MHT CET · Physics · Thermal Properties of Matter
At \(273^{\circ} \mathrm{C}\), the emissive power of a perfectly black body is \(R\). It emissive power at \(0^{\circ} \mathrm{Cis}\)
- A \(\frac{R}{4}\)
- B \(\frac{R}{8}\)
- C \(\frac{R}{16}\)
- D \(\frac{R}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{R}{16}\)
Step-by-step Solution
Detailed explanation
From Stefan's law, the total radiant energy emitted per second per unit surface area of a black body is proportional to the fourth power of the absolute temperature \((T)\) of the body \(E=\sigma T^4\)
Given, \(T_1=273^{\circ} \mathrm{C}=273+273 \mathrm{~K}=564 \mathrm{~K}\)
\(T_2=0^0 \mathrm{C}=273 \mathrm{~K}\)
\(\begin{aligned}
& \therefore \frac{E_1}{E_2}=\frac{T_1^4}{T_2^4} \\
& \Rightarrow E_2=\frac{T_2^4}{T_1^4} E_1=\frac{(273)^4}{(564)^4} R=\frac{R}{16}
\end{aligned}\)
Given, \(T_1=273^{\circ} \mathrm{C}=273+273 \mathrm{~K}=564 \mathrm{~K}\)
\(T_2=0^0 \mathrm{C}=273 \mathrm{~K}\)
\(\begin{aligned}
& \therefore \frac{E_1}{E_2}=\frac{T_1^4}{T_2^4} \\
& \Rightarrow E_2=\frac{T_2^4}{T_1^4} E_1=\frac{(273)^4}{(564)^4} R=\frac{R}{16}
\end{aligned}\)
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