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JEE Main & Advanced Physics Thermometry, Calorimetry & Thermal Expansion Question Bank Critical Thinking

  • question_answer
    Water of volume 2 litre in a container is heated with a coil of \[1\,kW\] at \[27\,{}^\circ C\]. The lid of the container is open and energy dissipates at rate of \[160\,J/s.\] In how much time temperature will rise from \[27\,{}^\circ C\] to \[77\,{}^\circ C\] [Given specific heat of water is \[4.2\,kJ/kg\]]                                 [IIT-JEE (Screening) 2004]

    A)            8 min 20 s                              

    B)            6 min 2 s

    C)            7 min                                       

    D)            14 min

    Correct Answer: A

    Solution :

                       Heat gained by the water = (Heat supplied by the coil) ? (Heat dissipated to environment) Þ \[mc\ \Delta \theta ={{P}_{Coil}}\ t-{{P}_{Loss}}\ t\] Þ \[2\times 4.2\times {{10}^{3}}\times (77-27)=1000\,t-160\ t\] Þ \[t=\frac{4.2\times {{10}^{5}}}{840}=500\ sec\ =8\ min\ 20\ sec\]


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