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14 January, 13:38

The molar heat of vaporization for liquid water is 40.6 kJ/mole.

How much energy is required to change 3.4 g of liquid water to steam if the water is already at 100oC?

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  1. 14 January, 16:35
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    7670 joules

    Explanation:

    Molar heat → 40.6 kJ / mol

    1 mol of water occupies 18 g, but we have only 3.4 g. Let's make a rule of three:

    18 g of H₂O release 40.6 kJ of heat

    Then, 3.4 g will release (3.4. 40.6) / 18 = 7.67 kJ

    If we convert the kJ to J → 7.67 kJ. 1000 J / 1 kJ = 7670 Joules

    We can also do this:

    If water is already at 100°C, we apply the latent heat (heat that is relased in a change of state)

    Q = 2256 J/g. 3.4 g → 7670.4 J
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