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Ocean Level Change Due to Temperature Increase

  1. Aug 25, 2008 #1
    1. The problem statement, all variables and given/known data

    The area of the Earth that is covered by water is 361×10[tex]^{6}[/tex] km[tex]^{2}[/tex], the volume of the water is 1.4×10[tex]^{9}[/tex] km[tex]^{3}[/tex], and the mass of the water is 1.4×10[tex]^{21}[/tex] kg. The density of liquid water, as a function of temperature, can be approximated by ρ = 1008 − T /2 kg/m[tex]^{3}[/tex], where T is in C. If the average temperature of the oceans increases by 2 C, and assuming that the area of the oceans remains roughly constant, calculate the expected rise in the level of the oceans due to the temperature change.

    2. Relevant equations

    P = Po + [tex]\rho[/tex]gh
    PoA +mg = PoA + [tex]\rho[/tex]Agh
    p = m/v

    3. The attempt at a solution

    I tried solving for T since there was no initial temperature provided using p = m/v. I got 1008 - T/2 = 1.4 x 10[tex]^{21}[/tex] / 1.4 x 10[tex]^{12}[/tex] (after correcting the units for volume). This gave me some ridiculous value for T which cannot be correct, and I honestly don't know what to do. No additional equations were provided, but that doesn't mean I can't use them. Any help is greatly appreciated!
  2. jcsd
  3. Aug 25, 2008 #2


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    I don't think solving for T is the right approach. The problem describes the effect that temperature has on density. Then they tell you temperature increases 2 degrees C. If the density changes but the mass remains the same - you know that conservation of mass thing - what does that do to the volume then?
  4. Aug 25, 2008 #3
    I think you should look at using the "ideal gas law".

    That being the case, water has a really good specific heat compared to just about everything else and because of how warm water can "float" on top of cold water, for the temperature of all of the water in the entire world to all warm by 2 degrees C, the temperature of the air would have to get a lot hotter, seriously, way, way hotter than humans could possibly withstand -- we're talking like "holocausts all over the world" temperatures. I know your teacher is trying to give you some good examples to work through, but I think they should have used a swimming pool or a glass of water or something else where that sort of temperature change is actually realistic. Just my two centa.

    Anyway, I'd use the ideal gas law and the molecular weight of water. Of course, if the water was to rise that high, then you'd have much faster surface evaporation. You also have almost 35 grams of salt in every liter of salt water and the salt will affect how quickly the amount of water will expand, since that salt won't expand during the temperature increase, but I think if you just use the molecular weight of water that you'll be fine. ;)
  5. Aug 26, 2008 #4


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    I don't think that is a useful approach, given that the problem given already apparently offers a solution.
  6. Aug 26, 2008 #5
    I know that the volume will change as well. If density decreases, shouldn't volume increase? I'm not sure what to do with this information though. Should I leave T unknown and substitute a T+2 in it's place for the temperature increase of 2 C?
  7. Aug 26, 2008 #6


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    You're asked for the change in height. With the surface area held constant then you know that the percentage change in density will result in the same percentage change in height don't you?

    Figure first the average height from the volume divided by the surface area.
    Then if you apply the percentage change of density to that you should be home free shouldn't you?

    What percentage change in density will a 2 degree C change make?
  8. Aug 28, 2008 #7
    I'm just going to point out a few things. First, you have to convert cubic kilometers into cubic meters. Second, you have to follow the order of operations. I don't know which part you messed up on, but with the figures that you're giving, the density will end up being the density of reguar water (funnily enough) and the temperature will be pretty cool, somewhere between 10 and 20 degrees Celcius --- figure it out yourself. ;)

    [tex]\rho =1008-\frac{T}{2}[/tex] subtract 1008 from both sides
    [tex]2(\rho -1008)=-T[/tex] multiply everything by 2
    [tex]2\rho -2016=-T[/tex] I don't like -T, so let's multiply everything by -1
    [tex]-1(2\rho -2016)=T[/tex] multiply the -1 through
    [tex]-2\rho +2016=T[/tex]
    Now we plug in the numbers, after having manipulated the variables and "solving" the equation, not before.
    [tex]-2(1000)+2016=T[/tex] so T is... ?

    Factor your temperature change back in and you'll find that the oceans will rise . . . well, when you remember to use significant digits, the oceans won't rise at all, according to your wild guesstimation of mass which was rounded off to a thousand, million, million, million places and the volume guesstimation which was only rounded off to a million, million, million places.

    By the way, your "conversion factor" of 1008 only works with pure water and ocean water is really, really, salty -- about 35 grams per liter. I know, that's a lot, but go take a sip, ocean water is incredibly salty. Also, the conversion factor for pure water should probably be:
    as you can't just plug a value into a formula and ignore its units of measurement, that's a good way to trip yourself up later on -- there has to be something that will remove the Celcius from the equation.
  9. Aug 28, 2008 #8


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    I'd observe that

    [tex]\rho_w_a_r_m = 1008-\frac{T+2}{2} = 1007-\frac{T}{2}[/tex]

    Then derive the percentage change in terms of T:

    [tex](\frac{\rho_c_o_o_l}{\rho_w_a_r_m} - 1)*Volume = \Delta Volume = (\frac{1008-\frac{T}{2}}{1007-\frac{T}{2}} - 1) *Volume[/tex]

    Noting the effect of T which is given in C and has a lower bound of 0 C (Otherwise lower is snowball earth?) and an upper bound to be outrageous of say 30 C.

    Given then a range of T of 0 < T < 30 the ratio varies only .000993 to .000978. Given the other rough assumptions of the problem the value .000985 should be plenty close to use to apply to the average height if the area is to be taken as invariant.
    Last edited: Aug 28, 2008
  10. Aug 30, 2008 #9

    Andrew Mason

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    The volume increases by a factor of roughly 1008/1007 or roughly one meter for every kilometre of ocean depth. What is the average depth? Add 1 m for every km of average depth. This necessarily ignores the fact, however, that as the ocean rises, the area of the ocean surface may increase. But that increase in area will be small compared to the total surface area of the oceans.

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