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

  • 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 \[20{}^\circ C\]. If 440 gm. of hot water at \[92{}^\circ C\]is poured in it, the final temperature, neglecting radiation loss, will be nearest to

    A) \[58{}^\circ C\]

    B)        \[68{}^\circ C\]

    C) \[73{}^\circ C\]            

    D)        \[78{}^\circ C\]

    Correct Answer: B

    Solution :

    [b] Let the final temperature be T. Then \[200\times 1\times \left( T-20 \right)+20\times  \left( T-20 \right)\] =440(92 - T) Solving it, we get \[\operatorname{T}=68{}^\circ C.\].


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