(Solved): is the calculation of emissivity correct? Emissivity: Based on the thermogram shown and the informa ...

Emissivity: Based on the thermogram shown and the information obtained, the emissivity can be calculated on the soda can. For this, we will have to compare the measured temperatures of the material with the known temperature of the object. From the information in the thermogram, we know that the electrical tape has an emissivity of 0.95 . Therefore, the temperature recorded at the known emissivity can be used to obtain the unknown emissivity. From the Stefan-Boltzman law, we know that radiation energy in gray bodies is computed with the equation: $q=\epsilon \sigma A{T}^4$ Where: $\begin{array}{l}q=\text{ heat transfer per unit time }(\text{ Watts })\\ \epsilon =\text{ emissivity coef ficient }\\ \sigma =5.6703\times 10^{-8}{\mathrm{Wm}}^{-2}{\mathrm{K}}^{-4}\text{ Stefan }-\text{ Boltzman Constant }\\ A=\text{ area of emitting body }\left({\mathrm{m}}^2\right)\\ T=\text{ temperature }(K)\end{array}$ Since the area of the can of soda is unknown, we will express the equation in terms of heat flux: $q^{\prime }=\epsilon \sigma T^4$ Where: $\begin{array}{l}{q}^{\prime }=\text{ heat flux }\left(\mathrm{W}/{\mathrm{m}}^2\right)\\ \epsilon =\text{ emissivity coefficient }\\ \sigma =5.6703\times 10^{88}{\mathrm{Wm}}^{-2}{\mathrm{K}}^{-4}\text{ Stefan - Boltzman Constant }\\ T=\text{ temperature }(\mathrm{K})\end{array}$ With the known data of the masking tape we can calculate the heat flux. $\begin{array}{l}{q}^{\prime }=\epsilon \sigma T^4\\ q^{\prime }=0.95\ast 5.6703\times 10^{-8}\ast (17.08+273.15{)}^4\\ q^{\prime }=382.207\mathrm{W}/{\mathrm{m}}^2\end{array}$ Then we can obtain the emissivity of the soda can: $\begin{array}{l}{q}^{\prime }=\epsilon \sigma T^4\\ \epsilon =\frac{q^{\prime }}{{\sigma }^{\ast }{T}^4}\\ \epsilon =\frac{382.207}{5.6703\times 10^{-8}\cdot (24.78+273.15{)}^4}\\ \epsilon =0.8555\end{array}$