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JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time will the temperature of water rise from \[27{}^\circ C\] to \[77{}^\circ C\]? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) 8 min 20 s                           

    B) 6 min 2 s

    C) 7 min                    

    D) 14 min

    Correct Answer: A

    Solution :

    From question, In 1 sec heat gained by water = 1 KW - 160 J/s = 1000 J/s - 160 J/s = 840 J/s Total heat required to raise the temperature of water(volume 2L) from \[27{}^\circ C\] to \[77{}^\circ C\] \[={{m}_{water}}\times sp.ht\times \Delta \theta \] \[=2\times {{10}^{3}}\times 4.2\times 50\][Q mass = density × volume] And, \[840\times t=2\times {{10}^{3}}\times 4.2\times 50\] or\[t=\frac{2\times {{10}^{3}}\times 4.2\times 50}{840}\]\[=500=8\,\min 20s\]


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