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MGIMS WARDHA MGIMS WARDHA Solved Paper-2009

  • question_answer
    A black body of mass 34.38 g and surface area \[19.2c{{m}^{2}}\]is at an initial temperature of 400 K. It is allowed to cool inside an evacuated enclosure kept at constant temperature 300 K. The rate of cooling is 0.04°C/s. The specific heat   of  body   is   (Stefan's   constant\[\sigma =5.73\times {{10}^{-8}}J{{m}^{-2}}{{K}^{-4}}\])

    A)  2800 J/kg-K                       

    B)  2100 J Ag-K

    C)  1400 J/kg-K                       

    D)  1200 J/ kg-K

    Correct Answer: C

    Solution :

                     Newton's law of cooling is \[mc=\frac{dT}{dt}=\sigma ({{T}^{4}}-T_{0}^{4})A\] \[\Rightarrow \]               \[c=\frac{\sigma ({{T}^{4}}-T_{0}^{4})A}{m\left( \frac{dT}{dt} \right)}\] \[\Rightarrow \] \[c=\frac{\begin{align}   & (5.73\times {{10}^{-8}})[{{(400)}^{4}}-{{(300)}^{4}}] \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\times 19.2\times {{10}^{-4}} \\ \end{align}}{(34.38\times {{10}^{-3}})\times 0.04}\] \[\Rightarrow \] \[c=1400J/kg-K\]


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