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JEE Main & Advanced Physics Transmission of Heat Question Bank Radiation Newton's Law of Cooling

  • question_answer
    The temperature of a liquid drops from \[365K\] to 361 K in 2 minutes. Find the time during which temperature of the liquid drops from \[344\ K\] to \[342K\]. Temperature of room is \[293\ K\]                                         [RPET 1997]

    A)            84 sec                                      

    B)            72 sec

    C)            66 sec                                      

    D)            60 sec

    Correct Answer: A

    Solution :

                       \[\frac{365-361}{2}\]\[=K\left[ \frac{365+361}{2}-293 \right]\]= 70 KÞ \[K=\frac{1}{35}\]  Again \[\frac{344-342}{t}=\frac{1}{35}\left[ \frac{344-342}{2}-293 \right]=\frac{10}{7}\] Þ \[t=\frac{14}{10}\]min = \[\frac{14}{10}\times 60=84\]sec.


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