A) \[40\,kcal\text{/(}{{m}^{2}}\text{-}\min )\]
B) \[80\,kcal\text{/(}{{m}^{2}}\text{-}\min )\]
C) \[160\,kcal\text{/(}{{m}^{2}}\text{-}\min )\]
D) \[320\,kcal\text{/(}{{m}^{2}}\text{-}\min )\]
Correct Answer: D
Solution :
Given: Initial radiated energy \[({{E}_{1}})=20\,kcal\text{/}{{m}^{2}}\text{-}\min ;\] Initial temperature \[({{T}_{1}})=T\] and final temperature\[({{T}_{2}})=2T\]. We know from the Stefans law that (E) = \[e\,\sigma \,{{T}^{4}}\propto {{T}^{4}}.\] Therefore \[\frac{{{E}_{1}}}{{{E}_{2}}}={{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{4}}={{\left( \frac{T}{2T} \right)}^{4}}=\left( \frac{1}{2} \right)=\frac{1}{16}\] or \[{{E}_{2}}={{E}_{1}}\times 16=20\times 16=320\,kcal\text{/}({{m}^{2}}\text{-}\min )\] (...... where\[{{E}_{2}}=\]Radiant Energy)
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