WB JEE Medical WB JEE Medical Solved Paper-2012

  • question_answer
    22320 cal of heat is supplied to 100 g of ice at \[0{{\,}^{o}}C.\] If the latent heat of fusion of ice is \[80\,\text{cal}\,{{g}^{-1}}\] and latent heat of vaporization of water is \[540\,\text{cal}\,{{g}^{-1}},\] the final amount of water thus obtained and its temperature respectively are

    A) \[8\text{ }g,\text{ }100{{\,}^{o}}C\]        

    B)  \[100\text{ }g,\text{ }90{{\,}^{o}}C\]

    C) \[92\text{ }g,\text{ }100{{\,}^{o}}C\]       

    D)  \[82\text{ }g,\text{ }100{{\,}^{o}}C\]

    Correct Answer: A

    Solution :

     Heat required to convert ice to water at \[100{{\,}^{o}}C\] \[Q=m\times L+ms\Delta T=18000\,cal\] Amount of heat left = 4320 cal \[m\times L=4320\] \[m=8\,g\]steam


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