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JCECE Engineering JCECE Engineering Solved Paper-2011

  • question_answer
    In the combustion of \[2.0\,\,g\] \[\text{of}\] methane, \[25\,\,kcal\] heat is liberated. Heat of combustion of methane would be

    A) \[150\,\,kcal\]                  

    B) \[200\,\,kcal\]

    C) \[250\,\,kcal\]                  

    D) \[350\,\,kcal\]

    Correct Answer: B

    Solution :

    When \[2.0\,\,g\] or\[\left( or\,\,\frac{2}{16}\,\,mol \right)\]methane is burnt, heat evolved\[=25\,\,kcal\]. \[\therefore \]When \[1\,\,mol\] methane is burnt, heat evolved will be                 \[=\frac{25}{(2/16)}=\frac{25\times 16}{2}=200\,\,kcal\]


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