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4 November, 00:37

2. A block of aluminum with a mass of 140g is cooled from 98.4oC to 62.2oC with a release of 1137J of heat. From these data, calculate the specific heat of aluminum.

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  1. 4 November, 00:49
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    Q = M * C * ΔT

    Q / ΔT = M

    Δf - Δi = 98.4ºC - 62.2ºC = 36.2ºC

    C = 1137 J / 140 * 36.2

    C = 1137 / 5068

    C = 0.224 J/gºC
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