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Specific & Latent Heat - Heat required to turn water in Aluminum Tray -> Ice

  1. Nov 8, 2011 #1
    Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice

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

    200 g Water (l) is contained in an Aluminum ice tray that has a mass of 340 g. Both is at 18°C. How much heat (Q) must be removed to turn the water into ice at -15°C?

    Aluminum mAl= 340g
    Water mW= 200g
    Ti= 18°C

    Tf= -15°C


    2. Relevant equations

    Specific Heat Q = mcΔT
    Latent Heat Q = mLf

    3. The attempt at a solution

    QNET= heat removed to bring Aluminum from 18°C to -15°C
    + heat removed to bring Waterl from 18°C to 0°C
    + heat removed from water to change phase (liquid to solid)
    + heat removed to bring ice from 0°C to -15°C

    QNET = (mcΔT)Al + (mcΔT)W(l) + (mLf)w + (mcΔT)W(s)

    QNET = [.34kg (900 [itex]\frac{J}{kg*°C}[/itex])(-15°C-18°C)] + [.2kg (4186 [itex]\frac{J}{kg*°C}[/itex])(0°C - 18°C)] +[.2kg (333*103 [itex]\frac{J}{kg}[/itex])] + [.2kg (2100 [itex]\frac{J}{kg*°C}[/itex]) (-15°C-0°C)
    QNET = 35132.4 J = 35.1324 kJ

    My answer is not one of the choices. What am I doing wrong? Please help!

    Thank you in advance. Any and all help is much appreciated!
     
  2. jcsd
  3. Nov 8, 2011 #2
    Re: Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice

    when I did it I got 97.9kJ. I can only imagine that something went wrong with all the - signs.
    For the Al I used temp change = 33 (18 to -15), for the water 18 and for the ice 15.
     
  4. Nov 8, 2011 #3

    ehild

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    Re: Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice

    I think you might omit 103 from the latent heat.

    ehild
     
  5. Nov 8, 2011 #4
    Re: Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice

     
  6. Nov 8, 2011 #5
    Re: Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice

    There is certainly nothing wrong with your method (as far as I can see) it must be a computational error somewhere.
     
  7. Nov 8, 2011 #6
    Re: Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice

    "For the Al I used temp change = 33 (18 to -15)"

    Q = mcΔT
    33 grams for the mass of the Aluminum tray?
    QAl= (.34 kg) (900 [itex]\frac{J}{kg*C°}[/itex])(-15°C - 18°C)
    did I lay that out wrong or did I use the wrong numbers?

    "for the water 18"
    Water in liquid form is from 18°C to 0°C

    "ice 15"
    Ice from 0°C to -15°C

    is that what you meant?
     
  8. Nov 8, 2011 #7
    Re: Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice

    Q(aluminum) = -10098 J
    Q (water from 18°C to 0°C) = - 150696 J
    Q (water to ice phase change) = 666000 J
    Q (ice from 0°C to -15°C) = -6300 J

    (-10098 J) + (-150696 J) + (666000 J) + (-6300 J) = 35,132.4 J = 35.13 kJ

    What am I doing wrong? Are the signs wrong?
     
  9. Nov 8, 2011 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Re: Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice

    The latent heat has to be taken with negative sign. It is also removed heat. Change + to minus.

    ehild
     
  10. Nov 8, 2011 #9
    Re: Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice

    my values are
    Q for aluminium =0.34 x 900 x 33 = 10098 J
    Q water 18 to 0 = 0.2 x 4186 x 18 = 15070 J
    Q water to ice = 0.2 x 333000 = 66600 J I think this is the difference!!!
    Q ice to -15 = 0.2 x 2100 x 15 = 6300 J
    These all add up to 10098 + 15070 +66600 +6300 = 98068 or 98kJ
    We must be getting close !!!! I want to go to bed soon
     
  11. Nov 8, 2011 #10
    Re: Specific & Latent Heat - Heat required to turn water in Aluminum Tray --> Ice

    Thank you both for your help!!! Much appreciated!
     
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