Question
A wall surface on a house is $30^{\circ} \mathrm{C}$ with an emissivity of $\varepsilon=0.7 .$ The surrounding ambient air is at $15^{\circ} \mathrm{C}$ with an average emissivity of $0.9 .$ Find the rate of radiation energy from each of those surfaces per unit area.
Step 1
The formula to convert Celsius to Kelvin is $K = C + 273.15$. So, the temperature of the wall $T_1$ in Kelvin is $30^{\circ}C + 273.15 = 303.15K$ and the temperature of the air $T_2$ in Kelvin is $15^{\circ}C + 273.15 = 288.15K$. Show more…
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