QueryAssignment4

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course Phy 232

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.004. `query 4

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Question:

query intro set substance, water, both temperatures and masses known, final temperature known, find spec ht

Explain how the final temperature of the system is combined with the known initial temperatures and masses of both substances to obtain the unknown specific heat

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Your Solution:

For an ideal substance, we know that the total thermal energy of an isolated system is equal to 0. Therefore, each substance involved must cancel out the other. The equation m1*c1*T1 = m2*c2*T2 best describes this scenario. It shows that as one substance gains, the other loses to make up for and balance out the equation. The net energy of the whole system must be zero.

confidence rating #$&*: 3

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Given Solution:

** The change in the thermal energy of an isolated system is 0. So assuming that the systems are isolated the thermal energy change of one object is equal and opposite to that of the other.

For an ideal substance the change in the thermal energy of an object is directly proportional to the change in its temperature and to its mass (i.e., more mass and more change in temperature implies more change in thermal energy). The specific heat is the proportionality constant for the substance. Using `dQ for thechange in the thermal energy of an object we can express this as

• `dQ = mass * specific heat * `dT.

(General College and University Physics students note that most substances do not quite behave in this ideal fashion; for most substances the specific heat is not in fact strictly constant and for most substances changes with temperature.)

For two objects combined in a closed system we have `dQ1 + `dQ2 = 0, which gives us the equation

• m1 c1 `dT1 + m2 c2 `dT2 = 0

or equivalently

• m1 c1 `dT1 = - m2 c2 `dT2.

That is, whatever energy one substance loses, the other gains.

In this situation we know the specific heat of water, the two temperature changes and the two masses. We can therefore solve this equation for specific heat c2of the unknown substance. **

Your Self-Critique: OK

Your Self-Critique Rating: OK

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Question: prin and gen problem 14.3: 2500 Cal per day is how many Joules per day? At a dime per kilowatt hour, how much would this cost?

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Your Solution:

One Calorie is equivalent to 4,200 Joules. Therefore if there are 2,500 Calories per day then there are 10,500,000 Joules.

A kilowatt hour is equivalent to 1,000 J/second because a watt is 1 J/second. Therefore to find how many Joules we need to do, 1,000 J/second*3600 seconds(make up an hour) = 3,600,000 Joules.

To find the number of kilowatt hours we must calculate 10,500,000 / 3,600,000 = 2.92 kW*hr. At a rate of 10 cents per kW*hr, it would cost a total of 29 cents approximately.

confidence rating #$&*:

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Given Solution:

One Cal (with capital C) is about 4200 Joules, so 2500 Cal is about 4200 * 2500 Joules = 10,500,000 Joules.

A watt is a Joule per second, a kilowatt is 1000 Joules / second and a kiliowatt-hour is 1000 Joules / second * 3600 seconds = 3,600,000 Joules.

10,500,000 Joules / (3,600,000 Joules / kwh) = 3 kwh, rounded to the nearest whole kwh.

This is about 30 cents worth of electricity, and a dime per kilowatt-hour.

Relating this to your physiology:

• You require daily food energy equivalent to 30 cents’ worth of electricity.

• It's worth noting that you use 85% of the energy your metabolism produces just keeping yourself warm.

• It follows that the total amount of physical work you can produce in a day is worth less than a dime.

Your Self-Critique: OK

Your Self-Critique Rating: OK

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Question: prin phy and gen phy problem 14.07 how many Kcal of thermal energy would be generated in the process of stopping a 1200 kg car from a speed of 100 km/hr?

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Your Solution:

The first step in this problem is to calculate the speed of the car in m/s. We can do this by converting 100 km/hr into m/s by changing 100 km to 100,000 m and 1 hr to 3,600 seconds. Dividing 100,000 / 3,600 = 27.8 m/s. To find the energy we use the equation K = (.5)(m)(v^2). K = (.5)(1200)(27.8^2) = 462,962 J.

Then what is left is to convert J to kcal. This is done by changing Joules to calories by dividing by 4.186 to get 110,597 cal. Then dividing that number by 1,000 to get kcal, which ends up being 110.6 kcal.

confidence rating #$&*:

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Given Solution:

NOTE: The given solution is based on student solutions for a previous edition of the text, in which the mass of the car was 1000 kg and its initial velocity 100 km/hr. The adjustments for the current mass and velocity (1200 kg and 95 km/hr as of the current edition) are easily made (see the student question below for a solution using these quantities).

