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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    A body cools in 7 min from\[60{}^\circ C\]to\[40{}^\circ C\]. What time (in min) does it take to cool from \[40{}^\circ C\]to\[28{}^\circ C,\]if surrounding temperature is\[10{}^\circ C\]? (Assume Newtons law of cooling)

    A)  3.5                                        

    B)  14                         

    C)  7                                            

    D)  10

    E)  21

    Correct Answer: C

    Solution :

    From Newtons law of cooling \[\frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}=\left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right)\] Therefore, \[\frac{60{}^\circ -40{}^\circ }{7}\propto \left( \frac{60{}^\circ +40{}^\circ }{2}-10{}^\circ  \right)\]                             ?. (1) \[\frac{40{}^\circ -28{}^\circ }{t}\propto \left( \frac{40{}^\circ +28{}^\circ }{2}-10{}^\circ  \right)\]                              ?. (2) \[\therefore \]  \[t=7\,s\]


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