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course PHY 232
9:47 am 2/4
110124What is the weight of a column of water 10 cm high, in a tube of diameter 3 mm?
The weight of the water is the force exerted on it by gravity. Gravity accelerates free objects at 980 cm/s^2 or 9.8 m/s^2. The net force on an object is the product of its mass and the acceleration of gravity. So the force exerted by gravity is the product of its mass and the acceleration of gravity.
The mass of the water is the product of its volume and its density. Its density is 1 g / cm^3 or 1000 kg / m^3.
The volume of the water is the volume of a circular cylinder of diameter 3 mm and altitude 10 cm.
The volume of the cylinder is the product of its cross-sectional area and its altitude.
The cross-sectional area of the cylinder is pi r^2, where r is the radius of the circular cross-section
Thus, working backwards through the above steps (note the resemblance to what should be the familiar process of back-substitution when solving systems of equations):
The cylinder has cross-sectional area pi r^2 and volume pi r^2 h, where h is its altitude.
Its radius is 1.5 mm and its altitude is 10 cm = 100 mm, so its cross-sectional area is pi r^2 = pi * (1.5 mm)^2 = 2.25 pi mm^2, about 7 mm^2.
Its volume is thus area of cross-section * altitude = 7 mm^2 * 100 mm = 700 mm^3, approx..
A cm is 10 mm, so a 1-cm cube can be filled with 10 layers, each of 10 rows, each row with 10 one-mm cubes.
Thus 1 cm^3 = 1000 mm^3, and the approximate volume can also be expressed as 700 mm^3 = .7 cm^3. [ 1 m^3 = 1 000 000 cm^3 (100 layers of 100 rows each with 100 one-cm cubes ), so the cylinder's volume can also be expressed as .0000007 m^3 = 7 * 10^-7 m^3. ]
The mass of .7 cm^3 of water is .7 grams (density is 1 gram / cm^3).
The weight of .7 grams is .7 grams * 980 cm/s^2 = 700 g cm/s^2, approx.. This can be expressed as 700 dynes or .007 Newtons. [ Recall that g cm/s^2 = (.001 kg) * (.01 m) / s^2 = .00001 kg m/s^2 = .00001 Newton, or 10^-7 Newton. ]
How much force is required to support that column?
The column is stationary, so its acceleration is zero.
An object with acceleration zero is in equilibrium.
An object in equilibrium experiences zero net force.
The net force on the column is the sum of all the forces acting on it.
Gravity exerts a downward force of 700 g cm/s^2 or 700 dynes or .007 kg m/s^2 or .007 Newtons.
Something else must therefore be exerting an equal and opposite upward force.
The water column is supported by whatever is underneath it. Whatever that is, it exerts an upward force of 700 g cm/s^2, or .007 Newtons, etc..
If the column is supported by water at the base of the tube, what must be the pressure at the base?
Pressure is force / area.
The base of the tube has cross sectional area 7 mm^2, which is equal to .07 cm^2 (because 10 rows each of 10 one-mm squares will exactly cover a 1-cm square), or since 10 000 cm^2 is one m^2, .000007 m^2 (7 * 10^-6 m^2).
So the pressure is 700 dynes / (.07 cm^2) = 10000 dynes / cm^2.
Alternatively the pressure is .007 Newtons / (7 * 10^-7 m^2) = 1000 Newtons / m^2.
If a bottle containing 500 ml of air and 100 ml of water is sealed at atmospheric pressure, with a tube of diameter 3 mm extending from the water through the cap (to which is it sealed) to a height of 60 cm, by how much must the volume of the air decrease in order to cause water to rise to the height of the tube? Assume that the volume of water that enters the tube is negligible compared to the volume of air in the bottle.
Atmospheric pressure is about 100 000 N / m^.
An analysis equivalent to that about shows that the pressure required to support a 60 cm column of water is about 6 000 N / m^2.
The water is the tube is being pushed down by atmospheric pressure, and upward by the water just under it, which is at the level of the water in the container.
So the surface of the water in the container must be at a pressure 6 000 N / m^2 higher than atmospheric pressure, or 106 000 N / m^2. This requires that the gas pressure in the bottle be 106 000 N / m^2.
Before squeezing the bottle is at pressure 100 000 N / m^2. Squeezing raises the pressure by decreasing the gas volume, resulting in more frequent collisions between gas molecules and the walls of the container (and the water surface).
