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BCECE Engineering BCECE Engineering Solved Paper-2003

  • question_answer
    Enthalpy change when \[1\,g\]water is frozen at \[(\Delta {{H}_{fus}}=1.435\,\text{kcal}\text{mo}{{\text{l}}^{\text{-1}}})\]

    A) \[0.0797\text{ }kcal~~\]                               

    B) \[-0.0797\text{ }kcal\]   

    C) \[1.435\text{ }kcal\]     

    D)  \[-1.435\text{ }kcal\]

    Correct Answer: B

    Solution :

    Key Idea: Enthalpy change per gram water is calculated by dividing\[\Delta H\] by molecular mass of water. Given    \[\Delta H=1.4354\,kcal\,mo{{l}^{-1}}\]                 \[=-\frac{1.4354}{18}=-0.0797\,kcal\,{{g}^{-1}}\]


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