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

  • question_answer
    The initial temperature of a body is 80°C.  If its temperature falls to 64°C in 5 minutes and in 10 minutes to 52°C then the temperature of surrounding will be          [MP PMT 2003]

    A)            26°C                                         

    B)            49°C

    C)            35°C                                         

    D)            42°C

    Correct Answer: B

    Solution :

                       According to Newton's law \[\frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}=k\left[ \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right]\] Initially, \[\frac{(80-64)}{5}=K\,\left( \frac{80+64}{2}-{{\theta }_{0}} \right)\] Þ \[3.2=K[72-{{\theta }_{0}}]\]  ... (i) Finally  \[\frac{(64-52)}{10}=K\,\left[ \frac{64+52}{2}-{{\theta }_{0}} \right]\] Þ \[1.2=K[58-{{\theta }_{0}}]\]... (ii) On solving equation (i) and (ii) \[{{\theta }_{0}}=49{}^\circ C\].


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