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JEE Main & Advanced Physics Transmission of Heat JEE PYQ-Transmission Of Heat

  • question_answer
    When 100 g of a liquid A at \[100{}^\circ C\]is added to 50 g of a liquid B at temperature\[75{}^\circ C\], the temperature of the mixture becomes\[90{}^\circ C\]. The temperature of the mixture, if 100 g of liquid A at \[100{}^\circ C\] is added to 50 g of liquid B at \[50{}^\circ C\]will be-                                                                                                     [JEE Main 11-Jan-2019 Evening]

    A)  \[70{}^\circ C\]                       

    B)  \[85{}^\circ C\]

    C)  \[60{}^\circ C\]           

    D)       \[80{}^\circ C\]

    Correct Answer: D

    Solution :

    [d] Case I: Heat lost = Heat gained
    \[100\times {{s}_{A}}\times (100-90)=50\times {{s}_{B}}\times (90-75)\]
    \[2{{s}_{A}}\times 10={{s}_{B}}\times 15\]or\[4{{s}_{A}}=3{{s}_{B}}\]                        ….(i)
    Case II: Heat lost = Heat gained
    \[100\times {{s}_{A}}\times (100-\theta )=50\times {{s}_{B}}\times (\theta -50)\]
    \[(100-\theta )=\frac{4}{3}\times \frac{1}{2}(\theta -50)\]
    \[\theta =\frac{400}{5}={{80}^{o}}C\]


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