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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2002

  • question_answer
    One kilogram of ice at \[\text{0}{{\,}^{\text{o}}}\text{C}\]is mixed with one kilogram of water at\[\text{80}{{\,}^{\text{o}}}\text{C}\text{.}\]The final temperature of the mixture is (take: specific heat of water \[=4200\,J\,k{{g}^{-1}}\,{{K}^{-1}},\] latent heat of ice\[=336\,kJ\,k{{g}^{-1}}\]).

    A) \[40{{\,}^{o}}C\]                              

    B) \[~60{{\,}^{o}}C\]

    C) \[~0{{\,}^{o}}C\]                             

    D) \[~50{{\,}^{o}}C\]

    Correct Answer: C

    Solution :

    \[{{T}_{1}}=200+273=473{{\,}^{o}}K\] \[{{T}_{2}}=0+273=273{{\,}^{o}}K\] \[{{\mu }_{1}}=1-\frac{{{T}_{2}}}{{{T}_{1}}}=42.2%\] For second \[{{T}_{1}}=0+273=273{{\,}^{o}}K\] \[{{T}_{2}}=-200+273=73{{\,}^{o}}K\] \[{{\mu }_{2}}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\] \[=73.2%\] \[\frac{{{\mu }_{1}}}{{{\mu }_{2}}}=\frac{42.2}{73.2}=\frac{1}{1.73}\] \[{{\mu }_{1}}:{{\mu }_{2}}1:1.73\]


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