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Work/energy problem

  1. Dec 8, 2015 #1
    1. The problem statement, all variables and given/known data
    A cubical cooler, 0.45 m on a side, has an R-value of 3.2 m2 K/W. The sides of the cooler are 0.030 m thick. The cooler is filled with water at a temperature of 2 degrees C and is placed in a room with a temperature of 19 degrees C. The heat capacity of water is 4190 J/kg K. How long does it take for the water to warm to 3 degrees C?

    2. Relevant equations
    Q/T=KA(T2-T1)/L
    Q=cm(delta)T

    3. The attempt at a solution
    Q/T=(0.14)(0.453)(19-2)/0.45
    also 6.5 W(4190)(19-2)m[/SUP][/SUP]

    Correct answer is 5.9E4 s. My attempts got nothing close to that.

    [mentor's note: edited to fix formatting]
     
    Last edited: Dec 8, 2015
  2. jcsd
  3. Dec 8, 2015 #2

    SteamKing

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    Use the Preview Button to check the formatting of your thread before you post.

    As far as your calculations are concerned, what the heck are you doing? You haven't even figured out the surface area of the cooler through which heat is going to enter from the surroundings. You haven't even calculated the amount of water in the cooler, let alone the amount of heat it takes to warm the contents 1° C. :frown:

    If nothing else, plug the units of the various quantities into your calculations to see if you are computing the right results. :wink:
     
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