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Kinetic Theory of Gases and Radiation — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsKinetic Theory of Gases and Radiation3 Marks
Question
A body of surface area $400 \mathrm{~cm}^2$ and absorption coefficient 0.5 radiates energy $1.5 \mathrm{kcal}$ in 2 minutes when the temperature of the body is kept constant. Find the temperature of the body. (Given : J = $4186 \mathrm{~J} / \mathrm{kcal}, \sigma=5.67 \times 10^{-8} \mathrm{~J} / \mathrm{s} \cdot \mathrm{m} \cdot \mathrm{K}^4$ )
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Answer


$
\begin{aligned}
& \text { Data }: \mathrm{A}=400 \mathrm{~cm}^2=400 \times 10^{-4} 4 \mathrm{~m}^2 \\
& =4 \times 10^{-2} \mathrm{~m}^2 \text {, absorption coefficient, a }=0.8 \\
& \text { But } \mathrm{a}=\mathrm{e} \therefore \mathrm{e}=0.8, \mathrm{~J}=4186 \mathrm{~J} / \mathrm{kcal} \\
& \mathrm{Q}=1.5 \mathrm{kcal}=1.5 \times 4186 \mathrm{~J}=6279 \mathrm{~J} \\
& \mathrm{t}=2 \text { minutes }=120 \mathrm{~s}, \sigma=5.67 \times 10^{-8} \mathrm{~J} / \mathrm{s} . \mathrm{m} . \mathrm{K}^4 \\
& \text { Energy radiated, } \mathrm{Q}=\sigma \mathrm{AeT}^4 \mathrm{t} \\
& \therefore 6279=\left(5.67 \times 10^{-8}\right) \times\left(4 \times 10^{-2}\right) \times 0.8 \times \mathrm{T}^4 \times 120 \\
& \therefore \mathrm{T}^4=\frac{6279 \times 10^8}{21.77}=288.4 \times 10^8 \\
& \therefore \mathrm{T}=4.121 \times 10^2 \mathrm{~K}
\end{aligned}
$
This is the temperature of the body.

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