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Heat problem (cooler at a beach)

  1. May 15, 2015 #1
    1. The problem statement, all variables and given/known data
    upload_2015-5-15_1-40-23.png

    2. Relevant equations

    Q = kAΔT/L

    3. The attempt at a solution

    so I need to find the volume of the cooler, and then the volume is the volume is ice in it?

    then I need to find the surface area of 1 side of the cooler, than multiply it by 6. then plugging in the total area in Q = kAΔT/L

    is this a good approach to the problem?
     

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    Last edited: May 15, 2015
  2. jcsd
  3. May 15, 2015 #2
    Hey, that L is not latent heat in equation but length.
     
  4. May 15, 2015 #3
    oh, sorry! good catch
     
  5. May 15, 2015 #4
    There is another equation for heat transfer what is that?
     
  6. May 15, 2015 #5
    Q = L m

    we equate them so we get

    L m = kAΔT/Length

    we have L

    we have k

    we have A and L

    I'm assuming ΔT is 29 C?
     
  7. May 15, 2015 #6
    How to bring time term here which is 3 hrs?
     
  8. May 15, 2015 #7
    i think the answer we get, if we did all the algebra, is the amount of heat per seconds.

    so we just convert 3 hours to seconds. then multiply by the heat per seconds

    correct?
     
  9. May 15, 2015 #8
    First of all the formula for heat conduction is
    Q = kAtΔT/L where k is constant , A is area, t is time, ΔT is temperature difference and L is length.
     
  10. May 15, 2015 #9

    CWatters

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    In case it's not obvious... In this case L is the thickness of the insulation (eg the length of the heat path)
     
  11. May 15, 2015 #10
    Q = (0.035) (0.02m x 0.4m) (10800 seconds) (29 C) / (0.4m) = 219.24 joules

    i believe we set Q = L m = 219.24 joules

    m should be 6.56x10-4kg

    correct?
     
  12. May 15, 2015 #11
    Yes, it is correct.
     
  13. May 15, 2015 #12

    CWatters

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    No "A" is the surface area. You were right when you wrote..

    L is the length of the heat path which is through the walls (eg 2cm).
     
  14. May 15, 2015 #13
    Yeah, sorry.
    The length to be taken is 2cm.
     
  15. May 15, 2015 #14
    oh, i forgot to multiply the area by 6, so it should be 6( 0.02m x 0.4m) = A

    so new answer should be 26300 joules

    i would have never thought the length was 2cm
     
  16. May 15, 2015 #15

    CWatters

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    kAΔT/L gives you the power (energy per second) flowing through the walls. If the walls are thicker (better insulated) then L is longer and less power flows. Make sense?
     
  17. May 15, 2015 #16
    ok, got the answer to be 1.57 kg, sounds much more reasonable now
     
  18. May 15, 2015 #17

    CWatters

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    Yes. I make the equation..

    Q(Joules) = (0.035) (6 * 0.4m x 0.4m) (10800 seconds) (29 C) / (0.2m)
     
  19. May 15, 2015 #18
    should be 0.02 m
     
  20. May 15, 2015 #19
    Yes you are right.
     
  21. May 15, 2015 #20

    CWatters

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    Yes sorry /0.02.
     
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