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JEE Main & Advanced Physics Thermodynamical Processes Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    The change in internal energy of a thermo- dynamical system which has absorbed 2 kcal of heat and done 400 J of work is \[(1\text{ }cal=4.2J)\]

    A) 2 kJ

    B) 8 kJ    

    C) 3.5 kJ

    D) 5.5 kJ

    Correct Answer: B

    Solution :

    [b] According to first law of thermodynamics \[Q=\Delta U+W\] Given: \[Q=2kcal=\text{ }2000\times 4.2=8400J\] \[W=400J\] \[\Delta U=Q-W=8400-400=8000J\]


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