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Cooling by thermal radiation

  1. Mar 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi! I need some help by my phisycs homework at universty. Here is the problem:
    We have a steel ball, with density of 7800kg/m^3, with speficic heat capacity 460J/kgK. Temperature of the ball is 1700K. How much time does the ball need to cool down to half of it's temperature (850K), if it cools only because of the thermal radiation. Stefan-Boltzman constant is 5,67×10^-8. There is no other radiation on the ball, from the surrounding area.
    The ball's mass is 1kg.

    2. Relevant equations


    3. The attempt at a solution
    I suppose, that the power of radiation (P = j × A = (s.b. constant) × T^4 × A) is decreasing as the temperature drops, so i tried to set up my equaton with differentials:

    dQ = Pdt
    mc dT = A × (sb const.) × T^4 dt

    i integrated the equation, and got this:
    mcT^-4dT = A(sb const)dt
    -1/3 mcT^-3= A(sb const) × t

    integral is from Ta/2 to Ta

    and i shrinked the expression and expressed time from the equation

    t= 8mc/(Ta)^3×A×(sb. const)

    ; c= spec. heat capacity
    sb const. = stefan boltzman constante = 5,67 × 10^-8
    Ta = temp. at start = 1700K

    the result i get is 51 secons. The solution should be 313 seconds. What am I doing wrong? Can you help me?
     
  2. jcsd
  3. Mar 25, 2013 #2
    Check your algebra... when solving the integral you should have a 7 instead than 8... anyway (using of course 7) I have the right result...
     
  4. Mar 25, 2013 #3
    I did the algebra again, I hope photo sharp enough
    ... i get the same result, I checked twice.

    51,14 seconds.

    The correct result should be 313 seconds :S
     

    Attached Files:

  5. Mar 25, 2013 #4
    I probably found your problem... when computing
    $$ \left(\frac{1}{T^3}\right)_{T/2}^{T}=\frac{1}{T^3/8}-\frac{1}{T^3}=\frac{8}{T^3}-\frac{1}{T^3}=\frac{1}{T^3}(8-1) $$
    instead you computed
    $$ \frac{1}{T^3/8 -T^3} $$
    which is clearly different because then you don't get a factor 1/8...
    Using the right formula you should have 313 more or less (I did it before and it was OK) :smile:
     
  6. Mar 25, 2013 #5
    Oh of course.... I have found the problem now... Thanks :)
     
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