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!

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Thermal Conduction
  1. Thermal Conductivity (Replies: 3)

  2. Thermal conductivity (Replies: 6)

  3. Thermal Conduction (Replies: 7)

  4. Thermal conductivity (Replies: 8)

  5. Thermal conductivity (Replies: 4)