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Thermal physics question

  1. Jun 14, 2003 #1
    Right I am so stumped on part b of this question so I will post part a too so that you guys may check to see if I did that correctly.

    a) object heated to 230 celius is placed inside a 25 celius room. Determine net rate of heat
    loss due to raiation if object has surface area of 68 cm^2 and an emissivity of
    0.780.

    b) estimate the temp of object after 1 min if its density is 3.60 *10^3 kg/m^3 and its
    specific heat is 768 J/kg*degress celius.

    My work for a)

    Actually I used a simliar equation to solve the first problem

    change in Q/time = q for limitations of kboard
    emissivity=e
    bolstman constant= o
    temp = T, Te= temp of environment To=temp of obj
    a=area

    q=eoA(Te^4-To^4)

    so my quation for a was

    q=.78(5.67*10^-8)(.68)(25^4-230^4)
    q= -84.1 w/s

    as for part b I have no idea how to work density into any of the equations I have at all.... driving me mad
     
  2. jcsd
  3. Jun 14, 2003 #2
    First of all, i am not really an expert so if this is very important maybe waiting for someone that knows more than me would be better.
    For part a :
    If the unit of the bolstman constant you are using is Watt/(m2.k4) then you made a mistake.
    You see, 68 cm2 does not equal 0.68 m2, it actually equals 0.0068 m2.
    Another thing, the final unit of the answer (since it is a rate of energy loose) will be Joule/S (or Watt) and not Watt/S in the SI system.
    Now for part b:
    I see now way to solve this part unless you are given one of two information :
    1-The shape of the object (a cube, sphere ... etc)
    2-The volume of the object
    If you are given neither of those, i see no way to solve it
     
  4. Jun 14, 2003 #3
    ah yes quite right the units were actually a typo meant to say J/s so that would be a watt. As for the .68m^2 I guess you are correct there, I see how that would be done now. The boltsman constant was given as that value from my text, I did not calculate it but I did not bother with units. I thought perhaps I could get mass through something like this:

    3.60*10^5kg/m^3 and .0068m^2, so .0068^.5 = m(meters)

    3.60*10^5kg/.0068^1.5
    m= 642006066.3 kg? well I guess not that mass is waaaay to huge.... Can anyone tell me how to get mass given density, specific heat capacity and area?? Or any equations even involving density?

    I have figured out how I could end up solving the final temp for b and my equation is this:

    Let answer of part a= Q

    Tf= ((Q*60)/(c*mass))+Ti

    Came from:

    cm(Tf-Ti)/t = eoA(Te^4-T^4) Note: I solved right side in part a and that equals Q.

    Anyway if I get mass this will not be a hard problem however as you said I am missing vital info and this is driving me mad, I have a test on this unit on monday and a final on wensday.

    So I have: Ti=230 celius (object), 25 celius room, surface area = 68 cm^2, emissivity = .78, time = 60 s, density = 3.60*10^3 kg/m^3 and specific heat capacity = 768 j?kg celius. Need mass....
     
  5. Jun 15, 2003 #4
    You see, the only connection between mass and density is volume.
    If D=Density, M=Mass, and V=Volume then :
    D=M/V
    Now, you will either be given the volume, or given a clue that will lead u to what the volume is.
    The only 'clue' that i can see is the area of surface, but this is not enough, since a certain area of surface does not lead to a certain volume unless you know the shape (i mean, you might have a cube with the area of suface 68 cm2 and a sphere with the same area of surface, but they will not both have the same volume !).

    (ok, you already know all of the following in small font, i don't know why i typed it !)
    Now, let's suppose you figured out the volume someway, and called it V. Then:
    M=D*V
    Also remmember that E(nergy)=P(ower)*t(ime)
    P=Q in our question, and t=60 sec
    so :
    E=60Q
    Now, the specific heat is : "the quantity of energy that 1kg of a certain material gains when it is heated 1 celsius degree"
    Remember that the heat energy is proportional with mass, and with [del]T (where T is temprature)
    So :
    E = Sheat*M*[del]T (you can simply memorize this rule, or figure it out)
    In our case :
    60Q = 768*(D*V)*[del]T
    (remember that E = 60Q (in our case) , M=D*V )
    Solve for [del]T
    [del]T = (60*Q)/(768*D*V)
    T2 - T1 = (60*Q)/(768*D*V)
    T2 = (60*Q)/(768*D*V) + 230
    (since T1 = 230)

    Good enough ?


    Back to where we got stuck, the volume !
    Let's suppose that the object is a cube.
    The area of surface of a cube = 8*l2 where l is the length of any side.
    The volume of the cube is l3
    0.0068 = 8*l2
    l2=0.00085
    l=~0.029154759474
    V=l3=~2.478154555309252e-5 m3
    This is just an example, IF the object was a cube.
    EDIT:
    I figured out there is another mistake in your first answer (sorry). The temprature should be first converted to Kelvin before plugging it in the Boltsman Law.
     
    Last edited: Jun 15, 2003
  6. Jun 15, 2003 #5
    Right I think this question is another of my teachers errors so I'm solivng as cube cause no toher shape is givin.

    .0068=8S^2
    S=.02915 m

    Kg/m^3=3600
    kg=3600m^3
    kg=3600(.0291)^3
    kg=0.089 approx

    cm(Tf-Ti)=eo(Ti^4-Te^4)A*t note t=time

    Tf= ((eo(Ti^4-Te^4)At)/c*m)+Ti
    Tf= 313 k = 39.8 celius (Assumming obj is a cube)

    And revised ans for part A was change in Q/s= 16.9 watts

    Btw this forum was a good help, will come back here if I have trouble in the future.
     
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