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26 December, 05:56

A 1.3 g sample of gold (specific heat capacity = 0.130 J/g °C) is heated using 86.9 J of energy. If the original temperature of the gold is 25.0°C, what is its final temperature?

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  1. 26 December, 08:34
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    556.95°C

    Explanation:

    The following were obtained from the question:

    M = 1.3g

    C = 0.130 J/g °C

    Q = 86.9J

    T1 = 25.0°C

    T2 = ?

    Q = MC (T2 - T1)

    86.9 = 1.3 x 0.13 (T2 - 25)

    Clear the bracket

    86.9 = 0.169T2 - 4.225

    Collect like terms

    89.9 + 4.225 = 0.169T2

    94.125 = 0.169T2

    Divide both side by 0.169

    T2 = 94.125/0.169

    T2 = 556.95°C
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