91Ó°ÊÓ

1. The length of the cube is $l=6.0×{10}^{−6}\text{m}$
2. The emissivity of the cube is$\mathrm{Î\mu }=0.75$
3. The temperature of the cube is$T=−100\text{°°äor}173\text{K}$
4. The temperature of the environment is $T_{\text{env}}=−150\text{°°äor}123\text{K}$.

Thermal radiation is the process of transferring heat through electromagnetic radiation that is produced by the thermal motion of matter particles. We can use the concept of thermal radiation. to find the net rate of thermal radiation. Using the Stefan-Boltzmann equation for the rate of radiation via thermal energy, we can get the required amount.

Formulae:

Rate of radiation by the body,$P_{\text{rad}}=\mathrm{σ}\mathrm{Î\mu }A{T}^4$…(¾±)

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

Rate of radiation by the body from its surroundings,$P_{\text{abs}}=\mathrm{σ}\mathrm{Î\mu }A{T}_{\text{env}}^4$…(¾±¾±)

Net rate of power radiation by the body,$P_{\text{net}}=P_{\text{abs}}−P_{\text{rad}}$ .…(¾±¾±¾±)

The net$P_{\text{net}}$ of energy exchange due to thermal radiation using equations (i), (ii) and (iii) is given by

$\begin{array}{c}{P}_{\text{net}}=\mathrm{σ}\mathrm{Î\mu }A{T_{\text{env}}}^4−\mathrm{σ}\mathrm{Î\mu }A{T}^4\\ =\mathrm{σ}\mathrm{Î\mu }A({T_{\text{env}}}^4−T^4)\\ =5.6704×{10}^{−8}\frac{\text{W}}{{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}}×0.75×6×{(6.0×{10}^{−6}\text{m})}^2×[{(123\text{K})}^4−{(173\text{K})}^4]\\ =−6.1×{10}^{−9}\text{W}\end{array}$

The negative sign indicates that energy is lost during radiation. Hence, the net thermal radiation rate is $−6.1×{10}^{−9}\text{W}$.