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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 2

  • question_answer
    2 kg of ice at \[-20{}^\circ C\] is mixed with 5 kg 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. (Given : Specific heat capacities of water and ice are \[1\,\,ca\operatorname{l}\,{{g}^{-1}}{}^\circ {{C}^{-1}}\]and \[0.5\text{ }cal\text{ }{{g}^{-}}^{1}{}^\circ {{C}^{-}}^{1}\] respectively. Latent heat of fusion of ice \[=80\,cal\,{{g}^{-1}})\]

    A) 7 kg

    B) 6 kg

    C) 4 kg        

    D) 2 kg

    Correct Answer: B

    Solution :

    [b]: Heat released by 5 kg of water when its temperature fall from \[20{}^\circ C\] to \[0{}^\circ C\] is \[{{Q}_{1}}={{m}_{water}}{{s}_{water}}\Delta T\] \[=5\times {{10}^{3}}\times 1\times (20-0)={{10}^{5}}cal\] Heat absorbed by 2 kg of ice at \[-20{}^\circ C\] to increase its temperature to \[0{}^\circ C\]is \[{{Q}_{2}}={{m}_{ice}}{{s}_{ice}}\Delta T\] \[=2\times {{10}^{3}}0.5\times 20=0.2\times {{10}^{5}}cal\] So, the temperature of mixture will be \[0{}^\circ C\] \[0{}^\circ C\]. The remaining heat \[Q={{Q}_{1}}-{{Q}_{2}}=0.8\times {{10}^{5}}cal\] The remaining heat will melt a mass m of ice \[\therefore \]\[m=\frac{Q}{{{L}_{f}}}=\frac{0.8\times {{10}^{5}}}{80}={{10}^{3}}g=1kg\] The final mass of water in the container \[=\left( 5+1 \right)kg=6kg\]


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