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How much heat is required to convert ice into steam?

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data
    An ice cube of volume 12.1 cm3 is initially at a temperature of -18.1°C. How much heat is required to convert this ice cube into steam?
    (the final answer must be in Joules)


    2. Relevant equations
    Q = m*c*deltaT
    Q = m*Lf (latent heat of fusion)
    Q = m*Lv (latent heat of vaporization)
    density = m/v


    3. The attempt at a solution
    Since I was only given the volume of the ice cube, I decided to use density = mass/volume to find the mass of the ice cube. Since the final answer must be in joules, I used the density of 9.17 kg/cm^3 as the density, and found the mass to be 1109570.

    I then plugged in that mass into
    Q = (1109570) (2090 J/kg°C the c of ice) (18.1) to convert ice to water
    Q = (1109570) (4180 J/kg°C the c of water) (100) to convert water to steam
    Q = (1109570) (334000 J - the Lf of water)
    Q = (1109570) (2200000 J - Lv of water)

    I added all the Qs together and found the final answer to be 5.09e14, which is incorrect.
    Finding the correct units was a bit tricky so I may have done something wrong there.
    If anyone has any suggestions, please let me know.
     
  2. jcsd
  3. Apr 25, 2010 #2
    Your approach is correct.

    Double check the units and value for the density of ice. Recall the density of water is about 1 gr per cubic cm, so the density of ice should be slightly less than that.

    Also, you've mixed units in the Lf and Lv for water. It looks like you might have switched from "cal / g degree' to 'J / g degree'. Don't forget to also change the g to kg.

    Make sure everything is in the same units - m, kg, and J are all MKS units, so stick with those.
     
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