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Thermodynamic Equilibrium and Pressure Questions

  1. Dec 6, 2006 #1
    Hey just two questions i need a little bit of homework help on...they just aren't clicking and maybe im just approaching them the wrong way. A good way to start or any help would be greatly appreciated for them. Thanks :)

    The pressure question: A 0.13 kg balloon is filled with helium (density = 0.179 kg/m3). If the balloon is a sphere with a radius of 4.6 m, what is the maximum weight it can lift?

    at first i was thinking P=F/A and that pressure was equal to density*9.8*height
    but i dont know what the height would be and i cant figure out what else would work???

    The thermo question:
    A ceramic coffee cup, with m = 105 g and c = 1090 J/(kgK), is initially at room temperature (24.0°C). If 175 g of 80.3°C coffee and 12.2 g of 5.00°C cream are added to the cup, what is the equilibrium temperature of the system? Assume that no heat is exchanged with the surroundings, and that the specific heat of coffee and cream are the same as the specific heat of water.

    i kno q=mc(tf-ti) and that the q would be equal so mctf-ti(coffeecup)= mctf-ti(coffee) + mctf-ti(cream)
    and then solving for change in temp to add to the temps already given. but i cant get a change in t.
    Last edited: Dec 6, 2006
  2. jcsd
  3. Dec 6, 2006 #2
    hi, sorry if i had this in the wrong area before but I really need help. Ive looked at these all day and Im not getting the right answers. ANY help would be amazing thanks :)
  4. Dec 6, 2006 #3
    You first want to find what it would take to get a helium balloon to stay still. You need to consider the forces acting on the balloon. If you let it be, you know from experience that the balloon will fly away; what would it take to keep the balloon where it is?

    Imagine a balloon tied to a string, the ballon is still; the force pushing it up is called buoyancy, it is pulled down by its weight and the tension in the string. Mathematically put: buoyancy-weight-tension=total force=0. (the minus signs arise because the weight and tension act in the same direction, which is opposite the buoyancy force).

    What you're really interested in is tension, you should be able to see that the balloon cannot lift anything that has a weight greater than or equal to the tension.

    tension=buoyancy-weight. T=B-mg

    The buoyancy is given by Archimedes's principle: the magnitude of the buoyancy force always equals the weight of the fluid displaced by the object.

    In this case the displace fluid is air, you need to work out how much air is displaced by the balloon
  5. Dec 6, 2006 #4
    im sorry, im still not understanding this that clearly...

    i understand everything up until T=B-mg

    So in solving for the buoyancy force i would need to do:

    so would that be (.13)*(9.8)= 1.274N

    now wouldnt weight be the same thing though?
  6. Dec 6, 2006 #5
    'B=m(air)*g' is right. Your next bit is wrong.

    You need the density of air, for free I'll tell you it's about 1.2 kg/m3. Then you need the volume of air displaced by the balloon.

    The mass of a parcel of air equals the density of air times the volume of the parcel.

    Hint: you might need to find the equation for the volume of a sphere.
  7. Dec 6, 2006 #6
    Thanks sooo much for replying by the way :)

    going from memory the volume of a sphere is 4/3pir^3 right?

    so using the density and radius given then i can multiply those two together to get the mass of the air. Multiply that by g. And that would be my buoyancy force.

    Take that and subract the 1.274 since that is the weight. Is that all correct now?
  8. Dec 6, 2006 #7
    Well I've got an answer here so if you tell me what you get I can tell you if I agree or not.
  9. Dec 6, 2006 #8
    ok i solved for the volume of the balloon and got 407.72
    next i multiplied the volume by the density given (.179) and got 72.98

    B-weight= 72.98- 1.274 = 71.7
  10. Dec 6, 2006 #9
    oh wait just realized i forgot to incorporate g

    so i would do 715-1274?? that doesnt sound right?
  11. Dec 6, 2006 #10
    what are you units for volume? I think you might have got that wrong.

    Also the weight of the balloon, given as 0.13 kg is the weight of the balloon before it is filled with helium. I know this to be true because the weight of helium within a volume as large as this (this is sort of "hotair balloon" scale we're dealing with here) is greater than 0.13 kg.

    The true weight of the balloon is the weight of the unfilled ballon plus the weight of the helium.
    Last edited: Dec 6, 2006
  12. Dec 6, 2006 #11
    volume is 4/3 pi (4.6m)^3 = 407.72m^3?

    so how do i find the weight of the balloon then?
    we seriously never even came close to doing anything like this question in class, i dont understand why they would assign it:(
  13. Dec 6, 2006 #12
    okay, forgive me that volume is right. ooops


    that should get you through now. try and get a final answer on your own and tell me how you got it, then i can help you (or hopefully just tell you it's right!)
    Last edited: Dec 6, 2006
  14. Dec 6, 2006 #13
    i keep getting 715-1274

    i dont understand the part a few posts up where i use the density of air. why would i use that and where?
  15. Dec 6, 2006 #14
    I can see where the 715 comes from, what do you think that is? I'll just tell you to save time, it's the weight of the helium in the balloon, add on the weight of the unfilled balloon to get the weight of the filled balloon.

    Where does the 1274 come from? Remember buoyancy equals the weight of the displace fluid. Do you understand the concept of displaced volume here?
    The displaced volume is simply equal to the volume of the submerged object, in this case the balloon. Multiply displaced volume by air density (I gave you this earlier) to find the mass of the displaced fluid, then multiply by g to get weight. Remember this will equal the buoyancy!

    I have to go to bed now!!
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