AP EAMCET · PHYSICS · Thermal Properties of Matter
The rate of radiation of a black body at \(0^{\circ} \mathrm{C}\) is \(E \mathrm{Js}^{-1}\). The rate of radiation of the black body at \(273^{\circ} \mathrm{C}\) will be
- A \(E \mathrm{Js}^{-1}\)
- B \(4 E \mathrm{Js}^{-1}\)
- C \(\frac{E}{2} \mathrm{Js}^{-1}\)
- D \(16 E \mathrm{Js}^{-1}\)
Answer & Solution
Correct Answer
(D) \(16 E \mathrm{Js}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, \(T_1=0^{\circ} \mathrm{C}=(0+273) \mathrm{K}=273 \mathrm{~K}\) \(T_2=273^{\circ} \mathrm{C}=(273+273) \mathrm{K}=546 \mathrm{~K}\) According to Stefan-Boltzmann's law, rate of radiation,…
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