JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    When \[{{M}_{1}}\]gram of ice at \[10{}^\text{o}C\] (specific heat \[=0.5\text{ }cal\text{ }{{g}^{1}}{}^\text{o}{{C}^{1}})\]is added to M2 gram of water at \[50{}^\text{o}C,\] finally no ice is left and the water is at 0ºC. The value of latent heat of ice, in cal \[{{g}^{-1}}\]is:                      [JEE Main Held on 12-4-2019 Morning]

    A) \[\frac{5{{M}_{1}}}{{{M}_{2}}}-50\]                  

    B) \[\frac{50{{M}_{2}}}{{{M}_{1}}}\]

    C) \[\frac{50{{M}_{2}}}{{{M}_{1}}}-5\]                  

    D) \[\frac{5{{M}_{2}}}{{{M}_{1}}}-5\]

    Correct Answer: C

    Solution :

    Heat lost = Heat gain \[\Rightarrow {{M}_{2}}\times 1\times 50={{M}_{1}}\times 0.5\times 10+{{M}_{1}}.{{L}_{f}}\] \[\Rightarrow {{L}_{f}}=\frac{50{{M}_{2}}-5{{M}_{1}}}{{{M}_{1}}}\]\[=\frac{50{{M}_{2}}}{{{M}_{1}}}-5\]


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