query 4

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course phy 242

june 15 around 9 am

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:

If the system is a close system , there will be a net change in thermal energy which will be 0 , because it is in fact a closed system so therefore there will be no heat transferred with the outside. And so we obtain an equation such as :0 = m1c1(Tf-Ti) + m2c2(Tf-Ti). Now if we know the final temp, intimal temp and teir masses along with one of their specific heat then we can find the specific heat of the other substance.

confidence rating #$&*: 9

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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 heatis the proportionality constant for the substance. Using `dQ for the change 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 c2 of the unknown substance. **

Your Self-Critique: my response is not as detailed as the solution.

Your Self-Critique Rating: 8 out of 10

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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:

For the person hiking : W= F*d = m*a*d= 100*9.81m/s^2*200m= 196,200 joules

For the boiling of water: Q=m*c*dT = 236g*4.186J/g*C* (100-20)C = 79,031.68 Joules

I used 236 g being about the mass of the cup.

Therefore, it requires the person more energy to climb the hill.

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Self-Critique Rating:ok

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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:

Qice= 500g*2.09J/g*C * 30C = 31,350 J

Qconcrete=8000g* 1.39J/g*C * 30C = 333,600 J

Qwater=4000g*4.18 J/g*C * 30C= 501,600 J

At none of the temperature would any two of these components gain the same amount of heat.

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Self-Critique Rating: ok

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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, (1.03 *10^5 Pa) 1 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:

PV= nRT

(1.03 *10^5 Pa)*(1.5*10^(-3))m^(3) =n* 8.314 J/mol*K * 380 K

n= 0.049 moles

0.049moles * 30g/m = 1.47 g

Now we set proportionality between temperature and volume.

V2/V1=T2/T1 : V2/1.5L= 380K/300K

V2= 1.9 L

The ratio between the initial volume and the increased volume time the mass of the substance will give us the remaining grams of ethane: 1.5L/1.9L * 1.47g = 1.2 g

Since all the values are constant except for Pressure and temperature we can set P and T proportionality and solve for final pressure : P1/T1 =P2/T2 1atm/300K= P2/380K P2= 1.27 atm

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 don’t understand why the proportion for the pressure and temperature is P2/P1=T1/T2 as opposed to P1/T1 =P2/T2 .

@&

Clearly the pressure will drop as the closed flask is cooled.

The given solution did not invert any of the proportionalities, but it did base the proportionalities on the fact that gas had escaped at the higher temperature.

I believe your solution attempted to relate the original state to the final state, without considering that n had changed.

*@

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:

PV = nRT, and the temp is constant, so the product PV for the air is always equal to a constant C.

At the beginning, this product is P0*A*h, where P0 is atmospheric pressure, A is the cylinder's cross sectional area, and h is the cylinder height

Then when the air is pressed down to a height x, the product is P*A*x, where P is the new pressure, therefore P0*A*h = P*A*x and P = P0*h/x.

Now let's examine the pressure that the mercury exerts down on the air.

The force downward on the air is F = F_g + F_a, where F_g is the weight of the mercury and F_a is the force of the surrounding atmosphere (atmospheric pressure).

Now F = P*A

And F_g = m*g = p*V*g = p*A*(h-x)*g

Where m, p, and V are the mass, density, and volume of the mercury

And F_a = P0*A

So by dividing everything by A, we have P = p*(h-x)*g + P0

We can set this pressure equal to the pressure we found earlier

P = P0*h/x = p*(h-x)*g + P0

or P0*h/x - P0 = p*(h-x)*g

or P0*(h/x - 1) = p*(h-x)*g

P0*(h-x) = x*p*(h-x)*g

P0 = x*p*g

Therefore x = P0/(p*g)

Now solving for the value: x= 101300 pa/(13600 kg / m^3*9.81m/s^(2))= 0.759 m

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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: 9

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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 .001 m/s; if ice how many molecules; if ice sphere what is diameter; is it visible?

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Your Solution:

KE = 3/2*k*T

½ *m*v^(2) = 3/2*k*T

m = 3/2*k*T/(v^(2)*1/2)

We can change the equation to allow us to solve for volume by dividing both side by the density of water: m/D = 3/2*k*T/(0.5*v^(2)*D)

Now we have V = 3*k*T/(v^(2)*D), in which we can set it equal to the volume of the sphere : 4/3* pi*r^(3) = 3/2*k*T/(0.5*v^(2)*D).

Since we are solving for diameter we must derive r = [ 9 k T / ( 4 v^2 rho) ] ^(1/3).

Now solving for kinetic energy, KE = 3/2 *k*T= 3/2*(1.38*10^(-23))J/K *273 K= 5.65*10^-21 J.

To find how many molecules there are we must solve for mass using the derived equation: m = 3/2*k*T/ (v^(2)*1/2)= (5.65*10^-21 J)/ (0.001m/s^2 *(1/2)) = 1.13 *10^-14 kg= 1.13*10^-11 g

Finding the amount the number of molecules: 1.13*10^(-11) g * (1mole /18 g)= 6.28*10^-13 moles. 6.28*10^-13 moles* (6.02*10^23) = 3.78*10^11 molecules.

Now about the diameter: density of ice at 273 K = 1000 kg/m^3, 1000 kg/m^3= (1.13 *10^-14 kg)/ V, V= 1.13 *10^-17 kg/m^2

V=4/3* pi*r^(3)

1.13 *10^-17 kg/m^2 = 4/3* pi*r^(3)

r = 1.39*10^-6 m

Diameter = 2r = 2.78*10^-6 m

This is not visible to the naked eye.

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: ok

Your Self-Critique Rating: 7

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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:

I have assumed that the water’s height on both side of the u-tube is 40 cm from the bottom.

P=D*g*h

A is what we are looking for.

P = P0+(Dw*g*h2)= P0+(DOil*g*A)

Dw*g*h2=DOil*g*A

So we are given the height of the water and the difference so: h2=A-3cm

Now solving for the derived equation:

Dw*g*h2=DOil*g*A

h2=Doil/Dw *A

Now substituting h2=A-3cm:

A-3cm = Doil/Dw *A

A-3cm= 900/1000 *A

0.1A =3 cm

A= 30cm

The height of the water will be 80 cm since from the beginning it was considered that the water level was 40 cm in both sides so therefore if it was moved to one side then another 40 cm would be added to the current 40 cm level giving 80 cm.

If the oil continues to be added then the water level would increase on one side with the bottom of the water level filled with the oil.

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Self-Critique Rating: ok

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Self-critique (if necessary):

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Self-critique rating:

&#This looks good. See my notes. Let me know if you have any questions. &#