**STUDENT SOLUTION for 1000 kg car at 100 km/hr WITH ERROR IN CONVERSION OF km/hr TO m/s:

The book tells that according to energy conservation

• initial KE = final KE + heat or (Q)

• 100km/hr *3600*1/1000 = 360 m/s

INSTRUCTOR COMMENT:

100km/hr *3600 sec / hr *1/ (1000 m / km) = 360 km^2 sec / ( m hr^2), not 360 m/s.

The correct conversion would be 100 km / hr is 100 * 1000 meters / (3600 sec) = 28 m/s or so.

STUDENT SOLUTION WITH DIFFERENT ERROR IN UNITS

Ke=0.5(1000Kg)(100Km)^2 = 5MJ

1Kcal=4186J

5MJ/4186J==1194Kcal

INSTRUCTOR COMMENT:

Right idea but 100 km is not a velocity and kg * km^2 does not give you Joules.

100 km / hr = 100,000 m / (3600 sec) = 28 m/s, approx.

so

KE = .5(1000 kg)(28 m/s)^2 = 400,000 J (approx.). or 100 Kcal (approx). **

STUDENT QUESTION:

The book and this problem originally states *1200kg* as the mass. The solution uses 1000kg, giving the answer as about 100kcal. The book uses 95km/hr, 1200kg, and gets an answer of 100kcal. This problem shows 100km/hr, with a mass of 1000kg in the solution and still an answer of 100kcal. I just want to make sure I am doing it correctly.

Where 26.39m/s should be used, I am using the conversion for this specific problem with 100km/hr = 1000m/1hr =

1hr/3600s = 27.78 ~28m/s.

KE = 1/2mv^2

= ½(1200kg)(28 m/s)^2 = 470,400kg*m/s^2 = 470,400 J

470,400J = 1 cal/4.186J = 1 kcal/1000cal = 112.37 kcal = 112kcal

INSTRUCTOR RESPONSE:

I apparently missed the change in the latest edition of the text; or perhaps I chose not to edit the student solutions posted here.

In any case your solution is good.

Your Self-Critique: OK

Your Self-Critique Rating: OK

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Question: query gen phy problem 14.13 .4 kg horseshoe into 1.35 L of water in .3 kg iron pot at 20 C, final temp 25 C, final temp.

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Your Solution:

Using the formula m1*c1*T1 + m2*c2*T2 = 0 is the easiest way to approach this problem. Except now we will need to add a third part because there are 3 parts to this problem. The equation then becomes m1*c1*T1 + m2*c2*T2 + m3*c3*T3 = 0. 1 can be the horseshoe, 2 can be the water, and 3 can be the pot. We are solving for the initial temperature of the horseshoe.

(.4)*(450)*(25-T1) + (1.35)*(4190)*(5) + (.3)*(450)*(5) = 0.

After calculations we get down to 25-T1 = -160.8. Therefore T1 = 185 C

confidence rating #$&*:

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Given Solution:

** STUDENT SOLUTION BY EQUATION (using previous version in which the amount of water was 1.6 liters; adjust for the present 1.35 liters):

M1*C1*dT1 + M2*C2*dT2 = 0 is the equation that should be used.

0.4kg * 450J/Kg*Celcius *(25celcius -T1) + 0.3kg * 450J/kg*celcius * (25 celcius - 20 celcius) + 1.6kg *4186J/kg*celcius * (25 celcius - 20 celcius) = 0

Solve for T1, T1 = 214.8 Celsius

Solution below is 189.8 C.

GOOD STUDENT SOLUTION:

This problem is very similar to the first video experiment. You can determine in the same way the thermal energy gained by the water. The pot gains a little thermal energy too,and you have to know how many Joules are gained by a kg of the iron per degree per kg, which is the specific heat of iron (give in text). You can therefore find the energy lost by the horseshoe, from which you can find the energy lost per degree per kg.

For this problem I think we are assuming that none of the water is boiled off ass the hot horse shoe is dropped into the water. First I will find the initial temperature of the shoe.

Since water's density is 1g/ml, 1 milliliter of water will weigh 1 gram. So 1.35 Liters of water will have a mass of 1.35 kg.

