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JEE Main & Advanced Physics Thermodynamical Processes Question Bank First Law of Thermodynamics

  • question_answer
    A system is provided with 200 cal of heat and the work done by the system on the surrounding is 40 J. Then its internal energy                                                        [Orissa PMT 2004]

    A)            Increases by 600 J              

    B)            Decreases by 800 J

    C)            Increases by 800 J              

    D)            Decreases by 50 J

    Correct Answer: C

    Solution :

                       \[\Delta Q=\Delta U+\Delta W\]                    \[\because \]\[\Delta Q=200cal=200\times 4.2=840J\]and \[\Delta W=40J\]            Þ\[\Delta U=\Delta Q-\Delta W=840-40=800J\]


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