Punjab Medical Punjab - MET Solved Paper-2001

  • question_answer
    The original temperature of a black body is\[{{727}^{o}}C\]. The temperature at which this black body must be raised so as to double, the total radiant energy, is

    A) \[971\,\,K\]                       

    B) \[1190\,\,K\]

    C) \[2001\,\,K\]                     

    D) \[1458\,\,K\]

    Correct Answer: B

    Solution :

    Using the Stefans law, the total radiant energy is                 \[\theta =eA\sigma {{T}^{4}}t\propto {{T}^{4}}\] hence\[\frac{{{E}_{2}}}{{{E}_{1}}}=\frac{T_{2}^{4}}{T_{1}^{4}}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{4}}\]here\[\frac{{{E}_{2}}}{{{E}_{1}}}=2\]                 \[{{T}_{1}}={{727}^{o}}C+273=1000\,\,K\] or            \[{{T}_{2}}=1000\times {{2}^{1/4}}=1000\times 1.19\]                      \[=1190\,\,K\]


You need to login to perform this action.
You will be redirected in 3 sec spinner