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26 December, 03:43

How much heat is required to convert 5.88 g of ice at - 12.0 ∘c to water at 27.0 ∘c? (the heat capacity of ice is 2.09 j/g∘c, δhvap (h2o) = 40.7kj/mol, δhfus (h2o) = 6.02kj/mol) ?

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  1. 26 December, 07:40
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    You have to calculate the heat for three separate processes:1) heat the ice from - 12°C to 0°C, 2) melt the ice at 0°C, and 3) heat the liquid water from 0°C to 27.0 °C.

    1) Heating the ice from - 12°C to 0°C

    Q1 = m * C * ΔT = 5.88g * 2.09 j/g°C * [0°C - (-12°C) ] = 147.47 j

    2) Melting the ice at 0°C

    Q2 = m * Δh fus

    Convert 5.88 g to moles = > 5.88 g / 18.0 g/mol = 0.327 moles

    Q2 = 0.327 moles * 6.02 kj / mol = 1.96653 kj = 1966.53 j

    3) Heating liquid water from 0°C to 27.0 °C

    Q3 = m * C * ΔT

    C = 4.1813 j/g°C

    Q3 = 5.88 g * 4.1813 j/g°C * (27.0°C - 0°C) = 663.82 j

    4) Total heat, Q

    Q = Q1 + Q2 + Q3 = 147.47j + 1966.53 j + 663.82 j = 2777.82 j ≈ 2778 j

    Answer: 2778 j
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