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Quantity of heat, phase change of ice to water

  1. Oct 4, 2016 #1
    1. The problem statement, all variables and given/known data
    A 1x106 kg piece of ice is placed into a lake. How much heat is taken from the lake to raise the temperature of the ice from 0 °C to 1x10-20 °C? How much volume does the lake increase by?
    Latent heat for water is 334x103 J/kg

    2. Relevant equations
    Found in my textbook,
    cice = 2100 J/kg*K
    Q = m c ΔT
    Q = mLf
    ρwater = 1.0x103 kg/m3

    3. The attempt at a solution
    First I found how much heat is required to raise the temperature of the ice from 0 °C to 1x10-20 °C.

    Q = mcΔT
    Q = (1x106 kg)(2100 J/kg*K)(1x10-20 °C - 0 °C)
    Q = 2.1x10-11 J

    Then I found how much ice would melt with that amount of heat.

    Q = mLf
    m = Q / Lf
    m = 2.1x10-11 J / 334x103 J/kg
    m = 6.287x10-17 kg

    Finally, I found the volume of the water, or ice that melted.

    ρ = m / V
    V = m / ρ
    V = 6.287x10-17 kg / 1.0x103 kg/m3
    V = 6.287x10-20 m3

    My understanding is the adding heat to ice doesn't increase its temperature and instead melts some of it into water that is at the same temperature. So I'm confused about the part asking how much heat is required to raise the temperature of the ice. I don't know if I took the right approach for this problem. Is it because the temperature change is so small?
     
  2. jcsd
  3. Oct 4, 2016 #2

    gneill

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    Staff: Mentor

    Hi Tafe,

    Welcome to Physics Forums.

    Yes, the temperature change they give is negligible and really quite indistinguishable from zero. I think the idea they want to convey is that even to raise the temperature of the ice by that insignificant amount you first have to melt all of it. So you need to convert the ice to liquid water as the first step. Only then can you raise the temperature of the water by the given amount.
     
  4. Oct 4, 2016 #3
    It makes so much more sense now! I think I got it.

    Heat required in phase change, ice at 0 °C to water at 0 °C:
    Q = mLf
    Q = (1x106 kg)(334x103 J/kg)
    Q = 3.34x1011 J

    Heat required in heating water at 0 °C to 1x10-20 °C:
    Q = mcΔT
    Q = (1x106 kg)(4190 J/kg*K)(1x10-20 °C - 0 °C)
    Q = 4.19x10-11 J

    The sum of these amounts is the total heat required to raise the temperature of the ice. The volume increase of the lake is the volume of the the entire piece of ice melted.
     
  5. Oct 4, 2016 #4

    gneill

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    Staff: Mentor

    The energy calculations look good.

    Shouldn't you give a numerical value for the lake's increase in volume?
     
  6. Oct 4, 2016 #5
    The ice is now water, so
    ρwater = 1.0x103 kg/m3

    ρ = m / V
    V = m / ρ
    V = (1x106 kg) / (1.0x103 kg/m3)
    V = 1000 m3
     
  7. Oct 4, 2016 #6

    gneill

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    Staff: Mentor

    Looks good.
     
  8. Oct 4, 2016 #7
    Thank you so much gneill. :)
     
  9. Oct 4, 2016 #8

    gneill

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    Staff: Mentor

    You're welcome! :smile:
     
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