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7 November, 04:20

How much energy is needed to change 44 grams of ice at - 8° to steam at 107°?

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  1. 7 November, 07:40
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    Q=m (c∆t + heat of fusion + heat of evaporation)

    m = 44g

    c = 4.186 J/g. C

    ∆t = 107 - (-8) = 115 C

    heat of fusion = 333.55 J/g

    heat of evaporation=2260 J/g

    Q=44 (4.186*115 + 333.55 + 2260)

    Q = 135297.36 J
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