1.35kg of water is heated by 5 degrees

• The specific heat of water is 4186 J/kg/degrees celsius so 4186 J / kg / C *1.35 kg * 5 C = 28255 J of energy is necessary to heat the water but since it is in equilibrium with the bucket it must be heated too.

mass of bucket = 0.30 kg

• specific heat of iron = 450 J/kg/degrees

• 450 J / kg / C * 0.30 kg * 5 C = 675 J to heat bucket

So it takes

• 675 J to heat bucket to 25 degrees celsius

• 28255 J to heat water to 25 degrees celsius

so the horse shoe transferred 675+28255 = 28930 J of energy.

Mass of horse shoe = 0.40 kg

• horse shoe is also iron

• specific heat of iron = 450 J/kg/degree

• energy transferred / mass = 28930 J / 0.40kg =72,326 J / kg

• 72 330 J / kg, at 450 (J / kg) / C, implies `dT = 72,330 J/kg / (450 J / kg / C) = 160.7 C, the initial temperature of the horseshoe.

A symbolic solution:

m1 c1 `dT1 + m2 c2 `dT2 + m3 c3 `dT3 = 0.

Let object 1 be the water, object 2 the pot and object 3 the horseshoe. Then `dT1 = `dT2 = + 5 C, and `dT3 = 25 C - T_03, where T_03 is the initial temperature of the horseshoe.

We easily solve for `dT3:

`dT3 = - (m1 c1 `dT1 + m2 c2 `dT2) / (m3 c3) so

`dT3 = - (m1 c1 `dT1 + m2 c2 `dT2) / (m3 c3) = - (1.35 kg * 4200 J / (kg C) * 5 C + .3 kg * 450 J / (kg C) ) / ((.4 kg * 450 J / (kg C) ) = -160 C, approx. so

25 C - T_03 = -160 C and

• T_03 = 160 C + 25 C = 185 C, approx..

STUDENT ERROR: MISSING COMMON KNOWLEDGE: Estimate 1.60L of water = 1KG. Could not find a conversion for this anywhere.

INSTRUCTOR RESPONSE: Each of the following should be common knowledge:

• 1 liter = 1000 mL or 1000 cm^3.

• Density of water is 1 gram/cm^3 so 1 liter contains 1000 g = 1 kg.

• Alternatively, density of water is 1000 kg / m^3 and 1 liter = .001 m^3, leading to same conclusion. **

Your Self-Critique: OK

Your Self-Critique Rating: OK

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Question: `q001. Which requires more energy, a 100 kg person climbing a hill 200 meters high or a cup of water heated from room temperature to the boiling point?

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Your Solution:

It takes more energy for the cup of water to boil because using the formula m1*c1*T1 we can find that (1 cup)(4190)(100 C) takes up more energy than a person climbing a 200 meter hill.

confidence rating #$&*:

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Self-Critique Rating: I do not know exactly how to calculate the person’s energy without given a time interval for the 200 meter climb.

@&

You can estimate a person's mass; any realistic estimate will lead to the same conclusion.

The energy required is equal to the potential energy change of that mass. Time has nothing to do with it.

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Question: `q002. A container holds 4 kilograms of water, 8 kilograms of concrete and 500 grams of ice, all at 0 Celsius. The system is heated to 30 Celsius.

Place in order the heat required to raise the temperature of the these three components of the system, given common knowledge and the facxt that the specific heat of concrete is about 1/3 that of water.

At about what temperature would two of these components have gained the same amount of heat?

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Your Solution:

Water: (4)*(4190)*(30) = 502,800 J

Concrete: (8)*(1396)*(30) = 335,200 J

Ice: [(.5)(3.3*10^5) + (.5)(4190)(30)] = 226,850 J

Ice and concrete would have used the same amount of heat at temperature 18.185 C. Setting them equal to find the common temperature.

confidence rating #$&*:

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Self-Critique Rating: OK

@&

Good, though the comparison could have been made without detailed calculations (e.g., twice the mass at 1/3 the specific heat would require less energy).

*@

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Question: query univ problem 18.61 / 18.60 (16.48 10th edition) 1.5 L flask, stopcock, ethane C2H6 at 300 K, atm pressure. Warm to 380 K, open system, then close and cool.

