14 Dec 07
Originally posted by RamnedIt's no surprise you get different results when you start with different values. The initial question stated the object's density as 0.93, but this post says it's 0.9. (Also, I think the correct density of ethanol is .789, not .806)
The answer is 532 cm^3. turns out I missed it.
I do not know where to head now. I've got the physics part of it down, it's just the math, or so I think! How much ethanol must be added to get the new total density = to the object's density of 900 kg/m^3 (needs converted.)?
So initially, the object's volume can be found looking at the volume it displaces ...[text shortened]... ried it, and also my Physics professor tried it. We all got differing results! I'm stumped.
14 Dec 07
Originally posted by richjohnsonaha maybe I keep using the wrong numbers! I am only 33 off...using 30 too low on density and .007 on ethanol? (I used ethyl alcohol - they are pretty much the same.)
It's no surprise you get different results when you start with different values. The initial question stated the object's density as 0.93, but this post says it's 0.9. (Also, I think the correct density of ethanol is .789, not .806)
I'll check.
18 Dec 07
4 edits
Here is how to do that problem. I used Density of ethyl alcohol, same as ethanol basically.
1. Vw = 500 cm^3 = 5e-4 m^3 = 500 g or .5 kg.
2. De = (Me / Ve) => Ve = Me/ De => Ve = (Me / .806e3)
V = Total mass / total density => V = M / .9e3
3. V = Ve + Vw
V = Ve + (5e-4)
(Substituting...)
(M / .9e3) = (Me / .806e3) + 5e-4
((Me + 5e-4)/.9e3) = (Me / .806e3) + 5e-4
(Solve. The above might come out in wrong units, I can't remember - just needs converted to cm^3.)
Another fluid problem
A 1.00 kg beaker containing 2.00 kg of oil (density = 916 kg/m^3) rests on a scale. A 2.00 kg block of iron is suspended from a spring scale and is completely submerged in the oil (suspended, not touching the beaker.) Find the equilibrium readings of both scales.
19 Dec 07
Originally posted by RamnedRamned, you used a different number for the density of the object than the one given in the question (930 kg/m3). Also, where did you get your value for the density of ethanol (same thing as ethyl alcohol) of 806 g/cm3? Wikipedia reports it as 789 g/cm3.
Here is how to do that problem. I used Density of ethyl alcohol, same as ethanol basically.
1. Vw = 500 cm^3 = 5e-4 m^3 = 500 g or .5 kg.
2. De = (Me / Ve) => Ve = Me/ De => Ve = (Me / .806e3)
V = Total mass / total density => V = M / .9e3
3. V = Ve + Vw
V = Ve + (5e-4)
(Substituting...)
(M / .9e3) = (Me / .80 ...[text shortened]... in the oil (suspended, not touching the beaker.) Find the equilibrium readings of both scales.
I stand by my original solution.
As for this question, we need the density of iron which Wikipedia reports as 7.86 g/cm3 near room temperature, which is 7860 kg/m3.
The block of iron displaces some of the oil in the beaker, which causes a buoyant force to be applied to the iron (reducing the weight registered on the spring scale). This buoyant force is balanced by an equal and opposite force acting down on the fluid in the beaker (increasing the weight on the flat scale). The magnitude of the buoyant force is given by:
Fb = rho(oil) * V(iron) * g
Now, the volume of the 2.00 kg iron block is simply:
V(iron) = m(iron) / rho(iron) = 2.00 / 7860 = 2.5445 x 10^-4 m3
So the buoyant force is:
Fb = 916 * 2.5445 x 10^-4 * 9.81 = 2.2865 = 2.29 N
The reading on the spring scale will be:
W = m(iron)*g - Fb = 2.00 * 9.81 - 2.2865 = 17.335 = 17.3 N
The reading on the flat scale will be:
W = m(beaker)*g + m(oil)*g + Fb = 9.81 * (1.00 + 2.00) + 2.2865 = 31.7165 = 31.7 N
19 Dec 07
1 edit
Originally posted by PBE6I stil say you have to take into account the drop in surface tension since ethanol has a surface tension value lower than water does.(almost 4 times lower)
Ramned, you used a different number for the density of the object than the one given in the question (930 kg/m3). Also, where did you get your value for the density of ethanol (same thing as ethyl alcohol) of 806 g/cm3? Wikipedia reports it as 789 g/cm3.
I stand by my original solution.
As for this question, we need the density of iron which Wikipedia r ...[text shortened]... e will be:
W = m(beaker)*g + m(oil)*g + Fb = 9.81 * (1.00 + 2.00) + 2.2865 = 31.7165 = 31.7 N
Ethanol will float on top of the water so the object will be "floating" on ethanol, not water. So any horizontal forces acting on the object will be reduced and therefore alter the buoyancy value.
http://www.surface-tension.de/
http://www.sciencedirect.com/[WORD TOO LONG]
19 Dec 07
Originally posted by uzlessSurface tension would only apply at the surface of the fluid, but if the object has the same density as the fluid it can float anywhere in the fluid without moving up or down. Also, ethanol is miscible with water so you wouldn't have a distinct layer, you would just have a uniform solution.
I stil say you have to take into account the drop in surface tension since ethanol has a surface tension value lower than water does.(almost 4 times lower)
Ethanol will float on top of the water so the object will be "floating" on ethanol, not water. So any horizontal forces acting on the object will be reduced and therefore alter the buoyancy value.
h ...[text shortened]... ew=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=9470428e97b2d564ed79d9459b7ef9dc
19 Dec 07
2 edits
Originally posted by PBE6Ah, wasn't aware that ethanol wouldn't separate from water. My bad....just figured with the lighter specific gravity ethanol would float.
Surface tension would only apply at the surface of the fluid, but if the object has the same density as the fluid it can float anywhere in the fluid without moving up or down. Also, ethanol is miscible with water so you wouldn't have a distinct layer, you would just have a uniform solution.
Those dam Carbon-Hydrogen atoms foul me up again. Fack you chemistry!
19 Dec 07
Originally posted by RamnedJust like to point out two things:
I rounded 930 to 900 since, when converting to kg /m^3, it makes such little a difference.
Same for using ethyl alcohol. Basically it is the same thing as ethanol (the densities are very similar).
V = Ve + Vw is the route to go. Substitute the values of V = M/D, Ve = Me / De, etc.
1) Ethanol is the IUPAC name for ethyl alcohol. They are the same compound - not similar - the same.
2) Rounding data given in the question is a no-no. It introduces systematic error into the solution.
However, I have to say that I enjoy your puzzles immensely. My criticism is given in the spirit of finding truth, and is not meant to be pedantic or harsh in any way. Keep the puzzles coming! 🙂