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Manipal Engineering Manipal Engineering Solved Paper-2012

  • question_answer
    A coffee cup calorimeter initially contains\[125g\]of water at a temperature of\[{{24.2}^{o}}C\]. After adding of\[10.5g\]of\[KBr\], the temperature becomes\[{{21.1}^{o}}C\]. The heat of solution is

    A) \[40\,\,J/g\]                      

    B) \[117\,\,J/g\]

    C) \[167.7\,\,J/g\]

    D)        \[420.05\,\,J/g\]

    Correct Answer: C

    Solution :

    \[\Delta G-ns\Delta T=(125+10.5)\times 1\times 3.1\]               \[=420.05cal=\frac{420.05}{10.5}cal/g\] \[=40cal/g=167.7\,\,J/g\]


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