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What is the thickness of the styrofoam?

  1. Jan 1, 2016 #1
    1. The problem statement, all variables and given/known data

    For your cabin in the wilderness, you decide to build a primitive refrigerator out of Styrofoam, planning to keep the interior cool with a block of ice that has an initial mass of 24.0 kg. The box has dimensions of 0.5 m x 0.8 m x 0.5 m. Water from melting ice collects in the bottom of the box. Suppose the ice block is at 0 °C and the outside temperature is 21 °C. If the top of the empty box is never opened and you want the interior of the box to remain at 5 °C for exactly one week, until all the ice melts, what must be the thickness of the Styrofoam?

    2. Relevant equations

    no equations especified

    3. The attempt at a solution

    heat required for melting the ice: 80.16 x 105 J
    time in seconds during one week: 6.048 x 105 s

    If I wish to mantain the temp at 5 °C until all the ice melts, then the rate at wich thermal energy enters in the box must be 80.16 x 105 J / 6.048 x 105 s = 13.25 W. Using the expression for the heat current I found the thickness of the styrofoam to be equal to 0.428 cm, but the correct answer is 2.5 cm.
     
    Last edited: Jan 1, 2016
  2. jcsd
  3. Jan 1, 2016 #2

    SteamKing

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    What expression? Please show your work.
     
  4. Jan 1, 2016 #3
    The expression is H = k.A.(TH - TC) / L, where H is the rate at wich energy flows, k the thermal conductivity of the material, A its area, TH the temp at the hotter side, TC the temp at the colder side and L the thickness of the material (length of the path). Just now I notice that if I multiply the result by 6 (the number of sides of the box), I get 2.57 cm, wich is close of 2.5 cm (the answer of the author). Also, for the variable A (the area), I've considered the box as a unique fat shape of area equal to the sum of the area of each side of the box. Does this have anything to do with the answer that I found? (sorry my bad english)
     
  5. Jan 1, 2016 #4

    SteamKing

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    I'm not sure what a 'unique fat shape' box is.

    You are given the dimensions of the box in the problem statement. It is not a cube, however, so you must calculate the area of the surface of the box, which will not be 6 times the area of anything.

    You also don't mention the value of k for the styrofoam.
     
  6. Jan 1, 2016 #5
    I definitely have no idea how to solve this problem. Please give me the solution.
     
  7. Jan 1, 2016 #6

    SteamKing

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    We can't give you the solution. That's against the rules of the HW forum.

    I don't understand. I asked 1.) a question about the value of the thermal conductivity of the styrofoam, and 2.) I pointed out that the container is not a cube, that you must calculate the surface area of the box from the dimensions given.

    Now, you apparently can't answer the simple question about a value you supposedly used in your calculations, you can't calculate the area of a rectangular box, and you are waving a white flag, saying you can't find the solution of the problem.

    Why can't you at least try to do 1.) and 2.) above without facing an existential crisis?
     
  8. Jan 1, 2016 #7
    Ok. That's because my english is poor and so I'm shy to post a comment.

    the thermal conductivity given in the book is 0.027 W / m.K
    the total area of the box I calculated is 2.1 m²
     
  9. Jan 2, 2016 #8
    Okay. I tried a new approach and this is the result.

    The heat needed to melt the ice is 80.16 x 105 J and it takes a time interval t1. The heat for increasing the temp from 0 to 5 °C is 5.028 x 105 J and it takes a time interval t2 - t1. The sum t1 + t2 must be equal to 6.048 x 105 s (one week), so using the expression for heat current, I found the thickness to be 2.65 cm. However I'm sure that it's still wrong.
     
  10. Jan 2, 2016 #9
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  11. Jan 2, 2016 #10

    SteamKing

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    You calculated a thickness of 2.65 cm and you say the answer key gives 2.5 cm. IMO, that seems to be about right. Remember, the data you were originally given has only two significant figures.
     
  12. Jan 2, 2016 #11
    yet, 2.6 is greater than 2.5 cm :frown:
     
  13. Jan 2, 2016 #12

    SteamKing

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    But it's not that much greater. Heat transfer calculations are usually subject to a lot of assumptions. IMO, if your answer is within 10% of the book answer, it should be OK, unless you made some glaring error in arithmetic.
     
  14. Jan 2, 2016 #13
    ohhh now I found 2.52 cm..... Lol
     
  15. Jan 2, 2016 #14
    80.16 x 105 / t1 = 21 kAL-1
    85.188 x 105 / (t - 2t1) = 16 kAL-1 ***
    solving for t1, we get 1.7 x 105s
    using this result in the expression for heat current and solving for the target variable I get L ≅ 2.52 cm.

    *** My doubt is in this part. It is correct to consider the heat entering in that time interval as the sum of the heat needed for increasing the temp and the heat needed for melting the ice? (sorry my english)
     
  16. Jan 2, 2016 #15

    SteamKing

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    I agree with the calculation above.

    The ice has already melted, so you should not be including the heat of fusion again here. The melt water is at a temperature of 0° C until all the ice has melted, then the temperature of the water increases to 5° C.
    I haven't checked your arithmetic here, because I think that while there are two different heat rates occurring while 1) the ice melts and 2) the water warms slightly,
    you have not applied these conditions correctly to solve for the thickness of the styrofoam.
    I think you have to consider the two different heating rates, as I explained above.

    I've tried working this problem several different ways, and I don't get the book answer of L = 2.5 cm. :frown:
     
  17. Jan 2, 2016 #16
    Yes, you're right, but it was the only way I found to solve this problem, and when I searched for other variables of this same problem, using these calculations I got the same answer as the author. See the problem 17.101 here https://www.coursehero.com/file/p50...each-rod-Hint-The-length-of-the-combined-rod/ and solve it using my above calculations. You should get 3.52 cm as the final result. (sorry my english)
     
  18. Sep 2, 2016 #17
    You can actually obtain 2.5 cm if you use k=0.01 W/m.k. Try using it instead of k=0.027 W/m.k. There'll be no need to multiply by 6.


    -ken
     
  19. Sep 3, 2016 #18

    gneill

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    Where does k=0.01 W/(m K) come from? According to the OP the given value for styrofoam is k=0.027 W/m K. You can't arbitrarily change given constants in order to adjust your calculation results.
     
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