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Homework Help: Rate of heat flow problem

  1. May 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A styrofoam cooler (k = 0.030 W/(m·°C) has outside dimensions of 0.190 m × 0.210 m × 0.340 m, and an average thickness of 2.2 cm. How long will it take for 3.60 kg of ice at 0°C to melt in the cooler if the outside temperature is 26.0°C?

    2. Relevant equations

    Delta Q / Delta t = [k A (T1-T2)] / l

    3. The attempt at a solution

    Delta Q = m L = 3.60 * 3.33*10^5 = 119880 cal.

    => Delta t = [Delta Q * l] / [k*A (T1 - T2)]

    Delta t = 119880 *2.2*0.01 / 0.030*0.190*0.210*0.340*26 = 692.34 h which is wrong.

    Please advise! Thanks!
     
  2. jcsd
  3. May 11, 2010 #2

    Mapes

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    Have you checked your units?
     
  4. May 11, 2010 #3
    Yes, I have checked the units but I have no idea where I am wrong!!!! ????
     
  5. May 11, 2010 #4

    Mapes

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    cal W-1 m-1 is not units of time. It looks like the calorie part is just a typo, but think carefully about what area is used in the conduction equation.
     
  6. May 11, 2010 #5
    Area = 0.190 m × 0.210 m × 0.340 m = 0.013566 m^3. Correct?

    And what's up with the calc? :(
     
  7. May 11, 2010 #6

    Mapes

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    Area isn't measured in cubic meters. What's the total cross-sectional area that the heat transfer occurs through?

    And I don't get your calculation of "119880 cal."
     
  8. May 11, 2010 #7
    Oh, Q = 1198800 cal

    I have no ideas about the cross-sectional area ???
     
  9. May 11, 2010 #8

    Mapes

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    This energy is still not correct, and I can tell that you really have not checked your constants and units for correctness. You're mixing up calories and Joules, and these are not equal.

    If the heat were only being transferred through one side of the box, what would be the appropriate area? Now generalize this to the six-sided box.
     
  10. May 11, 2010 #9
    Got ya, so:

    Q = 1198800 J = 286520.076 calories.

    and A = 6 *0.190 m × 0.210 m × 0.340 m = 0.081396 m^2.

    Yes???
     
  11. May 11, 2010 #10

    Mapes

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    Still looks like a volume to me (measured in cubic meters). Try finding the area of each side individually and adding them together.
     
  12. May 11, 2010 #11
    Ok, let me try:

    0.190 m × 0.210 m + 0.210 m × 0.340 m + 0.340m x 0.190m = 0.1759 m^2. Yes?
     
  13. May 11, 2010 #12

    Mapes

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    Closer...
     
  14. May 11, 2010 #13
    ???? what do u mean closer?
     
    Last edited: May 11, 2010
  15. May 11, 2010 #14
    2 * 0.1759 = 0.3518 m^2 = area, correct?
     
  16. May 11, 2010 #15
    Still stuck! Any one?
     
  17. May 11, 2010 #16

    Mapes

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    Agreed.
     
  18. May 11, 2010 #17
    Delta t = 1198800 *2.2*0.01 / 0.030*0.3518*26 = 96112.3 s = 1602 mins = 26.7 h which is wrong. How come ???
     
  19. May 11, 2010 #18

    Mapes

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    Perhaps they want you to be a bit more precise with the area. Instead of using the area calculated from the outside of the container, try using the area calculated from the midpoint of the container walls. The answer is about 18% larger.

    EDIT: Fixed typo.
     
    Last edited: May 11, 2010
  20. May 11, 2010 #19
    Sorry! I dont know how to calculate from the midpoint :(

    I mean how?
     
  21. May 11, 2010 #20
    You mean:

    [0.190 m × 0.210 m + 0.210 × 0.340 m] / 2 = 0.05565 * 2 = 0.1113 , correct?
     
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