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Heat loss from a house

  1. Apr 18, 2014 #1
    1. The problem statement, all variables and given/known data


    I'm struggling to understand question 1.C
    How do I incorporate Tw and later eliminate this?


    2. Relevant equations

    -k*A*(ΔT/ΔX)

    3. The attempt at a solution

    For 1.a I used this equation: -k*A*((Tinside-Toutside)/thickness wall) I thought to put T inside first as they ask for the inward heat flow and that will be negative (heat loss)....

    For 1.c I'm really stuck...
     

    Attached Files:

  2. jcsd
  3. Apr 18, 2014 #2

    mfb

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    Staff: Mentor

    You get two heat flow equations, one for the styrofoam and one for the bricks. Using both together, you can elimitate Tw.
     
  4. Apr 18, 2014 #3
    Thanks for your response.


    Js=-ks*A*((Tinside-Tw)/thickness styrofoam)
    Jb=-kb*A*((Tw-Toutside)/thickness brick wall)

    How do I now eliminate Tw?
     
  5. Apr 18, 2014 #4

    mfb

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    Staff: Mentor

    With the standard ways to manipulate equations.

    Why did you use two different J?
     
  6. Apr 18, 2014 #5
    I thought that I had two different J as they both represent a different number....
    Im really not sure how to continue from here... Im sorry.... Can you give me a hint?
     
  7. Apr 18, 2014 #6
    J is the same for both because the heat goes through both walls in succession. The walls are in series.

    Chet
     
  8. Apr 18, 2014 #7
    But won't the flow through styrofoam be different from the flow through brick as they have a different k? And one Goes from inside to the brick wall and the other from the brick wall to outside.
    I understand that there is also a 'total' flow from in to outside.

    And how do I combine these 2 equations. I might be very dumb, but really trying to see it!
     
  9. Apr 18, 2014 #8
    This is very much analogous to an electric circuit with two unequal resistors in series. The overall voltage drop is analogous to the overall temperature difference, and the current is analogous to the heat flow. The current flow through each of the resistors is the same. The temperature at the interface between the styrofoam and the brick is analogous to the voltage in the wire between the two resistors. So:

    I = ΔV1/R1

    and

    I = ΔV2/R2

    ΔV=ΔV1+ΔV2=I (R1 + R2)

    I = ΔV/(R1 + R2)

    Chet
     
  10. Apr 18, 2014 #9

    I really appreciate your effort!
    I understand now that J is the same, but still don't know what to do with the to equations.... Sorry!
     
  11. Apr 18, 2014 #10
    Do the same thing we did here with the two voltage drops. Solve for the two temperature differences, and then add them to eliminate Tw.

    Chet
     
  12. Apr 18, 2014 #11
    I got it!!
    Thank you so much!!
     
    Last edited: Apr 18, 2014
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