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VMMC VMMC Medical Solved Paper-2015

  • question_answer
    A vessel containing 100 g water is thermally isolated at 0°C. The air above the water is pumped out, some of the water freezes and some evaporates at 0°C itself. Mass is left inside the vessel is. [latent heat of vaporisation of water at 0°C \[=2.1\times {{10}^{6}}J\,k{{g}^{-1}}\] latent heat of fusion of ice\[=3.36\times {{10}^{5}}J\,k{{g}^{-1}}\]]

    A) 86g                                        

    B) 103g

    C) 77g                                        

    D)  63g

    Correct Answer: A

    Solution :

    Given, total mass of water =M= 100 g \[{{L}_{1}}=2.1\times {{10}^{5}}\text{J}\,\text{k}{{\text{g}}^{-1}}\] \[{{L}_{2}}=3.36\times {{10}^{5}}\text{J}\,\text{k}{{\text{g}}^{-1}}\] Suppose mass of ice formed = m  the mass of water evaporated = M ? m amount of heat taken by water evaporate \[=(M-M){{L}_{1}}\] On freezing heat given by water\[=m{{L}_{2}}\] As, \[M{{L}_{2}}=(M-m){{L}_{1}}\] \[\Rightarrow \]\[m=\frac{M{{L}_{1}}}{{{L}_{1}}+{{L}_{2}}}\]\[\Rightarrow \]\[m=\frac{100(21\times {{10}^{5}})}{(21+3.6)\times {{10}^{5}}}=86g\]


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