What is the final pressure of the ethane and how many grams remain? Explain the process you used to solve this problem.

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Your Solution:

To find the mass, we can use the ideal gas law formula. First, we must find the number of moles. This can be calculated by rearranging the formula to n = PV/RT. Plugging in the numbers gives us the answer, n = (1.03 *10^5)*(1.5 * 10^-3) / [(8.31)*(380 K)] = .048 mol.

Knowing the molecular weight of carbon and hydrogen helps us find the mass., which totals 12(2) + 1(6) = 30 grams / mol. Now multiplying .048*30 gives us = 1.44 grams.

I am unsure of how to solve for the pressure of the ethane.

confidence rating #$&*:

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Given Solution:

** use pV = nRT and solve for n.

• n = p V / (R T) = (1.03 *10^5 Pa )(1.5 * 10^-3 m^3 ) / [ (8.31 J / (mol K) )(380 K) ] = .048 mol, approx..

If the given quantities are accurate to 2 significant figures, then calculations may be done to 2 significant figures and more accurate values of the constants are not required.

The atomic masses of 2 C and 6 H add up to 30.1, meaning 30.1 grams / mol.

So total mass of the gas is initially

• m(tot) = (.048 mol)(30.1 g/mol)

• m(tot) = 1.4 g

Now if the system is heated to 380 K while open to the atmosphere, pressure will remain constant so volume will be proportional to temperature. Therefore the volume of the gas will increase to

• V2 = 1.5 liters * 380 K / (300 K) = 1.9 liters.

Only 1.5 liters, with mass 1.5 liters / (1.9 liters) * 1.4 grams = 1.1 grams, will stay in the flask.

• The pressure of the 1.1 grams of ethane is 1 atmosphere when the system is closed, and is at 380 K.

As the temperature returns to 300 K volume and quantity of gas will remain constant so pressure will be proportional to temperature.

• Thus the pressure will drop to P3 = 1 atm * 300 K / (380 K) = .79 atm, approx.. **

Your Self-Critique: I now see how the parts were used to calculate the pressure of the ethane. I didn’t take the extra steps to solve for that in my original work.

Your Self-Critique Rating: OK

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Question: univ phy query problem (publisher has omitted this problem from the 12th edition) 18.62 (16.48 10th edition)

A uniform cylinder is .9 meters high, and contains air at atmospheric pressure. It is fitted at the top with a tightly sealed piston.

A little bit of mercury (density 13600 kg / m^3) is poured on top of the piston, which increases the force exerted by the piston. The piston therefore descends, compressing the confined air until the pressures equalize. Mercury continues to be added, further lowering the piston and compressing the air.

If this continues long enough, mercury will spill over the top of the cylinder. How high is the piston above the bottom of the cylinder when this occurs?

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Your Solution:

First we need to find the pressure that the mercury exerts by using rho g y. This comes to equal 13,600*9.8*y = 133,280 N/m^3 * y. Now finding the altitude of the air column, we use atmospheric pressure / total pressure *y. Therefore, plugging in we get 101,000 / (101,000 + 133,280) * y = .433*y. The point at which the mercury spills over the edge is defined as .9m-y. So setting that equation and the previous equation equal, (101,000) / (101,000 + 133,280) * y = .9m*y gives us a final y value of approximately 1.5 m.

confidence rating #$&*:

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Given Solution:

** Let y be the height of the mercury column.

Since

• T and n for the gas in the cylinder remain constant we have P V = constant, and

• cross-sectional area remains constant V = A * h, where h is the height of the air column,

we have P * h = constant. Thus

• P1 h1 = P2 h2, with

P1 = atmospheric pressure = Patm and

h1 = .9 m, P2 = Patm + rho g y.

Mercury spills over when the depth of the mercury plus that of the air column is .9 m, at which point h2 = h1 - y. So the equation becomes

• Patm * h1 = (Patm + rho g y) * (h1 - y).

We can solve this equation for y (the equation is quadratic).

We obtain two solutions:

• one solution is y = 0; this tells us what when there is no mercury (y = 0) there is no deflection below the .9 m level.

• The other solution is

y = (g•h1•rho - Pa)/(g•rho) = .140 m,

which tells us that .140 m of mercury will again bring us to .9 m level.

