KCET · Physics · Thermal Properties of Matter
If a black body emits \(0.5 \mathrm{~J}\) of energy per second when it is at \(27^{\circ} \mathrm{C}\), then the amount of energy emitted by it when it is at \(627^{\circ} \mathrm{C}\) will be
- A \(40.5 \mathrm{~J}\)
- B \(162 \mathrm{~J}\)
- C \(13.5 \mathrm{~J}\)
- D \(135 \mathrm{~J}\)
Answer & Solution
Correct Answer
(A) \(40.5 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
According to Stefan's law.
\(E \propto T^{4}\)
\(\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{4}\)
\(\frac{\mathrm{E}_{1}}{0.5}=\left[\frac{273+627}{273+27}\right]^{4}\)
\(\mathrm{E}_{1}=0.5\left(\frac{900}{300}\right)^{4}\)
\(\Rightarrow \quad \mathrm{E}_{1}=40.5 \mathrm{~J}\)
\(E \propto T^{4}\)
\(\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{4}\)
\(\frac{\mathrm{E}_{1}}{0.5}=\left[\frac{273+627}{273+27}\right]^{4}\)
\(\mathrm{E}_{1}=0.5\left(\frac{900}{300}\right)^{4}\)
\(\Rightarrow \quad \mathrm{E}_{1}=40.5 \mathrm{~J}\)
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