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Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    The molar heat capacity of rock salt at low temperature varies with temperature according to Debyes \[{{T}^{3}}\] law Thus \[C=k\frac{{{T}^{3}}}{{{\theta }^{3}}}\] where k = 1940 J mol\[Jmo{{l}^{-1}}{{k}^{-1}}\] Calculate how much heat is required to raise the temperature of 2 moles of rock salt from 10 K to 50 K

    A)  800 J                                    

    B)  373 J

    C)  273 J                    

    D)         None of these

    Correct Answer: C

    Solution :

                    \[dQ=nCdT\]                 \[dQ=nk\frac{{{T}^{3}}}{{{\theta }^{3}}}dT\]                 \[Q=\frac{nk}{{{\theta }^{3}}}\int_{{{T}_{1}}}^{{{T}_{2}}}{{{T}^{3}}dT}\]                 \[=\frac{nK}{{{\theta }^{3}}}\left( \frac{T_{2}^{4}-T_{1}^{4}}{4} \right)\]                 \[=\frac{2\times 1940\times ({{50}^{4}}-{{10}^{4}})}{{{(281)}^{3}}\times 4}\]                 \[=272.79\,\,J=273\,\,J\]


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