To find the required volume we use P V = n R T. Air cannot enter or leave the system so n is constant. The temperature will be assumed constant, though your hand might increase the temperature by a couple of degrees. Thus n R T is constant and P V = n R T is constant.
We conclude that P1 * V1 = P2 * V2, so that V2 = V1 * (P1 / P2) = 500 ml * (100 000 N/m^2) / (106 000 N/m^2) = 470 ml, approx..
If the water was to be forced up the tube by changing the temperature of the air, by how much would the temperature have to change? Assume the initial temperature to be 300 Kelvin. Assume at first that the change in volume of the air due to water displaced into the tube is negligible.
In this case V and n are constant, so
P / T = n R / V is constant.
Thus P1 / T1 = P2 / T2 and
T2 = T1 * P2 / P1 = 300 K * 106 000 Pa / (100 000 Pa) = 318 K.
Answer the preceding question without assuming that the volume of water that enters the tube is negligible.
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The volume of water in the tube is 4.24 cm^3, easily obtained from problem statement. My approach is to add this to V1 since the final pressure state is based on this 'final' initial volume. The answer makes sense too:
V2=V1(P1/P2) = 504.24(100000/106000) = 476 mL
So it needs 6 mL less change in volume.
@& Now we need to achieve a pressure of 106 000 Pa, starting at 100 000 Pa and temperature 300 K, while displacing sufficient water to fill the tube.
The volume of 60 cm of tube is pi * (.15 cm)^2 * 60 cm 4.5 cm^3, very approximately.
In this case P, V and T are all changing. Since P V / T = n R, with n constant, P V / T is constant and
P1 V1 / T1 = P2 V2 / T2
so that
T2 = T1 * (P2 / P1) * (V2 / V1)
= 300 K * (106 000 Pa / (100 000 Pa) ) * (504.5 mL / (500 mL) = 321 K, approx..*@
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If a 'pressure tube' was used to indicate the pressure of the system, by what percent would the length of the air column in that tube change between the sealing of the bottle and the raising of the water? Would the answer depend on whether the water column was raised by squeezing the bottle or by heating it?
The air column is separated from the bottle by a 'plug' of water in the tube. When the bottle is sealed both the air column and the air inside the bottle are at atmospheric pressure.
When the bottle is squeezed the force exerted on the 'plug' from the direction of the bottle is greater than the pressure exerted on it by the air column, and it accelerates toward the end of the air column.
This decreases the volume of the air column, causing more frequent collisions between the air in the column and the water column (as well as the sides and end of the tube), resulting in more force on the end of the water column.
The final result is that the pressure in the air column equalizes with the pressure in the bottle, now at 106 000 Pa rather than the original 100 000 Pa.
No air enters or leaves the tube, which remains at room temperature, keeping the air inside at room temperature. So again pressure and volume will be the only variables and we conclude that
V2 = 100 000 Pa / (106 000 Pa) * V1 = .94 V1, approx..
This is about a 6% decrease in volume.
The cross-sectional area of the tube doesn't change, so this entails a 6% decrease in the length of the air column.
Note that 106 000 Pa is 6% higher than 100 000 Pa; however V2 is 100 / 106 = .944, which is only 5.6% less than the original volume. So a 6% increase in pressure results in about a 6% decrease in volume, but the percents aren't quite equal. The volume actually decreases by 5.6%.
At atmospheric pressure, how many moles of air are in the bottle?
P = 100 000 Pa, V = 500 mL = .0005 m^3 and T = 300 K. R = 8.31 Joules / (mol Kelvin) so n = P V/ (R T) = ... = .02 moles.
The translational kinetic energy of the molecules in a mole of gas is 3/2 n R T. How much translational kinetic energy is there in the air within that bottle before it is heated? How much after it is heated?
n = .02 mol, so
KE_trans = 3/2 n R T = 3/2 * .02 mol * 8.31 J / (mol K) * 300 K = 75 J, approx.
After heating the temperature is 318 K and we easily find that the translational KE of the air particles has increased by about 4.5 Joules.
By how much does the gravitational PE of the system change when the water is raised in the tube?
A 60 cm column of water in a tube of diameter 3 mm is easily found to weigh about .04 N.
Its center of mass is 30 cm above the water level in the container.
So its gravitational PE is .04 N * .3 m = .012 Joules.
