BCECE Engineering BCECE Engineering Solved Paper-2004

  • question_answer
    At \[127{}^\circ C\], radiated energy is \[2.7\times {{10}^{-3}}\text{J/s}\text{.}\] At what temperature radiated energy is 4.32    \[4.32\times {{10}^{6}}\text{J/s?}\]?

    A) 400 K                                    

    B)        4000 K

    C)        80000 K                                

    D)         40000 K

    Correct Answer: C

    Solution :

    According to Stefans law, the total radiant energy emitted per second from unit Surface area of a black body is proportional to the fourth power of its absolute temperature, that is \[E\propto {{T}^{4}}\] or            \[\frac{{{E}_{2}}}{{{E}_{1}}}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{4}}\]                       ?(i) Given,   \[{{T}_{1}}=127+273=400\,K,\]                 \[{{E}_{1}}=2.7\times {{10}^{-3}}\,J/s\]                 \[{{E}_{2}}=4.32\times {{10}^{6}}\,J/s\] Substituting the values in Eq. (i), we get \[\frac{4.32\times {{10}^{6}}}{2.7\times {{10}^{-3}}}={{\left( \frac{{{T}_{2}}}{400} \right)}^{4}}\] or            \[{{T}_{2}}=400{{\left( \frac{4.32\times {{10}^{6}}}{2.7\times {{10}^{-3}}} \right)}^{1/4}}\] or            \[{{T}_{2}}=400{{(16\times {{10}^{8}})}^{1/4}}\] or            \[{{T}_{2}}=400\times 2\times {{10}^{2}}=80000\,K\]


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