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Thermal Conduction

  1. Jan 12, 2009 #1
    1. The problem statement, all variables and given/known data
    The temperature in an electric oven is 158°C. The temperature at the outer surface in the kitchen is 51°C. The oven (surface area = 1.8 m2) is insulated with material that has a thickness of 0.023 m and a thermal conductivity of 0.045 J/(s·m·°C).

    (a) How much energy is used to operate the oven for 7 hours?

    2. Relevant equations
    Q = (k A deltaT time)/L

    3. The attempt at a solution
    k = .045
    A = 1.8
    deltaT = 107
    time = 7
    L = .023

    I got 2637.78 but it says that's wrong. What did I do wrong?
  2. jcsd
  3. Jan 12, 2009 #2


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    Check units. I suspect a mistake in the units of k = 0.045 J/(s·m·°C). Perhaps m squared?
    Also your 7 hours should be expressed in terms of the time unit in k so that the time cancels out in the calculation.
  4. Jan 12, 2009 #3
    The units of the k is correct. I double checked it. What do you mean by the 7 hours?
  5. Jan 12, 2009 #4


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    You can't multiply hours by J/(s·m·°C).
  6. Jan 12, 2009 #5
    So convert it to seconds? I did that and still got the wrong answer.
  7. Jan 12, 2009 #6


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    Conductivity has units W/mK
    Power through the oven wall dQ/dt = k * Area * temperature_difference / thickness
    So total energy Q = k * Area * temperature_difference * time / thickness
    Check the units, J = J/smK * m^2 * K * s / m

    k = 0.045 W/mK
    A = 1.8 m^2
    deltaT = 107 K
    time = 7*3600 = 25,200s
    L = 0.023m
  8. Jan 12, 2009 #7
    Why is your k in W/mK? Shouldn't it be in J/(smC)?

    BTW, did you get 9496017.3913 J as your answer?
  9. Jan 12, 2009 #8


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    It's the same thing, W = J/s and it is conventional to quote temperature differences in kelvin rather than deg C. They are the same size so it doesn't change the value - it's just a detail thing.

    Also note the number of figures in your answer.
    You are only told the area to two significant figures so you cannot possibly give an answer with 12digits of accuracy. You should put 9.5 x106 or 9,500,000 J
  10. Jan 12, 2009 #9
    It worked! Thank you for the help.
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