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Heat flow with two materials

  1. Apr 9, 2005 #1
    A wall is composed of two materials. Material 1 has a thickness of 7.21 cm and a thermal conductivity of 0.1 while material 2 has a thickness of 6.57 cm and a conductivity of 1. If the temperature difference inside to outside is 25 Co and the wall has an area of 10 m2, what is the energy loss per second to the nearest watt?

    My work:
    Q1=kA (deltaT/L)t
    =(.1)(10)(25/0.0721)t
    =346.7t

    Q2=kA (deltaT/L)t
    =(1)(10)(25/0.0657)t
    =3805.2

    Qtotal = Q1+Q2
    = 346+3805
    =4151W

    The answer is 318. Any help would be greatly appreciated.
     
  2. jcsd
  3. Apr 9, 2005 #2
    I think I found the equation i was supposed to use:

    Q/t=(A*delta T)/(L/k)

    But it still didn't work out. :yuck: Please help!
     
  4. Apr 10, 2005 #3

    FredGarvin

    User Avatar
    Science Advisor

    Make sure your units are correct. I am going to have to assume which ones you are using since you do not state them.

    First, find the equivilent thermal conduction resistance, U by using:
    [tex]U=\frac{1}{\frac{L_A}{K_A} + \frac{L_B}{K_B}}[/tex]

    [tex]U= \frac{1}{\frac{.0721}{0.1} + \frac{.0657}{1.0}}[/tex]

    [tex]U= 1.271 \frac{W}{m^2 * C}[/tex]

    Now use Newton's Law of Cooling in the form of:

    [tex]q_x=UA \Delta T[/tex]

    [tex]q_x=(1.271 \frac{W}{m^2 *C})(10 m^2)(25 C)[/tex]

    [tex]q_x = 317.8 W = 318 W[/tex]
     
    Last edited: Apr 10, 2005
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