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Using thermal conductivity to melt ice

  1. Jan 25, 2014 #1
    1. The problem statement, all variables and given/known data
    An ice cube at 0.00 °C measures 14.9 cm on a side. It sits on top of a copper block with a square cross section 14.9 cm on a side and a height of 18.1 cm. The bottom of the copper block is in thermal contact with a large pool of water at 92.5 °C. How long does it take the ice cube to melt? Assume that only the part in contact with the copper liquefies, that is, the cube gets shorter as it melts. The density of ice is 0.917 g/cm3.

    Tice = 273 K
    Twater = 365.5 K
    LHeat of fusion = 334 kJ/kg
    Vice = (14.9cm)3 = 3,307.95 cm3
    h = 0.181 m
    κCu = Thermal conductivity of Copper = 385 W/(m*K)
    A = face area of Cu = (0.149m)2 = 0.0222 m2


    2. Relevant equations
    ρ = m/V
    Q = mLHeat of fusion
    PCond = Q/t = Aκ(ΔT/L)


    3. The attempt at a solution
    I did this is 3 steps.

    1)
    Used density to obtain the volume of the ice cube

    ρi = mi/Vi

    mi = ρiVi

    mi = (0.917 g/cm3)(3,307.95 cm3)

    mi = 3,033.39 g

    2)
    Used the mass to obtain amount of energy required to melt the ice, Qi.

    Qi = miLHeat of fusion

    Qi = (3,033.39g)(334 J/g)

    Qi = 1,013,152.3 J

    3)
    Used Qi and the conduction rate equation to calculate the time t.

    PCond = Qi/t = ACuκCu(ΔT/LCu)

    QiLCu = (ACuκCuΔT)t

    QiLCu/(ACuκCuΔT) = t

    t = (1,013,152.3 J)(0.181 m)/(0.0222 m2)(385 W/(m*K))(92.5 K)

    t = 231.952 J/W = 231.952 (W*s)/W = 231.952 s

    t [itex]\approx[/itex] 232 s

    I submitted this answer to my online homework and it told me I was incorrect. I need to make sure that I am correctly analyzing the situation and accounting for everything I need to.

    Any help is appreciated.

    Please don't be rude. I will gladly provide more info on my calculations. Thanks.
     
  2. jcsd
  3. Jan 25, 2014 #2

    haruspex

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    I can think of nothing you are doing wrong, and I get the same answer.
    There is one piece of information missing: we don't know the initial temperature profile of the block. But taking it to start with a linear profile with the top and bottom at 0 and 92.5C seems reasonable.
     
  4. Jan 25, 2014 #3
    Ok, thanks! The professor told us that one of the twelve homework problems was going to give a wrong answer but he didn't tell us which so that we would still attempt all of them. This may be the problem he was talking about. I have two other problems on my homework that I haven't answered so I couldn't be sure that this was the one.
     
  5. Feb 5, 2014 #4
    So looking at the solution that the proffessor posted on this problem, the only thing that I see that he did differently is that he used kCu=390 W/(m*K) instead of kCu=385 W/(m*K).

    That gives me an answer of 228.97 s
     
  6. Feb 5, 2014 #5

    haruspex

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    For the temperature range in question,
    http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html gives 400
    http://hyperphysics.phy-astr.gsu.edu/hbase/tables/thrcn.html gives 385
    http://en.wikipedia.org/wiki/List_of_thermal_conductivities gives four estimates from 385 to 401.
    So any answer in the range 223 to 232 should have been accepted.
     
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