Question

A cube of edge length $\mathbf{6}\mathbf{.0}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{\text{m}}$, emissivity $\mathbf{0}\mathbf{.75}$, and temperature$\mathbf{-}\mathbf{100}\mathbf{\text{°C}}$ floats in an environment at $\mathbf{-}\mathbf{150}\mathbf{\text{°C}}$.What is the cube’s net thermal radiation transfer rate?

Short answer

The cube’s net thermal radiation transfer rate is $-6.1\times {10}^{-9}\text{W}$.

Step-by-step solution

Stating thegiven data

1. The length of the cube is $l=6.0\times {10}^{-6}\text{m}$
2. The emissivity of the cube is$\epsilon =0.75$
3. The temperature of the cube is$T=-100\text{°Cor}173\text{K}$
4. The temperature of the environment is $T_{\text{env}}=-150\text{°Cor}123\text{K}$.

Understanding the concept of rate of thermal radiation

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,$P_{\text{rad}}=\sigma \epsilon A{T}^4$…(i)

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

Rate of radiation by the body from its surroundings,$P_{\text{abs}}=\sigma \epsilon A{T}_{\text{env}}^4$…(ii)

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

Calculation of the net radiation rate of the cube

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

$\begin{array}{c}{P}_{\text{net}}=\sigma \epsilon A{T_{\text{env}}}^4-\sigma \epsilon A{T}^4\\ =\sigma \epsilon A({T_{\text{env}}}^4-T^4)\\ =5.6704\times {10}^{-8}\frac{\text{W}}{{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}}\times 0.75\times 6\times {(6.0\times {10}^{-6}\text{m})}^2\times [{(123\text{K})}^4-{(173\text{K})}^4]\\ =-6.1\times {10}^{-9}\text{W}\end{array}$

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