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A pizza floating in space?

  1. May 18, 2012 #1
    1. The problem statement, all variables and given/known data

    A large, hot pizza floats in outer space after being jettisoned as refuse from a spacecraft. (Suppose the pizza is 60 cm in diameter and 2.0 cm thick, sizzling at 100°C. There's a lot of water in the cheese, but a lot of air in the crust, so we estimate a specific heat and density for the pizza between that of water and that of air. Suppose its specific heat is c = 3000 J/kg · °C and its density is 500 kg/m3. Assume it is dark in the infrared, with emissivity of 0.8.)

    What is the order of magnitude of its rate of energy loss?

    (b) What is the order of magnitude of its rate of temperature change?

    2. Relevant equations




    3. The attempt at a solution

    P= eσ(∏r2)(T4-Tc2)

    e=0.8

    ∏r2= ∏(0.30)2

    P=0.8(σ)(∏(0.30)2)((100+273.15)4-(273.15)4)

    P=177.3

    so magnitude of heat loss= ~10W

    and for b) ~104 as per eqn.

    is this right because there was a lot of information not used?
     
  2. jcsd
  3. May 19, 2012 #2

    gneill

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    Staff: Mentor

    There may be extraneous information for the particular question asked, but it appears that you didn't make full use of the information that you did choose to use :smile:

    Remember that a pizza is a three dimensional object -- it has more than just a top(pings) surface. Also, the background radiation temperature of space is not 0C; It's a tad cooler than that!
     
  4. May 19, 2012 #3

    Shouldn't the background temp be -273.15K (absolute zero)?
    because thats what I had

    But since working in kelvins I had to change the pizza temp from 100°C to (100+273.15)K
    is this right?

    and for the Pizza shape can I use that formula for a volume?

    Because I was not sure since it only had A for surface area.

    If I can it should be more like...

    eσ(∏r2h)(T4-Tc4)

    and did part b seem right?
     
  5. May 19, 2012 #4
    Since you don't have the temperature of the surroundings in space (it varies greatly depending on the position of the Sun), I'll just assume you're finding out the rate of heat loss of the pizza, not the NET rate of heat loss.

    So ignore the temperature of the surroundings and use the formula:

    eσAT^4

    Well then for the rate of heat loss, like gneill said, the pizza is 3D. So treat it as a cylinder and use the surface area in the calculations, not the area of a circle.
     
  6. May 19, 2012 #5

    gneill

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    Staff: Mentor

    Absolute zero is, well, zero in degrees K. -273.15 is its value in Celsius. (Although, if you want to be pedantic you could say that the background temperature of space is 2.7K, the Cosmic Microwave Background temperature, disregarding aberrations due to local sources).
    Yes, that's correct. That's is value in Kelvins.
    Surface area is what you want. It has a top surface, a bottom surface, and sides...
    Part b is where the other information comes into play. The temperature of the pizza depends upon its heat content, specific heat, and mass. The rate of change of the heat content is the radiant heat loss as obtained in part a. So write an expression for the temperature of the pizza with respect to its heat content. Then decide how that temperature will change as the heat content changes (and you have the rate of change of the heat content).
     
  7. May 19, 2012 #6
    So I just have...

    eσ(2∏r(r+h))T4?? ---> where T=373.15

    going from titanium pen's post?
     
  8. May 19, 2012 #7

    gneill

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    Investigate the surface area of a closed cylinder. It has a circular top and bottom, and a cylinder for the sides.
     
  9. May 19, 2012 #8
    I thnk it is 2∏rh+2∏r2?? i don't see how this is wrong?
     
  10. May 19, 2012 #9

    gneill

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    My apologies. I didn't pay close enough attention to the fact that you had factored the expression in your previous post.

    Yes, that's the surface area.
     
    Last edited: May 19, 2012
  11. May 19, 2012 #10
    ok thanks gneill, so this expression is now correct.?
    and for b)

    let ∂= the answer of part a)

    so ∂=mCpizzaΔT

    find the mass by finding volume of pizza and use the density of pizza (given) to find its mass.


    But I have a feeling it involves an integral....

    perhaps evaluate the integral from Ti →Tc

    so maybe ∫mCpizzaT, evaluate with given limits above??
     
  12. May 20, 2012 #11

    gneill

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    The expression for the heat loss rate looks fine.

    No integral will be involved; you're looking for the instantaneous rate of temperature change. That means you want the rate of change of T given the current heat content.

    Start with the expression for the heat content of the pizza (for the given temperature). Then make the heat content and temperature into differential elements. Essentially this is a differentiation with respect to t, or in other words, the rate of change of heat content with respect to the rate of change of temperature. But since you know the rate of change of heat content from (a) ...
     
  13. May 20, 2012 #12
    ohh I think I get it

    Qpizza=mCpizzaT

    so dQ/dT=mCpizzadT.

    where dQ/dT=answer from (a)

    we know Ti=373.15

    solve for Tf

    then see what the difference between Ti and Tf is.?
     
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