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27 June, 01:03

How much heat is required to convert 5.53 g of ice at - 12.0 ∘C to water at 24.0 ∘C? (The heat capacity of ice is 2.09 J / (g⋅∘C), ΔHvap (H2O) = 40.7kJ/mol, and ΔHfus (H2O) = 6.02kJ/mol.)

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  1. 27 June, 01:28
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    2.55 * 10³ J = 2.55 kJ

    Explanation:

    Specific heat capacity of ice = 37.8 J / mol °C

    Specific heat capacity of water = 76.0 J / mol °C

    Ice at - 12 °C is converted to ice at 0 °C by absorbing heat Q₁

    Ice at 0°C melts to water at 0 °C. Let Heat absorbed during this phase change be Q₂.

    Let heat absorbed to raise the temperature of water from 0 C to 24°C be Q₃.

    Total heat = Q = Q₁ + Q₂ + Q₃

    Q₁ = (37.8 j/mol C) (5.53 g / 18.01532 g / mol) (0 - (-12)) = 139.23749 j

    Q₂ = (5.53 g/18.01532 g H₂O / mol) (6.02 x10³ j) = 1847.905 j

    Q₃ = (76 j/mol C) ((5.53 g/18.01532 g H₂O / mol) (24-0) = 559.8968 j

    Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j

    = 2547.039 j = 2.55 * 10³ J = 2.55 kJ
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