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

  • question_answer
    A cube of side 5 cm made of iron and having a mass of 1500 g is heated from \[25{}^\circ C\] to \[400{}^\circ C.\]The specific heat for iron is \[0.12\text{ }cal/g{}^\circ C\]and the coefficient of volume expansion is \[3.5\times {{10}^{-5}}/{}^\circ C,\]the change in the internal energy of the cube is (atm pressure \[1\times {{10}^{5}}N/{{m}^{2}}\])

    A) 320 kJ              

    B) 282 kJ

    C) 141 kJ  

    D)        423 kJ

    Correct Answer: B

    Solution :

    [b] \[Q=mC\Delta r=1.5\times 0.12\times 4200\times \left( 400-25 \right)\] \[=2.83\times {{10}^{5}}J\] \[W=P\left( \Delta V \right)=P(V\Delta T)\] \[=105\times \left( 5\times 10-2 \right)3\times 3.5\times 10-5\times 375=0.164\text{ }J\]\[\text{Thus }Q=\Delta U+W\] \[or\text{ }2.83\times {{10}^{5}}=\Delta U+0.164;\Delta U=282kJ\]


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