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, $P_{rad}=\sigma ϵA{T}^4$ …(i)

where,$\sigma$is the Stefan–Boltzmann constant and is equal to$(5.67\times {10}^{-8}{\text{\hspace{0.17em}W/m}}^{\text{2}}{\text{K}}^{\text{4}})$

The surface area of the sphere is given: $A=4\pi r^2$ …(ii)

(a) Calculation of the rate of the emitted energy of the sphere

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

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

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

(b) Calculation of the rate of the absorbed energy by the sphere

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

$\begin{array}{c}{P}_{abs}=(5.67\times {10}^{-8}\frac{\text{W}}{{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}})\left(0.850\right)\left(4\pi \right){\left(0.500\text{\hspace{0.17em}m}\right)}^2{\left(350.15\text{\hspace{0.17em}K}\right)}^4\\ =2.28\times {10}^3\text{W}\end{array}$

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

(c) Calculation of the net rate of energy exchange of the sphere

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

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

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