We might assume that this level corresponds to the level at which mercury begins spilling over. To completely validate this assumption we need to show that the level of the top of the column will be increasing at this point (if the height is not increasing the mercury will reach this level but won’t spill over).

• The level of the top of the mercury column above the bottom of the cylinder can be regarded as a function f (y) of the depth of the mercury.

• If mercury depth is y then the pressure in the cylinder is Patm + rho g y and the height of the gas in the cylinder is Patm / (Patm + rho g y ) * h1. The level of the mercury is therefore

f(y) = Patm / (Patm + rho g y) * h1 + y

The derivative of this function is f ' ( y ) = 1 - Patm•g•h1•rho/(g•rho•y + Patm)^2, which is a quadratic function of y.

Multiplying both sides by (rho g y + Patm)^2 we solve for y to find that y = sqrt(Patm)•(sqrt(g* h1 * rho) - sqrt(Patm) )/ (g•rho) = .067 m approx., is a critical point of f(y).

The second derivative f '' (y) is 2 Patm•g^2•h1•rho^2/(g•rho•y + Patm)^3, which is positive for y > 0.

This tells us that any critical point of f(y) for which y > 0 will be a relative minimum.

So for y = .0635 m we have the minimum possible total altitude of the air and mercury columns, and for any y > .0635 m the total altitude is increasing with increasing y.

This proves that at y = .140 m the total height of the column is increasing and additional mercury will spill over.

To check that y = .140 m results in a total level of .9 m:

• We note that the air column would then be .9 m - .140 m = .760 m, resulting in air pressure .9 / .760 * 101300 Pa = 120,000 Pa.

• The pressure due to the .140 m mercury column is 19,000 Pa, which when added to the 101,300 Pa of atmospheric pressure gives us 120,000 Pa, accurate to 3 significant figures.

The gauge pressure will be 19,000 Pa.

A more direct but less rigorous solution:

The cylinder is originally at STP. The volume of the air in the tube is inversely proportional to the pressure and the altitude of the air column is proportional to the volume, so the altitude of the air column is inversely proportional to the pressure.

If you pour mercury to depth y then the mercury will exert pressure rho g y = 13,600 Kg/m^3 * 9.8 m/s^2 * y = 133,000 N / m^3 * y.

Thus the pressure in the tube will thus be atmospheric pressure + mercury pressure = 101,000 N/m^2 + 133,000 N/m^3 * y. As a result the altitude of the air column will be the altitude of the air column when y cm of mercury are supported:

• altitude of air column = atmospheric pressure / (atmospheric pressure + mercury pressure) * .9 m =101,000 N/m^2 / (101,000 N/m^2 + 133,000 N/m^3 * y) * .9 m.

At the point where mercury spills over the altitude of the air column will be .9 m - y. Thus at this point

• 101,000 N/m^2 / (101,000 N/m^2 + 133,000 N/m^3 * y) * .9 m = .9 m - y.

This equation can be solved for y. The result is y = .14 m, approx.

The pressure will be 101,000 N/m^2 + 133,000 N/m^3 * .14 m = 120,000 N/m^2.

The gauge pressure will therefore be 120,000 N/m^2 - 101,000 N/m^2 = 19,000 N/m^2. **

Your Self-Critique: OK

Your Self-Critique Rating: OK

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Question: query univ phy 18.79 was 18.75 (16.61 10th edition) univ phy problem 16.61 for what mass is rms vel .1 m/s; if ice how many molecules; if ice sphere what is diameter; is it visible?

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Your Solution:

I understand what the question is asking for however I cannot seem to put the pieces together to solve it completely. I am not certain of or familiar with RMS velocity.

confidence rating #$&*:

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Given Solution:

** We can solve this problem knowing that ave KE per particle is 3/2 k T so the .5 m v^2 = 3/2 k T, where v is RMS velocity. Thus

• m = 3 k T / v^2.

From the density of water and the mass of the particle we can determine its volume, which is equal to 4/3 pi r^3. From this we find r.

• We obtain volume m / rho = 3 k T / (v^2 rho), where rho is the density of water.

• Setting this equal to 4/3 pi r^3 we get the equation 4/3 pi r^3 = 3 k T / (v^2 rho). The solution is

r = [ 9 k T / ( 4 v^2 rho) ] ^(1/3).

