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JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Self Evaluation Test - Thermal Properties of Matter

  • question_answer
    A kettle with 3 liter water at \[27{}^\circ C\] is heated by operating coil heater of power 2 kW. The heat is lost to the atmosphere at constant rate 130 J/sec, when its lid is open. In how much time will water heated to \[97{}^\circ C\] with the lid open? (specific heat of water =4.2kJ/kg)

    A) 472 sec             

    B) 693 sec

    C) 912 sec             

    D) 1101 sec

    Correct Answer: A

    Solution :

    [a] By the law of conservation of energy, energy given by heater must be equal to the sum of energy gained by water and energy lost from the lid. \[\operatorname{Pt}=ms\Delta \theta  + energy lost\] \[2000t=3\times 4.2\times {{10}^{3}}\times \left( 97-27 \right)+130t\] \[\Rightarrow t=472sec\]


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