1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help! Heat Loss

  1. Oct 18, 2005 #1
    Help! Heat Loss!!!

    A cup of coffee (with a lid) is enclosed in an insulated cup 0.5cm thick in the shape of a cube 13.1cm on a side. The thermal conductivity of the cup is 0.0002 cal/s*cm*DegreeC. The temperature of the coffee is 87C and the temperature of the surroundings is 14C. Find the heat loss due to conduction. Answer in units of J/s.

    I've tried P=kA(Th-Tc)/L ------> Energy transfer by conduction but somehow it's wrong?

    I see no other way!

    I know that:
    k = 0.0002cal/s*cm*DegC which = 0.0008372J/s*cm*DegC
    L=0.5cm
    A=13.1cm*13.1cm (because cube)
    A=171.61 cm^2
    Th=87C
    Tc=14C

    ANY IDEAS?!
     
    Last edited: Oct 18, 2005
  2. jcsd
  3. Oct 18, 2005 #2

    lightgrav

    User Avatar
    Homework Helper

    You're actually using kA(Th - Tc)/L , right?

    Are you leaking Heat thru all 6 sides?

    Did you convert from cal to J?

    Are you using 13.1 cm as inside measure, or outside?
    How do you know it's wrong?
     
  4. Oct 18, 2005 #3
    I know its wrong because my teacher said so :frown:

    Yes im using that equation.

    I think I'm leaking on all sides

    I did convert from cal to J.

    And I'm using 13.1cm as an outside measure I believe.
     
    Last edited: Oct 18, 2005
  5. Oct 18, 2005 #4

    lightgrav

    User Avatar
    Homework Helper

    Did you multiply the area of each side by 6 sides?
     
  6. Oct 18, 2005 #5
    Why x6 we are looking for overvall heat loss from this thermos like apparatus.
     
  7. Oct 18, 2005 #6

    lightgrav

    User Avatar
    Homework Helper

    you think Heat only leaks out the top?
    How many sides does a cube have?
     
  8. Oct 18, 2005 #7
    Ok i understand but how do i fit it into the equation? k(6A)(Th-Tc)/L ?
     
    Last edited: Oct 18, 2005
  9. Oct 18, 2005 #8

    lightgrav

    User Avatar
    Homework Helper

    yea, or A = 6 s^2
     
  10. Oct 18, 2005 #9
    Correct! Thank you very much!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help! Heat Loss
  1. Heat loss (Replies: 3)

  2. Heat Loss (Replies: 7)

  3. Heat loss/Heat Gained (Replies: 4)

Loading...