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WB JEE Medical WB JEE Medical Solved Paper-2009

  • question_answer
    19 g of water at \[\text{30}{{\,}^{\text{o}}}\text{C}\]and 5 g of ice at \[-\text{20}{{\,}^{\text{o}}}\text{C}\]are mixed together in a calorimeter. What is the final temperature of the mixture? (Given specific heat of \[\text{ice}\,\text{=}\,\text{0}\text{.5}\,\text{cal}\,{{\text{g}}^{-1}}{{({{\,}^{o}}C)}^{-1}}\]and latent heat of fusion of \[\text{ice = 80}\,\text{cal }{{\text{g}}^{-1}}\])

    A)  \[\text{0}{{\,}^{\text{o}}}\text{C}\]                                      

    B)  \[-\text{5}{{\,}^{\text{o}}}\text{C}\]

    C) \[\text{5}{{\,}^{\text{o}}}\text{C}\]                                       

    D)  \[10{{\,}^{\text{o}}}\text{C}\]

    Correct Answer: C

    Solution :

                     Let the final temperature of mixture be \[\text{t}{{\,}^{\text{o}}}\text{C}\text{.}\]Heat lost by water in calories, \[{{\text{H}}_{1}}=19\times 1\times (30-t)\] \[=570-19t\] Heat taken by ice, \[{{H}_{2}}=m{{s}_{i}}\Delta t+mL+m{{s}_{w}}t\] \[=5\times (0.5)20+5\times 80+5\times 1\times t\] According to principle of calorimetry, \[{{H}_{1}}={{H}_{2}}\] \[5\times (0.5)\times 20+5\times 80+5t=570-19t\]                 \[\Rightarrow \]               \[24t=570-450=120\]                 \[\Rightarrow \]               \[t=5{{\,}^{o}}C\]


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