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JEE Main & Advanced Physics Thermodynamical Processes JEE PYQ-Thermodynamical Processes

  • question_answer
    A thermally insulated vessel contains 150g of water at \[0{}^\circ C\]. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at \[0{}^\circ C\] itself. The mass of evaporated water will be closest to: (Latent heat of vaporization of water \[=2.10\times {{10}^{6}}Jk{{g}^{1}}\] and Latent heat of Fusion of water \[=3.36\times {{10}^{5}}Jk{{g}^{1}}\])                                                                                                                         [JEE Main 8-4-2019 Morning]

    A)  130 g

    B)                   35 g

    C)  20 g    

    D)       150 g

    Correct Answer: C

    Solution :

    [c] Suppose 'm' gram of water evaporates then, heat required
    \[\Delta {{Q}_{req}}=m{{L}_{\text{v}}}\]
    Mass that converts into ice = (150 - m)
    So, heat released in this process
    \[\Delta {{Q}_{rel}}=(150-m){{L}_{f}}\]
    Now,
    \[\Delta {{Q}_{rel}}=\Delta {{Q}_{req}}\]
    \[(150-m){{L}_{f}}=m{{L}_{V}}\]
    \[m({{L}_{f}}+{{L}_{\text{v}}})=150{{L}_{f}}\]
    \[m=\frac{150{{L}_{f}}}{{{L}_{f}}+{{L}_{\text{v}}}}\]
    \[m=20g\]


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