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1 April, 07:36

Exactly 20.0 mL of water at 32.0 °C is added to a hot iron skillet. All of the water is converted to steam at 100.0°C. The mass of the skillet is 1.15 kg. What is the change in temperature of the skillet?

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  1. 1 April, 08:55
    0
    98°C

    Explanation:

    Given dа ta:

    Volume of water = 20.0 mL

    Temperature = 32°C

    Mass of skillet = 1.15 kg

    Change in temperature = ?

    Solution:

    Q = m. c.ΔT

    ΔT = 100°C - 32°C

    ΔT = 68°C

    Q = m. c.ΔT

    Q = 20 g * 4.184 j/g.°C * 68°C

    Q = 5690.24 j

    Heat of vaporization of water = 2257 j/g

    q = 2257 j/g * 20 g

    q = 45140 j

    Total heat gained by water = 5690.24 j + 45140 j

    Total heat gained by water = 50830.24 j

    Number of moles of iron:

    Number of moles = mass / molar mass

    Number of moles = 1150/55.85

    Number of moles = 20.59 mol

    Molar heat capacity of iron is = 25.19 j/mol.°C

    50830.24 j / 25.19 j/mol.°C/20.59 mol

    98°C
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