From the mass, Avogadro's number and the mass of a mole of water we determine the number of molecules.

The following analysis shows the intermediate quantities we obtain in the process. Some of the calculations, which were done mentally, might be in error so you should redo them using precise values of the constants.

At 273 Kelvin we have ave KE = 3/2 k T = 5.5 * 10^-21 Joules.

mass is found by solving .5 m v^2 = 3/2 k T for m, obtaining m = 3/2 k T / (.5 v^2) = 5.5 * 10^-21 J / (.5 * (.001 m/s)^2 ) = 1.2 * 10^-14 kg.

The volume of the sphere is therefore 1.2 * 10^-14 kg / (1000 kg / m^2) = 1.2 * 10^-17 m^3.

Setting this equal to 4/3 pi r^3 we obtain radius r = ( 1.2 * 10^-17 m^3 / 4.2)^(1/3) = ( 2.8 * 10^-18 m^3)^(1/3) = 1.4 * 10^-6 m. Diameter is double this, about 2.8 * 10^-6 m. This is only 3 microns, and is not visible to the naked eye, though it could easily be viewed using a miscroscope.

A water molecule contains 2 hydrogen and 1 oxygen molecule with total molar mass 18 grams = .018 kg.

The 1.2 * 10^-14 kg mass of particle therefore consists of 1.2 * 10^-14 / (.018 kg / mole) = 6.7 * 10^-13 moles. With about 6 * 10^23 particles in a mole this consists of

6.7 * 10^-13 moles * 6 * 10^23 particles / mole = 4* 10^11 particles (about 400 billion water molecules). **

STUDENT COMMENT:

I'm still not sure about the 'visible' thing.

INSTRUCTOR COMMENT:

In any case, visible light has a wavelength between about .4 microns and .7 microns. Nothing smaller than this is visible even in principle, in the sense that its image can't be resolved by visible light.

If we mean 'visible to the naked eye', that limit occurs between 10 and 100 microns.

So this object is in principle visible (wouldn't be hard to resolve with a microscope), but not to the naked eye.

Your Self-Critique: I am unsure about the rho part of the equation. I know we established that it is the density of water, but is it always representative of the density of something? I am also struggling to navigate the DVDs to find this material in the course. Do you know where to find this area of notes so I can study and learn those? After looking through the problem I now see how to find the answer in the given solution.

Your Self-Critique Rating: OK

@&

Just as not every topic is discussed in a typical class, the same is the case with the DVD's. The class notes are intended to introduce the main ideas.

I haven't indexed the DVD's, rather expecting students to treat them as they would a live class, taking notes, etc.. A good set of notes will constitute an index, and more.

It is prefectly legitimate for you to list the principles and techniques you think apply to a situation or a problem, and ask me to critique and/or extend your list.

In this case, though, the material on the DVD's is pretty much out of sequence with parts of this chapter. The material most relevant to this problem is in Class Notes #11.

*@

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Question: `q003. A long U-tube, open at both ends, holds water to a level 40 cm above the bottom of the tube. Oil of density 900 kg / m^3 is added to one side. The oil floats on top of the water, forming an oil column on top of the water column on that side.

• What will be the height of the oil column when the difference in levels of the two sides is 3 centimeters?

• What will be the height of the water column when the water has been completely displaced from one side?

• What will happen if oil continues to be slowly added?

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Your Solution:

If the oil is continually added, then it will continue to push down on the water on the side in which it is added. This will cause the water level on the opposite side to rise more and more.

@&

The addition of the oil on one side increases the pressure at the bottom of the tube. The water on the other side must rise in order to increase the pressure at the bottom of the tube by an equal amount.

When you add the oil it pushes the water on that side a little lower, which causes the water on the other side to rise a little bit. So the level of the water on the side without the oil will be higher than on the side with the oil.

Let point A be at the level of the water in the side with the oil. From the level of point A to the bottom of the tube, how does the change in pressure on one side compare to the change in pressure on the other?

If x is the height of the water above the level of point A on the side without the oil, what are your expressions for the pressure at level A on each side?

Using your expressions, write an equation you can solve for x.

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confidence rating #$&*:

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Self-Critique Rating: I am not quite sure how to calculate the height of the air column in this problem.

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Self-critique (if necessary):

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Self-critique (if necessary):

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Self-critique rating:

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