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30 April, 06:01

A 4.45 kg block of ice at 0.00? c falls into the ocean and melts. the average temperature of the ocean is 3.70? c, including all the deep water.

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  1. 30 April, 06:36
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    The ocean does not change temperature but it does lose some entropy (he gives heat to melt the ice and to warm it to 3.70° C).

    I) For the ice:

    1) For melting the ice:

    Q = m · Lf = 4.45 kg · 334 · 10³ J/kg = 1,486,300 J

    Δ S = Q / T = 1,486,300 J / 273.15 K = 5.441 · 10³ J/K.

    2) To warm the melted ice to 3.70° C:

    Q = m c Δ T = 4.45 kg · 4,190 J / kgK · 3.70 K = 68,988.35 J

    Δ S = m c ln (T2 / T1) = 4.45 kg · 4,190 J/kgK · ln (276.85 / 273.15) =

    = 18,645.5 · ln (1.01354) = 18.645.5 · 0.013454 = 250.8692 J/K

    II) For the ocean:

    Δ S = Q / T = ( - 68,988.35 + 1,486,300) / 276.85 = - 5,617.8 J/K

    The net entropy change:

    Δ S = ΔS ice + ΔS ocean = 5,441.1 + 250.8692 - 5,617.8 = 74.1692 J/K

    Answer: 74 J/K.
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