Question

A solid cube of edge length r, a solid sphere of radius r, and a solid hemisphere of radius r, all made of the same material, are maintained at temperature 300 K in an environment at temperature 350 K. Rank the objects according to the net rate at which thermal radiation is exchanged with the environment, greatest first.

Short answer

The ranking of the given objects according to the net rate at which thermal radiation is exchanged with the environment is $\mathrm{sphere}>\mathrm{solid}\mathrm{hemisphere}>\mathrm{cube}$.

Step-by-step solution

The given data

1. The edge length of the cube is r
2. The radius of the solid sphere is r
3. The radius of the solid hemisphere is r
4. The temperature of area A is$\mathrm{T}=300\mathrm{K}$for all objects
5. The temperature of the environment is$350\mathrm{K}$

Understanding the concept of radiation rate

The emission or transmission of energy as waves or particles across space or a material medium is known as radiation. By using the formula for the net rate of radiation at which thermal radiation is exchanged with the environment, we can find the ranks of the given objects according to the net rate at which thermal radiation is exchanged with the environment.

Formula:

The net rate at which thermal radiation is exchanged with the environment,

${\mathrm{P}}_{\mathrm{rad}}={\mathrm{\sigma \epsilon AT}}^4$ …(i)

where, $\mathrm{\sigma }=5.6703\times {10}^{-8}\mathrm{W}/{\mathrm{m}}^2{\mathrm{K}}^4$is the Stefan-Boltzmann constant, $\mathrm{\epsilon }$is the emissivity of the object whose values is in between 0 and 1, A is the surface area of the object, and T is the temperature of that area in Kelvin.

Calculation of the rank of the objects according to rate of radiation

If the material isthe same, then$\mathrm{\sigma }$and$\mathrm{\epsilon }$isthe same for all objects, and temperature of area A is$\mathrm{T}=300\mathrm{K}$for all objects.

Thus, the net rate${\mathrm{P}}_{\mathrm{rad},\mathrm{c}}$ at which thermal radiation is exchanged with the environment of cube is given by:${\mathrm{P}}_{\mathrm{rad},\mathrm{c}}=6{\mathrm{\sigma \epsilon r}}^2{\mathrm{T}}^4$

Where,${\mathrm{A}}_{\mathrm{C}}$is the surface area of the cube and${\mathrm{A}}_{\mathrm{c}}=6{\mathrm{r}}^2$

Similarly, the net rate${\mathrm{P}}_{\mathrm{rad},\mathrm{s}}$ at which thermal radiation is exchanged with the environment of solid sphere is given by:

$\begin{array}{rcl}{\mathrm{P}}_{\mathrm{rad},\mathrm{s}}& =& 4{\mathrm{\pi \sigma \epsilon r}}^2{\mathrm{T}}^4\\ & =& \left(12.57\right){\mathrm{\sigma \epsilon r}}^2{\mathrm{T}}^4\end{array}$

Where, A_s is the surface area of the solid sphere and${\mathrm{A}}_{\mathrm{s}}=4{\mathrm{\pi r}}^2$.

And the net rate${\mathrm{P}}_{\mathrm{rad},\mathrm{h}}$ at which thermal radiation is exchanged with the environment of solid hemisphere is given by:

$\begin{array}{rcl}{\mathrm{P}}_{\mathrm{rad},\mathrm{h}}& =& 3{\mathrm{\pi \sigma \epsilon r}}^2{\mathrm{T}}^4\\ & =& \left(9.42\right){\mathrm{\sigma \epsilon r}}^2{\mathrm{T}}^4\end{array}$

Thus, we get the rate of radiation as follows:${\mathrm{P}}_{\mathrm{rad},\mathrm{s}}>{\mathrm{P}}_{\mathrm{rad},\mathrm{h}}>{\mathrm{P}}_{\mathrm{rad},\mathrm{c}}$

Therefore, the ranking of the objects according to the net rate at which thermal radiation is exchanged with the environment is $\mathrm{solid}\mathrm{sphere}>\mathrm{solid}\mathrm{hemisphere}>\mathrm{cube}$.