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28 December, 11:56

A 111.6 g sample of iron was heated from 0°C to 20°C. It absorbed 1004 J of energy. What is the specific heat capacity of iron?

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  1. 28 December, 14:53
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    Formula

    H = m*c*ΔT

    Givens

    H = 1004 J

    m = 111.6

    ΔT = 20 - 0

    ΔT = 20

    c = ?

    Solution

    1004 = 111.6 * c * 20 Multiply the right.

    1004 = 2232*c Divide to get c

    1004/2232 = c Do the division

    0.4498 = c

    Specific heat of iron = 0.4498
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