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2 September, 20:07

If you start with 163 g of water at 29◦C, how much heat must you add to convert all the liquid into vapor at 100◦C? Assume no heat is lost to the surroundings.

Answer in units of kJ.

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Answers (1)
  1. 2 September, 20:25
    0
    48.37514 kj

    Explanation:

    Given dа ta:

    Mass of water = 163 g

    Initial temperature = 29°C

    Final temperature = 100°C

    Heat added = ?

    Solution:

    Specific heat capacity:

    It is the amount of heat required to raise the temperature of one gram of substance by one degree.

    Specific heat capacity of water is 4.18 j/g.°C

    Formula:

    Q = m. c. ΔT

    Q = amount of heat absorbed or released

    m = mass of given substance

    c = specific heat capacity of substance

    ΔT = change in temperature

    ΔT = Final temperature - initial temperature

    ΔT = 100°C - 29°C

    ΔT = 71°C

    Q = 163 g * 4.18 j/g.°C * 71°C

    Q = 48375.14 j

    Joule to Kj conversion:

    48375.14 / 1000 = 48.37514 kj
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