Chemistry

Question

How much energy is required to vaporize 10 grams of water at its boiling point? In Joules.

2 Answer

  • Answer:

    On the other hand, the molecules in liquid water are held together by relatively strong hydrogen bonds, and its enthalpy of vaporization, 40.65 kJ/mol, is more than five times the energy required to heat the same quantity of water from 0 °C to 100 °C

  • Answer: The energy required to vaporize the given amount of water is 22600 J

    Explanation:

    The conversion involved in this process are:

    [tex]H_2O(l)(100^oC)\rightarrow H_2O(l)(100^oC)[/tex]

    To calculate the amount of heat required at constant temperature, we use the equation:

    [tex]q=m\times L_{vap}[/tex]

    where,

    [tex]q[/tex] = amount of heat absorbed = ?

    m = mass of water = 10 g

    [tex]L_{vap}[/tex] = latent heat of vaporization = 2260 J/g

    Putting all the values in above equation, we get:

    [tex]q=10g\times 2260J/g=22600J[/tex]

    Hence, the energy required to vaporize the given amount of water is 22600 J

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