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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    When 0.2 g of 1-butanol was burnt in a suitable apparatus, the heat evolved was sufficient to raise the temperature of 200 g water by\[5{}^\circ C\]. The enthalpy of combustion of 1-butanol in kcal\[mo{{l}^{-1}}\]will be

    A) + 37                                       

    B) + 370

    C) \[-370\]               

    D)        \[-740\]

    E) \[-14.8\]

    Correct Answer: C

    Solution :

    Heat required to raise the temperature of 200 g water by\[5{}^\circ C\] \[=m.\text{ }c.\text{ }\Delta t\] \[=200\times 1\times 5=1000\text{ }cal=1\text{ }kcal\] \[\therefore \]0.2 g 1-butanol on burning liberates = 1 kcal heat \[\therefore \]74 g (1 mol) 1-butanol on burning will liberate                 \[=\frac{1}{0.2}\times 74\] \[=370\] Since, the heat is liberated,\[\Delta H=-370\text{ }kcal\].


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