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JEE Main & Advanced JEE Main Solved Paper-2015

  • question_answer
    A solid body of constant heat capacity \[1J/{}^\circ C\] is being heated by keeping it in contact with reservoirs in two ways : (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat. (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat. In both the cases body is brought from initial temperature \[100{}^\circ C\] to final temperature \[200{}^\circ C\]. Entropy change of the body in the two cases respectively is : [JEE Main Solved Paper-2015 ]

    A)  ln2,2ln2                              

    B) 2ln2,8ln2

    C) ln2,4ln2                               

    D) ln2,ln2

    Correct Answer: D

    Solution :

                    \[\Delta s=\int_{{}}^{{}}{\frac{dQ}{T}}=\int\limits_{100}^{200}{mc\frac{dT}{T}=mc\int\limits_{100}^{200}{\frac{dT}{T}}}\]or\[\Delta s=mc\ln 2\] \[=\ln 2(\because mc=1)\]


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