CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    The surface area of a black body is\[5\times {{10}^{-4}}{{m}^{2}}\]and its temperature is\[727{}^\circ C\]. The energy radiated by it per minute is: \[(\sigma =5.67\times {{10}^{-8}}J/{{m}^{2}}-s-{{k}^{4}})\]

    A)  \[1.7\times {{10}^{3}}J\]             

    B)         \[2.5\times {{10}^{2}}J\]

    C)  \[8\times {{10}^{3}}J\] 

    D)         \[3\times {{10}^{4}}J\]

    E)  none of these

    Correct Answer: A

    Solution :

    From Stefans law \[E=\sigma {{T}^{4}}A\] Given, \[T=727{}^\circ C=(727+273)=1000\text{ }K,\] \[A=5\times {{10}^{-4}}{{m}^{2}}\] \[\therefore \]Energy \[=(5.67\times {{10}^{-8}})\] \[{{(1000)}^{4}}(5\times {{10}^{-4}})60\] \[\therefore \]  \[E=1.7\times {{10}^{3}}J\]


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