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Thermal conductivity

  1. Dec 24, 2013 #1
    1. The problem statement, all variables and given/known data


    In a domestic heating system, a room is warmed by a 'radiator' through which water passes at a rate of 0.12kg s-1. The steady-state difference between the inlet and outlet temperatures of the water is 6.0 K.

    The radiator is made of iron of thermal conductivity 80 W m-1 K-1 and has an effective surface area of 1.5 m2 with walls 2.0 mm thick.
    (a) At what rate is heat supplied to the room?
    (b) What is the mean temperature difference between the inner and outer surfaces of the radiator walls?
    [Specific heat capacity of water = 4.2 x 103 J kg-1 K-1]

    2. Relevant equations

    a)q/t = k a ΔT / L
    is it true use this equations?

    b) q = m c ΔT

    3. The attempt at a solution

    a)
    i put all the data to the formula, but the answer is 360000 w
    my teacher give me the answer is 3024 w
    which one is true?
     
  2. jcsd
  3. Dec 24, 2013 #2
    Your teacher is right. Show us your work.

    Chet
     
  4. Dec 24, 2013 #3
    k = 80
    A = 1.5 m2
    L = 2 x 10^-3
    delta T = 6 K

    q/t = 80 x 1.5 x 6 / (2x10^-3)

    so q/t = 360000

    what's wrong with thiss? :'(
     
  5. Dec 24, 2013 #4
    What's wrong is that the 6 C is the temperature rise of the water in passing through the radiator, not the temperature difference across the thickness of the radiator iron. How much heat does the water lose per second in passing through the radiator?

    Chet
     
  6. Dec 24, 2013 #5
    oh i see
    so shall i use q/t= m c deltaT ?
     
  7. Dec 24, 2013 #6
    Yes, if m is the mass flow rate and q/t is the rate of heat transfer.

    Chet
     
  8. Dec 24, 2013 #7

    CWatters

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    There are two parts to the question.

    For part..

    a) Apply conservation of energy. In steady state Power going into the rad = Power coming out. Power going in is (m/t)cΔT where (m/t) is the flow rate in kg/s^-1, c = specific heat capacity, and ΔT the temperature difference between input and output pipes (the output pipe isn't at absolute zero so some power is returned to the furnace/boiler)

    b) A different equation will apply. Have a go first.
     
  9. Dec 24, 2013 #8

    CWatters

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    Actually there is a better way to explain equation in part a...

    Power emitted to the room = Power into the rad - Power returned to the furnace

    = mcTflow - mcTreturn
    = mcΔT
     
  10. Dec 25, 2013 #9
    okay thanks a lot! i have found the answer
     
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