What is the change in the PE of the system as a percent of the change due to heating in the KE of the air molecules?
.012 Joules is .012 J / (4.5 J) = .003 times the change in translational molecular KE, or about 0.3 %.
We didn't get much PE from the thermal energy we had to put into the system.
Estimate how much force it would require to raise the water to a height of 60 cm by squeezing the bottle. An average adult can exert, with one hand, a maximum squeezing force equal to about 1/4 of his or her weight. Naturally you are stronger than average, in the same way that all the children are above average, so be sure to account for this.
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Not certain about this but this is my reasoning:
The change in pressure is ∆P= 6000 Pa and the area over which that force has to be distributed is the size of your hand. Mine is about 150 cm^2 or 0.0150 m^2 and so the force is 90 N. This doesn't seem like much since it's only 4 lbs and the question did mention being above average in strength.
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@& 90 N is about 20 lb, not 4 lb. A reasonable estimate, and well reasoned.*@
Estimate through what average distance that force is exerted as you tighten your grip, and estimate how much work you require to raise that water. How does your result compare with the energy required to increase the temperature of the air in the bottle?
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My harder squeezes have been about 2-3 cm on each side of the bottle. So W= F*d= 90 * 0.03= 2.7 J. This is just over half of the 4.5 J the system gained after being heated.
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Suppose the plastic walls of the bottle are .07 cm thick. Suppose the contents of the bottle are originally at a stable 300 Kelvin, and the bottle is immersed in water whose temperature is 330 Kelvin. Assuming that for at least a short time after immersion the inside of the bottle remains at 300 Kelvin, while the outside is raised to 330 Kelvin. What is the temperature gradient across the wall of the bottle?
The temperature gradient is just the rate at which temperature changes with respect to position.
Temperature changes by 30 K, while position changes by .7 mm.
What therefore is the temperature gradient, in K / mm?
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30/0.7 = 42.9 K/mm
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What would be the temperature gradient in K / m?
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30/0.0007= 42900 K/m
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What do you estimate to be the surface area of a bottle?
Detail your own estimate. The bottle is approximately cylindrical with diameter 6 cm and altitude 20 cm. That's in the rough neighborhood of 500 cm^2, but get your own result.
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With a volume of 600 mL or 600 cm^3 a reasonable guess would be 8 cm diameter and 12 cm height giving SA≈ 400 cm^2. This is based on my bottle that holds 1 L and is 8 cm by 20 cm.
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If the contents of the bottle are observed to gain 10 000 Joules of energy in 50 seconds, then at what rate with respect to clock time is energy flowing across the walls of the bottle?
The requested rate is by definition equal to change in energy / change in clock time = 10 000 Joules / 50 seconds = 200 Joules / second.
At what rate is energy flowing, per second, per square centimeter?
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(10000)/50/400= 0.5 J/s/cm^2
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At what rate is energy flowing, per second, per square centimeter, per unit of temperature gradient?
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0.5/ (30/0.07)= 0.0012
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@& Good.*@
That last rate is called the thermal conductivity of the plastic. What in the blazes are the units of that quantity? What are its SI units?
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J/s/cm^2/(k/cm) or (J/s)/k/cm based on our calculations which is close to the SI unit of watts/ kelvin-meter since J/s is watts and cm to m is easy to do.
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@&
Rate of energy flow is in Joules / second.
The rate per square centimeters is in
(Joules / second) / cm^2.
The rate per unit of temperature gradient (which to keep units consistent we will express in Kelvins / cm) is therefore
( (J / s) / cm^2 ) / ( K / cm)
= ( (J / s) / cm^2 ) * ( K / cm)
= J / (K s cm)
or
watts / (K cm) .
Watts being an SI unit we're better off expressing our distance in meters, in which case our units are
watts / (K m) = (kg m^2 / s^2) / (K m) = kg m / (s K).
The conductivity calculated from the given data would be
.0012 J / (K s cm),
which is easily converted to SI units:
.0012 J / (K s cm) = .0012 J / (K s (.01 m) ) = .12 J / (K s m).
This result is actually just a little below the range of units for various plastics, most of which have thermal conductivities in the range from .13 to .40 J / (K s m), or watts / (K m).
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Look up the thermal conductivity of plastic. Compare with your result (which, even if you do the problem right, isn't likely to be very close, because that 10 000 J in 50 sec is probably not accurate at all).
