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Heat loss experiment

  1. Nov 22, 2015 #1
    1. The problem statement, all variables and given/known data
    We recently did an experiment in class and we were asked to get some ideas together to try and explain our results. We measured the time it took water to fall from 70 degrees to 65 degrees C. The water was placed in a shiny steel container (which had a volume of about 400ml). We half filled it each time and wrapped different materials around it. The first time we used no material (so just the metal can). Then we put bubble wrap around it, then black paper and finally aluminium foil. The results we found are in order of quickest to slowest fall from 70 to 65 degrees: no material, aluminium foil, black paper, bubble wrap.

    2. Relevant equations
    None.

    3. The attempt at a solution
    By doing some research I have some ideas. I think they all should have the same amount of convection loss from the top of the can. I think when no material was used it cooled quickest because the metal can conducted the heat to the outside of the can the quickest which meant a bigger loss from convection around the outside of the can. I'm confused about why the foil was still quick because athletes used foil blankets to reflect the infrared back to their body so shouldn't the foil wrapped around the can have worked in a similar way. Foil has a real low emissivity (0.1) so shouldn't it have retained 90% of the infrared heat from the water? I thought black had a really high emissivity which is why they use it for radiators so shouldn't it have emitted a lot of the heat away from the can?
     
  2. jcsd
  3. Nov 22, 2015 #2

    OmCheeto

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    Excellent experiment!

    Pretty much.

    This sounds logical.

    Sounds very logical, also.
    Three questions:
    A. How tightly do athletes wrap these blankets around themselves, compared to how tightly you wrapped your foil around the container?
    B. Are these blankets really just only foil?
    C. What's the thermal conductivity of aluminum, and how does that compare to its emissivity?

    This is an excellent question.
    You might try googling; Why do they paint radiators black?
     
  4. Nov 22, 2015 #3
    Thanks for your guidance - much appreciated!
    A. I would guess I would have wrapped it tighter than an athlete would. So does this mean that an athlete would have a layer of air in between them and the blanket to serve as an insulator. Therefore it reflects radiation with little conduction. So if it is tighter in the experiment then are you trying to say it will be similar to when there is no material because you are just adding another conducting layer?
    B. Not sure - couldn't find much.
    C. Thermal conductivity of aluminium is very high. Emissivity is very low (0.04).

    I'm still a bit confused though. If the foil has a really low emissivity then very little heat will be radiated away. However, since it is a good conductor there will be a lot of heat loss from around the outside of the can as mentioned earlier. So I guess the loss from conduction to the air is much higher that is doesn't really matter that it has low emissivity? Could you say the same about the black card. The conduction is so slow through the card that not much is lost to convection around the outside and the high emissivity doesn't really contribute much loss? It's so confusing!
     
  5. Nov 22, 2015 #4

    OmCheeto

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    Yes.
    hint: We used to call them "Space Blankets".
    My guess is that your guess is correct, but I've actually never done a calculation of radiant heat loss.
    I don't think your instructor would like that answer; "Everyone unanimously guesses that we are correct".

    But last week I did an experiment that very similar to yours.
    From the data you collected, you should be able to determine, at least roughly, the amount of energy that was lost via radiation.
    I will do the same. (My setup used a glass container and 100 ml of water, heated to boiling.)

    Like your athlete in the aluminum blanket, anywhere the material is not in contact with the steel jar is going to have an air space. And "air" is an exceptional insulator.
     
  6. Nov 23, 2015 #5
    Thank you once again for your help! Could you kindly explain how you would work out the radiant loss? I'm not sure how you isolate this from the convection loss from the outside of the can?
     
  7. Nov 23, 2015 #6

    OmCheeto

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    As I said yesterday, I've never studied radiant heat loss, so everything I am about say, I've learned in the last 12 hours. (Which means the Mentors and HW Helpers should keep a very close eye one me! :biggrin: )

    From wiki's entry on "Thermal Radiation", I found an equation: P = ε ⋅ σ ⋅ A ⋅ T4 (grey body radiation)
    I interpret this as Power(P) = emissivity(ε) ⋅ Stefan-Boltzmann constant(σ) ⋅ area(A) ⋅ temperature(T, in Kelvin)

    From my setup, I was able to determine that more than half of the energy from my system was being lost via radiation!
    Needless to say, I was quite surprised.

    Now you, as I, have a bunch of data which gives the temperature of our setups at specific moments in time.
    From that data, we can determine P(power).
    And given that we have "Power" and "time" we can figure out energy.

    And beings that someone in the past said; "Energy is conserved", I concluded that "total energy loss" - "theoretical radiative energy loss" = "energy loss due to convection".
    Well.... I hope you see it.


    ps. Jimmy87, I've checked out the threads you've started, and have decided that you should be teaching me.
    pps. Mentors, as always: ok to delete, infract, and ban. Or send me a polite reminder, not to have so much fun, attempting to solve student's problems. :redface:
     
    Last edited: Nov 23, 2015
  8. Nov 23, 2015 #7
    I may be being a bit slow but I thought radiative heat loss is the heat loss through infrared radiation (linked to emissivity) and convection is convection and completely different. You have said that radiative energy loss = energy loss due to convection - I don't see how that works? I see it as:

    total heat loss = convection from outside of the can + heat lost from infrared radiation + convection from the top of the can (+ conduction to the table)
     
  9. Nov 23, 2015 #8

    OmCheeto

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    Yes.
    Yes.
    I'm not seeing where I said that.
    ???
    I think we need to slow down here.
    I said; "total energy loss" - "theoretical radiative energy loss" = "energy loss due to convection"
    which is equivalent to
    "total energy loss" = "theoretical radiative energy loss" + "energy loss due to convection".

    Which is what you've basically said. (sans the conduction, et al, of course).
    convection from outside of the can + convection from the top of the can = energy loss due to convection.

    ps. You forgot to mention the heat capacity of your vessel. My vessel accounted for 1/11th of the energy change of my system, as far as I can tell:
    836 joules: water
    84 joules: container

    So I threw that out.
     
  10. Nov 23, 2015 #9
    Ha, what a clown I am! Sorry I didn't see the minus sign! So to find the energy lost to radiation I simply do the following:

    surface area of the can x Boltz constant x emissivity of can's material x temperature to the power of 4 (in Kelvin) of the can.

    I'm guessing the temperature T is the change in temperature? So you then multiply this power by the time it took to create that temperature change to give to total radiative loss? How do you work out the total energy loss from the system? Is it just the energy loss from the water which you calculate using specific heat capacity?
     
  11. Nov 23, 2015 #10

    OmCheeto

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    Yes, and thanks. I fixed the error.*
    No.
    Yes and no, as ambient temperature also comes into play.
    Th-Tc, mass and composition of the system, and a bunch of maths.
    No.

    Sorry to be so succinct, but it's time for my nap.

    ----------------------
    *This will probably be the last homework problem I get involved in. My error rate is way too high. Thank god I've got 6 year old grand-kids.
     
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