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

  • question_answer
    A bucket full of hot water cools from \[{{75}^{o}}C\] to \[{{70}^{0}}C\] in time \[{{T}_{1}}\], from \[{{70}^{o}}C\] to \[{{65}^{o}}C\] in time \[{{T}_{2}}\] and from \[{{65}^{o}}C\] to \[{{60}^{o}}C\] in time \[{{T}_{3}}\], then              [NCERT 1980; MP PET 1989; CBSE PMT 1995; KCET 2003; MH CET 1999]

    A)            \[{{T}_{1}}={{T}_{2}}={{T}_{3}}\]                                     

    B)            \[{{T}_{1}}>{{T}_{2}}>{{T}_{3}}\]

    C)            \[{{T}_{1}}<{{T}_{2}}<{{T}_{3}}\]                                     

    D)            \[{{T}_{1}}>{{T}_{2}}<{{T}_{3}}\]

    Correct Answer: C

    Solution :

                       According to Newton's law of cooling Rate of cooling µ Mean temperature difference Þ \[\frac{\text{Fall in}\,\text{temperature}}{\text{Time}}\propto \left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right)\] Q \[{{\left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2} \right)}_{1}}>{{\left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2} \right)}_{2}}>{{\left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2} \right)}_{3}}\] Þ \[{{T}_{1}}<{{T}_{2}}<{{T}_{3}}\]


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