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Manipal Engineering Manipal Engineering Solved Paper-2010

  • question_answer
    1 \[c{{m}^{3}}\] of water at its boiling point absorbs 540 cal of heat to becomes steam with a volume of 1671 \[c{{m}^{3}}\]. If the atmospheric pressure \[=1.013\times {{10}^{5}}\,N/{{m}^{2}}\]and the mechanical equivalent of heat =4.19 J/cal, the energy spent in this process in overcoming intermolecular forces is

    A)  540 cal                                 

    B)  40 cal

    C)  500 cal                 

    D)         zero

    Correct Answer: C

    Solution :

    According to first law of thermodynamics                 \[\Delta Q=\Delta U+\Delta W\] \[\therefore \]  \[\Delta U=\Delta Q-\Delta W\]                 \[=540-\frac{p({{V}_{2}}-{{V}_{1}})}{J}\]                 \[=540-\frac{1.013\times {{10}^{5}}\times [(1671-1)\times {{10}^{-6}}]}{4.2}\]                 \[=540-40=500\,\,cal\]


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