BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer
    1 g ice absorbs 335 J of heat to melt at \[\text{0}{{\,}^{\text{o}}}\text{C}\text{.}\] The entropy change will be

    A) \[1.2\text{ J}{{\text{K}}^{-1}}\text{mo}{{\text{l}}^{-1}}\]             

    B)  \[335\text{ J}{{\text{K}}^{-1}}\text{ mo}{{\text{l}}^{-1}}\]

    C)        \[22.1\text{ J}{{\text{K}}^{-1}}\text{mo}{{\text{l}}^{-1}}\]          

    D)  \[0.8\text{ J}{{\text{K}}^{-1}}\text{ mo}{{\text{l}}^{-1}}\]

    Correct Answer: C

    Solution :

    \[\Delta {{S}_{f}}=\frac{\Delta {{H}_{f}}}{{{T}_{f}}}\] \[=\frac{(335\times 18)}{273}=22.1\,J{{K}^{-}}mo{{l}^{-1}}\]


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