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8 March, 01:28

If a system has 425 kcal of work done to it, and releases 5.00 * 102 kJ of heat into its surroundings, what is the change in internal energy of the system?

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  1. 8 March, 05:03
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    Given:

    The work done to the system = 425 kcal

    The heat released by the system = 5 x 10² kJ

    Note that 1 cal = 4.184 J

    Therefore,

    Heat input to the system is

    Q = - 500 kJ

    Work done by the system is

    W = - 425 kcal

    = - (425,000 cal) * (4.184 J/cal)

    = - 1.7782 x 10⁶ J

    = - 1.7782 x 10³ kJ

    According to the 1st law, the change in internal energy of the system is

    ΔU = Q - W

    = - 500 - (-1.7782 x 10³) kJ

    = 1278.2 kJ

    Answer: 1278 kJ
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