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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Morning)

  • question_answer
    Ice at \[-20{}^\circ C\] is added to 50 g of water at \[40{}^\circ C\]. When the temperature of the mixture reaches \[0{}^\circ C,\] it is found that 20 g of ice is still un melted. The amount of ice added to the water was close to (Specific heat of water \[=4.2\text{ }J/g/{}^\circ C\] Specific heat of Ice \[=2.1\text{ }J/g/{}^\circ C\] Heat of fusion of water at \[0{}^\circ C=334\text{ }J/g\]) [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

    A) 60 g                                         

    B) 50 g

    C) 40 g                 

    D)                  100 g

    Correct Answer: C

    Solution :

    Let m be the mass of ice added to water. \[(m-20)L+mC\Delta {{T}_{1}}=50{{C}_{1}}\Delta {{T}_{2}}\] \[\Rightarrow \]\[(m-20)(334)+m\times 2.1(20)=50\times 4.2\times (40)\] \[\Rightarrow \]\[m(334+42)=5400+6680\] \[\Rightarrow \]\[m=40.1g\]


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