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Heat energy transferred in converting ice into steam

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data

    Calculate the heat energy required to convert 0.10 kg of ice at -10 degrees Celcius into steam at 100 degrees Celcius.




    2. Relevant equations

    E = m c dt (1)

    where

    E = heat energy (kJ)

    m = unit mass (kg)

    c = specific heat capacity (kJ/kg oC)

    dt = temperature change (Kelvin or degrees Celsius)





    3. The attempt at a solution

    Given in the question:

    m = 0.1 kg
    dt = 100-(-10) = 110 degrees Celcius
    c = 2100 J Kg^-1 K^-1 (specific heat capacity of ice)


    So what I did was using the equation:

    E = m c dt

    E = (0.1)(2100)(110)

    E = 23100 J

    which would be 23.1 kJ



    The problem is that the answer section of the book says 301.6 KJ which is way above my result.
    What's the mistake I'm doing?

    Help appreciated.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 7, 2011 #2
    Ice must be heated to 0C; ice must then be converted into liquid water at 0C;
    Water at0C must be heated to 100C; water must be converted to vapor (steam) at 100C.
    There are 4 stages, each requiring heat energy
     
  4. Dec 7, 2011 #3
    Thanks for the hint.


    For Ice -> water:

    m = 0.1kg c=2100 J kg^-1 K^-1 dt = 10 degrees (-10 -> 0)

    therefore E= mc dt = 0.1(2100)(10) = 2100 J

    Energy required to melt the ice:

    E= specific latent heat of fusion of ice * mass
    = 325000(0.1)
    = 32500 J

    and then for water to steam:

    m = 0.1 kg c= 4200 J kg^-1 K^-1 (for water) dt= 100 degrees (0->100)

    => E= 0.1 (4200) (100) = 42000 J

    Energy required to vaporize 0.1 kg of water:

    E = specific latent heat of vaporization of water * mass
    = 225000(0.1)
    = 22500 J

    then i add them

    2100 + 32500 + 42000 + 225000

    and get 301600 J which is 301.6 kJ.


    Thank you.
     
  5. Dec 7, 2011 #4
    That looks good.
    Well done
    Do you see that by far the greatest amount of energy is in converting liquid water to steam?
     
  6. Dec 7, 2011 #5
    SORRY.... misprint, you have the latent heat of vaporisation to convert water to steam out by factor of 10
    Latent heat of vaporisation = 2300000 J/kg
    I have only just noticed.... fortunately you made a mistake in your multiplication so your answer is correct6.
    should always double check calculations !!!!!
     
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