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

  • question_answer
     A liquid cools down from \[{{70}^{o}}C\] to \[{{60}^{o}}C\] in 5 minutes. The time taken to cool it from \[{{60}^{o}}C\] to \[{{50}^{o}}C\] will be [MP PET 1992, 2000; MP PMT 1996]

    A)            5 minutes                              

    B)            Lesser than 5 minutes

    C)            Greater than 5 minutes   

    D)            Lesser or greater than 5 minutes depending upon the density of the liquid

    Correct Answer: C

    Solution :

                       According to Newton's law of cooling Rate of cooling µ mean temperature difference. Initially, mean temperature difference \[=\left( \frac{70+60}{2}-{{\theta }_{0}} \right)=(65-{{\theta }_{0}})\] Finally, mean temperature difference \[=\left( \frac{60+50}{2}-{{\theta }_{0}} \right)=(55-{{\theta }_{0}})\] In second case mean temperature difference decreases, so rate of fall of temperature decreases, so it takes more time to cool through the same range.


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