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Homework Help: Heat transfer: ice in water bath

  1. Oct 12, 2005 #1
    During one hot summer, a physics grad student went to the local toy store and purchased a plastic child's swimming pool. Upon returning home, he filled it with 200 liters of water at 25 C. Realizing that the water would probably not be cool enough, he threw ice cubes from his refrigerator, each of mass 30g, into the pool. (The ice cubes were originally at 0C.) He continued to add ice cubes, until the temperature stabilized to 16C. He then got in the pool.

    How many ice cubes did he add to the pool to get the temperature to 16 C? (Consider the pool and ice cubes an isolated system.)

    ok..i tried doing the following:

    heat lost by water= heat gained by ice cubes

    Q water = m c delta T

    from the fact d=m/v, i got mass of water as 200 kg

    so Q water = 200 (4186) 9 = 7534800 J

    then i'm not so sure as to where i go from here....

    I've been told to find the heat gained by ice cubes including the melting....how should i do that???
  2. jcsd
  3. Oct 12, 2005 #2


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    I'm sure your textbook discusses the heat of fusion for water.
  4. Oct 12, 2005 #3
    but the problem is i'm not sure how to set it up...

    i try something like this, and i know it's wrong....

    Q water =7534800 J

    each ice cube is .3 kg,

    Q ice = m c delta T: .3(2090)(16) = 10032 J

    melting? Q = 200 x (33.5 e 4) = 6.7 e 7

    i'm just confused...
  5. Oct 13, 2005 #4
    i'm sorry folks, i'm really lost with this one? can someone help me out?
  6. Oct 13, 2005 #5


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    I'll have to give it some thought. It's unclear to me why they introduce the notion of "hot day" and then don't provide any additional information about the "ambient temperature." Given that, apparently they want you to consider only the melting and heat transfer without regard to the ambient temperature.

    I think the basic idea is that the mass of the ice decreases while the mass of liquid water increases. Energy is required to melt the ice (latent heat of fusion) and to warm it up the to the temperature of the liquid water while the liquid water supplies the heat required to do that at a rate proportional to the temperature difference between the liquid water and the ice.

    From the statement of the problem, it would appear that at some particular ratio of ice to liquid that the temperature difference is "stationary" (16 C) - that's what I need to ponder!

    In the meantime, anyone else should feel free to jump in - i.e. the problem, not the pool!
  7. Oct 13, 2005 #6

    Andrew Mason

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    I think you have to assume that all the ice cubes melt. Otherwise there is no equilibrium and the pool keeps getting colder (assuming no heat is transferred from the surroundings). So:

    [tex]\Delta Q_{water} = mc\Delta T_1[/tex]

    where [itex]\Delta T_1 = 9 ^oC; m = 200 kg.; c = 1 cal./^oC g = 4.187 kJ./^oC kg.[/itex]

    [tex]\Delta Q_{ice} = m_{ice}(h_f + c\Delta T_2)[/itex]

    where [itex]\Delta T_2 = 16 ^oC; h_f = 334 kJ/kg. [/itex]

    Equate the two heats. Since the only unknown is the ice mass, you should be able to work it out.
    Last edited: Oct 13, 2005
  8. Oct 13, 2005 #7

    Q water = 200 (4.186)(9) = 7534.8

    Q ice = m (334 + (2.093 x 16)) = 367.488 m

    Q water = Q ice

    7534.8 / 367.488 = m of ice = 20.5 kg...

    since each cube weighs .03 kg..... 20.5/.03 = 684 cubes???

    i tried that didn't work....unless the units i'm working with are wrong?
  9. Oct 13, 2005 #8

    Andrew Mason

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    For Q ice you have to use 4.186 kJ/kg. as the specific heat (the specific heat of water) because it goes from 0 - 16 degrees as water. I get 626 ice cubes.

  10. Oct 13, 2005 #9
    ok...that works.....i was thinking since it's heat gained by ice cubes...i'll use the specific heat for ice....

    then may i ask under what kind of situation do i use specific heat for ice??

    conditions where..ice...is still in temp... below zero??
  11. Oct 14, 2005 #10

    Andrew Mason

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    When it is ice and not liquid water.

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