A) \[120\,\,W\]
B) \[240\,\,W\]
C) \[304\,\,W\]
D) \[320\,\,W\]
Correct Answer: D
Solution :
From Stefans-Boltzmann law \[E=\sigma ({{T}^{4}}-T_{0}^{4})\] \[\frac{{{E}_{2}}}{{{E}_{1}}}=\frac{(T_{2}^{4}-T_{0}^{4})}{(T_{1}^{4}-T_{0}^{4})}\] \[\Rightarrow \] \[{{E}_{2}}=\left( \frac{T_{2}^{4}-T_{0}^{4}}{T_{1}^{4}-T_{0}^{4}} \right){{E}_{1}}\] ? (i) Here\[{{E}_{1}}=60W,\,\,{{T}_{0}}={{227}^{o}}C=500K\], \[{{T}_{1}}={{727}^{o}}C=1000K\], \[{{T}_{2}}={{1227}^{o}}C=1500K\] From Eq. (i), we get \[\therefore \] \[{{E}_{2}}=\frac{{{(1500)}^{4}}-{{(500)}^{4}}}{{{(1000)}^{4}}-{{(500)}^{4}}}\times 60\] \[=\frac{{{(500)}^{4}}[{{3}^{4}}-1]}{{{(500)}^{4}}[{{2}^{4}}-1]}\times 60\] \[=\frac{80}{15}\times 60=320\,\,W\]
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