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

  • question_answer
    A cup of tea cools from 65.5o C to 62.5 o C in one minute in a room of 22.5 o C. How long will the same cup of tea take, in.............. minutes, to cool from 46.50 o C to 40.5 o C in the same room ? (choose nearest value)                [Kerala PMT 2004]

    A)            1    

    B)            2

    C)            3    

    D)            4

    Correct Answer: D

    Solution :

                       \[\frac{65.5-62.5}{1}=K\left( \frac{65.5+62.5}{2}-22.5 \right)\]Þ \[K=\frac{3}{41.5}\]                    And again \[\frac{46.5-40.5}{t}=\frac{3}{41.5}\left( \frac{46.5+40.5}{2}-22.5 \right)\]            Þ \[\frac{6}{t}=\frac{3}{41.5}\times 21\] Þ \[t=\frac{82}{21}\tilde{-}4\]minute.


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