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Time for block of ice to reach 20 C in an isolating box.

  1. Dec 22, 2012 #1
    Hi,
    1. The problem statement, all variables and given/known data
    I am trying to find how long it will take 8 kg of ice whose temperature is -10 C, kept in a sealed isolating box placed in room temperature (25 C), to turn into water whose temp. is 20 C. The box is a cube whose side is 20 cm, the thickness of the box is 2 cm, and κ = 0.04 W/m*K, CW = 4190 J/Kg*K, Ci = 2000 J/Kg*K, Lf = 3.34*105 J/Kg, Lv = 2.256*106 J/Kg.


    2. Relevant equations



    3. The attempt at a solution
    1. Warming the ice to melting point (0 C)
    2. Melting ALL the ice AT 0 C
    3. Warming the water to 20 C

    These will all be achieved through heat conduction, from the surrounding to the box.

    I believe stages 1 and 3 ought to be calculated via T(t) = Tc + (TH - Tc)*e-t/τ, where T(t) - Tc denotes the temp. difference between the box and its surrounding through a separating medium (in this case, one of the walls of the box), TH denotes the temp. of the body giving off heat at t=0, and τ = mCLf / κA.
    I believe stage 2 needs to be calculated via t = Lfm/H, where H = κA(TH - Tc)/L, where L is thickness of isolating layer.

    I'd truly appreciate comments on the validity of this approach.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 22, 2012 #2

    haruspex

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    Yes, except that you need to interpret TH and Tc carefully. The TH / Tc notation in the equation you quote assumes a hot body in (infinite) cold surroundings, so everything ends up at Tc. In the present problem, you need to set TH as the box contents temperature and Tc as the ambient.
    The Lf looks out of place, but there should be something in there for box thickness. You just meant L, I guess.
    Looks good.
     
  4. Dec 22, 2012 #3
    Hi haruspex,
    Thank you for replying. I did mean L (and not Lf).
    I would like to ask you, in view of the above, to kindly review my evaluation thereof:
    Stage 1: 25-0 = (25+10)e-t1/200,000, hence t1 = 67294.41 sec
    Stage 2: t2 = [3.34*105*8]/[(25-0/0.02)*0.04*0.2*0.2] 1336000 sec
    Stage 3: 25-20 = (25-0)e-t3/419,000, hence t3 = 674354.49 sec

    Could the final answer indeed be (approx.) 24 days? Seems a bit too much, wouldn't you say?
     
  5. Dec 22, 2012 #4

    haruspex

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    How many faces does a cube have? Other than that, your calculations look OK, and I'm not surprised it would take days. In practice it might take rather longer; this calculation assumes the ice conducts well.
     
  6. Dec 22, 2012 #5
    It's actually an isolating container "in the shape of a square whose side is 20 cm".
    Does anything still require emendation?
    PS I wasn't surprised it would take days, simply nearly a month seemed rather off.
     
  7. Dec 22, 2012 #6
    I wish to get this straight, please. Do I now need to multiply all the above calculations involving the surface area by a factor of 6?
     
  8. Dec 22, 2012 #7

    gneill

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    Presumably heat will enter the cube via all the sides, not just one. To find the thermal conductivity or resistance you'll need to take into account the entire surface area through which the heat enters. One 20x20 cm face is only, well, one face of the cube...
     
  9. Dec 22, 2012 #8
    You need to take into account transient heat conduction within the block of ice. The ice can't be treated as having a uniform temperature throughout its mass. Furthermore, once the surface of the ice rises to 0C, the ice at the surface of the block will begin to melt. Then things get more complicated because, in a gravitational environment, (a) the ice will float to the top of the ice chest, and (b) natural convection can occur within the water. If you assume that the ice chest is in outer space, away from gravitational attraction, these effects can be neglected. In this case, there will be heat conduction from the inner surface of the chest to the ice-water interface. The interface will be at 0C until all the ice is melted. So the transient heat conduction problem continues once the ice starts to melt, but in this case there is a moving melt front between the water and the unmelted block. At the interface between the ice and the water, there will be a discontinuity in the heat flux equal to the velocity of the interface times the density times the latent heat of fusion. This is a complicated moving boundary problem. Eventually, after all the ice is melted, the transient heat conduction problem will continue on the water, and the water will still be at a non-uniform temperature. Even when the average water temperature rises to 20C, the water temperature will be non-uniform. Even if this were a transient heat conduction problem on a non-melting solid, the 3D geometry of the cube would still have to be contended with. I'm not saying that this is not a solvable problem. I'm just saying that it is much more complicated problem than can be solved using the implicit simplifying assumptions invoked in the previous posts.
     
  10. Dec 22, 2012 #9
    Okay, thanks! I realise it is far more intricate, and yet supposing I am asked to regard the temperature within the container as uniform and continuous at all time, without any gravitational effects, will the sole required emendation now be to replace the A in my calculations with 6A?
     
  11. Dec 22, 2012 #10

    haruspex

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    Yes. If the problem as presented to you doesn't quote any numbers for conductivity through water or ice, that seems a reasonable assumption.
     
  12. Dec 22, 2012 #11
    Yes. I agree with Haruspex. But the answer you would get that way would only be a lower bound to the true amount of time it would take to reach 20 C.
     
  13. Dec 23, 2012 #12
    I thank you! :-)
     
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