JCECE Engineering JCECE Engineering Solved Paper-2006

  • question_answer
    A body cools from\[{{75}^{o}}C\] to\[{{70}^{o}}C\] 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 timer\[3\], then:

    A) \[{{t}_{3}}>{{t}_{2}}>{{t}_{1}}\]                

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

    C)  \[{{t}_{2}}>{{t}_{1}}={{t}_{3}}\]               

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

    Correct Answer: A

    Solution :

    From Newton's law of cooling                 \[\frac{dH}{dt}=K\left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right)\] where \[{{\theta }_{0}}\] is temperature of surrounding, \[\frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\]the temperature of body. Hence\[,\]          \[{{t}_{3}}>{{t}_{2}}>{{t}_{1}}\].


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