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13 May, 10:23

How much energy (heat) is required to convert 52.0 g of ice at - 10.0°C to steam at 100°C? Specific heat of ice 2.09 J/g • °C Specific heat of water 4.18 J/g • °C Specific heat of steam 1.84 J/g • °C Molar heat of fusion 6.02 kJ/mol Molar heat of vaporization 40.7 kJ/mol

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  1. 13 May, 10:25
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    Answer: The energy (heat) required to convert 52.0 g of ice at - 10.0°C to steam at 100°C is 157.8 kJ

    Explanation:

    Using this formular, q = [mCpΔT] and = [nΔHfusion]

    The energy that is needed in the different physical changes is thus:

    The heat needed to raise the ice temperature from - 10.0°C to 0°C is given as as:

    q = [mCpΔT]

    q = 52.0 x 2.09 x 10

    q = 1.09 kJ

    While from 0°C to 100°C is calculated as:

    q = [mCpΔT]

    q = 52.0 x 4.18 x 100

    q = 21.74 kJ

    And for fusion at 0°C is called Heat of fusion and would be given as:

    q = n ΔHfusion

    q = 52.0 / 18.02 x 6.02

    q = 17.38 kJ

    And that required for vaporization at 100°C is called Heat of vaporization and it's given as:

    q = n ΔHvaporization

    q = 52.0 / 18.02 x 40.7

    q = 117.45 kJ

    Add up all the energy gives 157.8 kJ
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