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30 August, 15:46

How many grams of water can be cooled from 33 ∘c to 15 ∘c by the evaporation of 50 g of water? (the heat of vaporization of water in this temperature range is 2.4 kj/g. the specific heat of water is 4.18 j/g⋅k.) ?

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  1. 30 August, 16:06
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    Answer: -

    1595 g

    Explanation: -

    Heat of vaporization = 2.4 Kj / g

    Mass of water to be vaporized = 50 g

    Heat released = Mass of water to be vaporized x Heat of vaporization

    = 50 g x 2.4 KJ / g

    = 120 KJ

    = 120000 J

    Initial temperature = 33+273 = 306 K

    Final temperature = 15+273=288 K

    Change in temperature required = T = 306 - 288 = 18 K

    specific heat of water is 4.18 J / g K

    Mass of water that can be cooled = Total heat / (specific heat of water x Change in temperature)

    = 120000 J / (4.18 J / g K x 18 K)

    = 1595 g
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