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It's 0.25 so about twice what was found above.
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How could we use such a system to measure the conductivity of the plastic that makes up the bottle?
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It'd be easy if you knew the exact temperatures and timed how long it took for the bottle and the water to reach equilibrium and measure the pressure in the bottle at that time and more accurately measure the dimensions of the bottle then you could come pretty close since even with our broad assumptions we were only off by 1/2 and we know the points at which we made decisions that would account for that 1/2 error.
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@& If the system comes to equilibrium the temperature gradient will vary, so k would become one of the constants in a differential equation.
Of course equilibrium occurs in the limit as time approaches infinity, and we don't have that long. So we would probably model the temperature flow between two nonequilibrium temperatures.
Or we would see how much energy is required to maintain some constant thermal gradient.*@
If after raising the temperature to 318 K to fill the tube, we raise the temperature another 20 K, how much water would we be able to collect at the 60 cm height?
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The pressure is 106,000/318=P2/338 --P2= 113000 thus ∆P= 7000
700= F/A =F/ (7x10^-6)--F= 0.049 N
F=rho*V*g ---> V= 0.05 cm^2
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@&
We would not allow P to increase. Rather we would allow the gas to expand.
The gas would expand, starting at 500 mL at 318 K and expanding at contant pressure until the temperature reached 338 K.
The new volume would be
338 K / (318 K) * 500 mL = 1.06 * 500 mL = 530 mL,
an increase of 30 mL.
Thus 30 mL of water (a mass of 30 grams of .03 kg) would be displaced 60 cm vertically.
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By how much would the gravitational PE of the water change?
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The average height through which the additional water travelled is 30 cm so:
∆PE= rho*V*g*y= 0.015 J
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@& 30 g of water has weight .030 kg * 9.8 m/s^2 = .3 N; raised 60 cm its PE therefore increases by .3 N * .6 m = .18 Joules.
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How much additional translational KE would be required?
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∆KE_trans = 3/2 nRT= 3/2 PV---∆KE_t= 0.85 J
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@& That would be 3/2 P `dV.
3/2 P V = 3/2 n R T is the total translational KE, not the change in translational KE.
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@& The .5 mL of gas contains about .02 moles.
`dT for the expansion is 20 Kelvin.
So 3/2 n R `dT = 3/2 * .02 mol * 8.31 J / (mol K) * 20 K = .5 J.
Similarly `dT for the first phase from 300 k to 318 K is 18 K, requiring .45 J to increase the KE of the molecules.
For the entire process the requirement is .95 J, for just the increase in molecule translational energy.
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What is the PE change as a percent of the additional translational KE?
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(∆PE/∆KE)*100= (0.015/0.85)*100= 1.8%
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Most of the air molecules in the container also have rotational KE, which turns out to be 2/3 of their translational KE, on the average. So how much thermal energy must be added to the system between the 300 K and 318 K temperatures, when the volume of the system didn't change by much?
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∆E= KE_tr + KE_rot = KE_tr + 2/3 KE_tr= 5/3 KE_tr
∆T= 18 K ---> KE_tr= 3/2 nRT = 3/2(20)8.31)18= 4487 J
∆E= 5/3(4487)= 7480 J
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@& Most of the air molecules in the container also have rotational KE, which turns out to be
.95 J was added to increase translational KE.
Rotational KE in this diatomic gas is 2/3 as great as translational KE, or about .62 J.
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When a gas is allowed to expand, it must do additional work against atmospheric pressure, equal to 2/3 of the additional translational KE. What therefore is the total thermal energy required to raise the temperature of the system from 318 K to 338 K?
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∆E= KE_tr+KE_rot+∆W_air= 7/3KE_tr = 3.5 nRT = 10500 J
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@& For the 20 degree increase at the end translational KE increased by about .5 J, so the work against pressure being 2/3 of this, the work of expansion would be about 2/3 * .5 J = .33 J.
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How much PE did we gain, as a percent of the thermal energy required to raise the temperature of our gas from 300 K to 318 K?
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1.4 x10^-4 %
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@& PE gain was about .18 J, energy requires was about 2 J, so `dPE is about .18 J / (2 J) = .09 or about 9%.
This actually seems too high, so check my numbers.
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Self-critique (if necessary):
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Self-critique rating:
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@& Good work, but in some cases your reasoning went a little astray. Check my notes (and don't believe my calculations).
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