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

  • question_answer
    2 kg of ice at \[\text{ }20{}^\circ C\] is mixed with 5kg of water at \[20{}^\circ C\] in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are \[1\text{ }kcal/kg\text{ }/{}^\circ C\] and \[0.5\text{ }kcal/kg/{}^\circ C\] while the latent heat of fusion of ice is 80 kcal/kg

    A) 7kg                  

    B) 6kg    

    C) 4kg                  

    D) 2kg

    Correct Answer: B

    Solution :

    Initially ice will absorb heat to raise its temperature to \[0{}^\circ C\] then it's melting takes place
    If \[{{m}_{1}}\]= Initial mass of ice, \[{{m}_{2}}\]' = Mass of ice that melts and \[{{m}_{W}}\]= Initial mass of water
    By Law of mixture
    Heat gained by ice = Heat lost by water
    \[\Rightarrow {{m}_{1}}\times \left( 20 \right)+{{m}_{1}}'\times L={{m}_{W}}{{c}_{W}}\left[ 20 \right]\]
    \[\Rightarrow 2\times 0.5\left( 20 \right)+{{m}_{1}}'\times 80=5\times I\times 20\]
    \[\Rightarrow {{\text{m}}_{\text{1}}}\text{ }\!\!'\!\!\text{ =1kg}\]
    So, final mass of water = Initial mass of water + Mass of
    Ice that melts = 5 +1 = 6kg


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