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

  • question_answer
     Hot water kept in a beaker placed in a room cools from \[{{70}^{o}}C\] to 60°C in 4 minutes. The time taken by it to cool from \[{{69}^{o}}C\] to \[{{59}^{o}}c\] will be                                            [JIPMER 1999]

    A)            The same 4 minutes         

    B)            More than 4 minutes

    C)            Less than 4 minutes          

    D)            We cannot say definitely

    Correct Answer: B

    Solution :

                       Rate of cooling \[=\frac{-d\theta }{dt}\propto \left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right)\]            In second case average temperature will be less hence rate of cooling will be less. Therefore time taken will be more than 4 minutes.


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