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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 \[{{80}^{0}}C\] to \[{{60}^{o}}C\] in one minute. The ambient temperature is \[{{30}^{o}}C\]. In cooling from \[{{60}^{o}}C\] to \[{{50}^{o}}C\] it will take         [MP PMT 1995; UPSEAT 2000;  MH CET 2002]

    A)            \[30\ \sec onds\]               

    B)            \[60\ \sec onds\]

    C)            \[90\ \sec onds\]               

    D)            \[50\ \sec onds\]

    Correct Answer: D

    Solution :

                       \[\frac{80-60}{1}=K\left( \frac{80+60}{2}-30 \right)\]    Þ \[K=\frac{1}{2}\]            Again \[\frac{60-50}{t}=\frac{1}{2}\left( \frac{60+50}{2}-30 \right)\]Þ \[t=0.8\times 60=48\]sec.


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