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NEET NEET SOLVED PAPER 2014

  • question_answer
    Steam at \[{{100}^{o}}C\] is passed into 20 g of water at\[{{10}^{o}}C\]. When water acquires a temperature of\[{{80}^{o}}C\], the mass of water present will be [Take specific heat of water = \[1\,cal{{g}^{-1}}{{\,}^{{}^\circ }}{{C}^{-1}}\] and latent heat of steam\[=540cal{{g}^{-1}}]\]  [AIPMT 2014]

    A)  24g

    B)  31.5g

    C)  42.5g

    D)  22.5g

    Correct Answer: D

    Solution :

      Heat lost by steam = Heat gained by water Let M? amount of heat converts into water. \[m'\times L=ms\,\Delta t\] \[m'\times 540=20\times 1\times (80-10)\] \[m'=\frac{20\times 70}{540}=2.5g\] Now, net water = 20 + 2.5 = 22.5g


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