JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Calorimetry

  • question_answer
    One kilogram of ice at 0°C is mixed with one kilogram of water at 80°C. The final temperature of the mixture is (Take : specific heat of water\[=4200\,J\,k{{g}^{-1}}\,{{K}^{-1}}\], latent heat of ice \[=336\,kJ\,k{{g}^{-1}})\]    [KCET 2002]

    A)            40°C                                         

    B)            60°C

    C)            0°C

    D)            50°C

    Correct Answer: C

    Solution :

                       \[{{\theta }_{\text{mix}}}=\frac{{{m}_{W}}{{\theta }_{W}}-\frac{{{m}_{\,i}}{{L}_{\,i}}}{{{c}_{W}}}}{{{m}_{\,i}}+{{m}_{W}}}\]            \[\because \,{{m}_{i}}={{m}_{W}}\]Þ\[{{\theta }_{mix}}=\frac{{{\theta }_{W}}-\frac{{{L}_{i}}}{{{c}_{W}}}}{2}\]\[=\frac{80+0-\frac{336}{4.2}}{2}=0{}^\circ C\]


You need to login to perform this action.
You will be redirected in 3 sec spinner