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

  • question_answer
    A beaker contains 200 gm of water. The heat capacity of the beaker is equal to that of 20 gm of water. The initial temperature of water in the beaker is 20°C. If 440 gm of hot water at 92°C is poured in it, the final temperature (neglecting radiation loss) will be nearest to       [NSEP 1994]

    A)            58°C                                         

    B)            68°C

    C)            73°C                                         

    D)            78°C

    Correct Answer: B

    Solution :

                       Heat lost by hot water = Heat gained by cold water in beaker + Heat absorbed by beaker                    Þ \[440(92-\theta )=200\times (\theta -20)+20\times (\theta -20)\]            Þ \[T=68{}^\circ C\]


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