JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Self Evaluation Test - Thermal Properties of Matter

  • question_answer
    300 gm of water at \[25{}^\circ C\]is added to 100 g of ice at\[0{}^\circ C\]. The final temperature of the mixture is

    A) \[-\frac{5}{3}{}^\circ \text{C}\]

    B) \[-\frac{5}{2}{}^\circ C\]

    C) \[5{}^\circ C\]  

    D) \[0{}^\circ C\]

    Correct Answer: D

    Solution :

    [d] \[{{\theta }_{mix}}=\frac{{{m}_{W}}{{\theta }_{W}}-\frac{{{m}_{i}}{{L}_{i}}}{{{S}_{W}}}}{{{m}_{i}}+{{m}_{W}}}\] \[=\frac{300\times 25-\frac{100\times 80}{1}}{100+300}=-1.25{}^\circ C\] Which is not possible. Hence \[{{\theta }_{mix}} = 0{}^\circ C\]


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