Assignment3_OpenQuery

course Phy 202

6/7 10pm

If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution:

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.

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Question: query intro set problem 14 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:

The specific heat of a substance, denoted as c, is the amount of energy it takes to raise one gram of the substance by one Celsius degree, and is usually given in the units Joules per gram per Celsius degree, or J/(g*C).

The amount of heat gained or lost by a substance is denoted as Q, and can be found by multiplying the mass of the of the substance, m, by the specific heat, c, and by the change in temperature, which is denoted by ‘dT and can be calculated by subtracting the initial temperature from the final temperature: ‘dT = Tf – T0. The equation is Q = m*c*’dT.

In a closed system, the amount of heat gained by a substance equals the amount of heat lost by the other substance in the system. As such, Q1 + Q2 = 0, and Q1 = -Q2. We can set the heat amounts for each substance equal but opposite to each other, as follows:

m1*c1*(Tf-T1) = -(m2*c2*[Tf-T2]), or m1*c1*’dT1 = -(m2*c2*’dT2)

Assuming the second substance lost heat, then ‘dT2 would be a negative, so we can apply the negative sign on the right to ‘dT2 to make it positive.

Normally, we would be given enough information for all except 1 of the variables, in this case c2. We can use the above equation, arrange it algebraically, and solve for c2 to determine the unknown specific heat:

c2 = (m1*c1*’dT1)/(m2*’dT2)

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 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: OK

Your Self-Critique rating #$&* OK

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Question: prin phy Ch 13.26. Kelvin temperatures corresponding to 86 C, 78 F, -100 C, 5500 C and -459 F.

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Your Solution: N/A

confidence rating #$&* N/A

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Given Solution:

The Kelvin temperature is 273 K higher than the Celsius temperature (actually 273.15 below, but the degree of precision in the given temperatures is not sufficient to merit consideration of tenths or hundredths of a degree).

• 86 C, -100 C and 5500 C are therefore equivalent to ( 86 + 273 ) K = 359 K, -100 + 273 K = 173 K, (5500 + 273) K = 5773 K.

The freezing point of water is 0 C or 32 F, and a Fahrenheit degree is 5/9 the size of a Celsius degree. Therefore

• 78 F is (78 F - 32 F) = 46 F above the freezing point of water.

• 46 Fahrenheit degrees is the same as (5/9 C / F ) * 46 F = 26 C above freezing.

• Since freezing is at 0 C, this means that the temperature is 26 C.

• The Kelvin temperature is therefore (26 + 273) K = 299 K.

Similar reasoning can be used to convert -459 F to Celsius

• -459 F is (459 + 32) F = 491 F below freezing, or (5/9 C / F) * (-491 F) = 273 C below freezing.

• This is -273 C or (-273 + 273) K = 0 K.

• This is absolute zero, to the nearest degree.

Your Self-Critique: N/A

Your Self-Critique rating #$&* N/A

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Question: prin phy and gen phy Ch 13.30 air at 20 C is compressed to 1/9 of its original volume. Estimate the temperature of the compressed air assuming the pressure reaches 40 atm.

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Your Solution:

Boyle’s law tells us that the pressure of a gas is inversely related to the volume of the gas, meaning that if the volume of the gas decreases, the pressure of the gas increases, if the amount and temperature of the gas are kept constant. The equation is described as P*V = constant.

Charles’ law tells us that the volume of the gas is directly related to the temperature of the gas, meaning that the volume of the gas will increase if the temperature increases and vice versa, keeping the amount and pressure of the gas constant. The equation is described as V/T = constant. I note here that T has to be measured in kelvins.

If the amount of gas is kept constant, combining the above laws indicates that the combination of the pressure, volume and temperature of the gas should be the same before and after any changes. Setting the two scenarios equal to each other, we can use the equation P1*V1/T1 = P2*V2/T2, and solve for the unknown, T2 = (P2*V2*T1)/(P1*V1).

The problem in the book states that the original pressure is atmospheric pressure, so I am estimating that P1 = 1.0 atm. Since the volume is compressed to 1/9 the original size, we can assume that V1 = 9*V2. The original temperature is 20 C + 273 K per C= 293 K. Substituting these and the remaining given values, we can estimate the final temperature to be the following:

T2 = (40 atm * V2 * 293 K) / (1 atm * 9*V2) V2 and atm cancel

T2 = 40 * 293 K / 9 = approximately 1300 K

The compressed air has a temperature of about 1300 K.

confidence rating #$&* 3

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Given Solution:

First we reason this out intuitively:

If the air was compressed to 1/9 its original volume and the temperature didn’t change, it would end up with 9 times its original pressure.

However the pressure changes from 1 atm to 40 atm, which is a 40-fold increase.

The only way the pressure could end up at 40 times the original pressure, as opposed to 9 times the original, would be to heat up. Its absolute temperature would therefore have to rise by afactor of 40 / 9.

Its original temperature was 20 C = 293 K, so the final temperature would be 293 K * 40/9, or over 1300 K.

Now we reason in terms of the ideal gas law.

P V = n R T.

In this situation the number of moles n of the gas remains constant. Thus P V / T = n R, which is constant, and thus P1 V1 / T1 = P2 V2 /T2.

The final temperature T2 is therefore

• T2 = (P2 / P1) * (V2 / V1) * T1.

From the given information P2 / P1 = 40 and V2 / V1 = 1/9 so

• T2 = 40 * 1/9 * T1.

The original temperature is 20 C = 293 K so that T1 = 293 K, and we get

• T2 = 40 * 1/9 * 293 K,

the same result as before.

Your Self-Critique: OK

Your Self-Critique rating #$&* OK

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Question: query gen phy ch 13.38 fraction of air released after tire temp increases from 15 to 38 C at 220 kPa gauge

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Your Solution:

A tire filled with air has the same volume, and we are intending to keep the pressure the same as well, so we can hold this constant for the scenario. The only variables changing are the number of moles and the temperature.

We can start with the ideal gas law, P*V = n*R*T, and solve algebraically so that the variables are on one side and the constants are on the other: n*T = P*V/R.

The equation n*T = constant indicates that if the temperature of the tire increases, then the number of moles need to decrease to maintain the same constant. As such, n1 * T1 = n2 * T2, and we solve for the final number of moles: n2 = n1*T1/T2.

The temperature of the tire increased from 15 C + 273 = 288 K to 38 C + 273 = 311 K. Substituting these temperatures in the equation above, we get:

n2 = n1 * 288K/311k = (288/311)*n1

If the temperature of the tire increases from 288 K to 311K, then (1 – 288/311) = 23/311, or about 7.3%, of the air needs to be released to maintain the same volume and pressure.

confidence rating #$&* 3

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Given Solution:

(Note that the given 220 kPa initial gauge pressure implies an absolute pressure of 311 k Pa; assuming atmospheric pressure of about 101 k Pa, we add this to the gauge pressure to get absolute pressure.

Remember that the gas laws are stated in terms of absolute temperature and pressure.

The gas goes through three states. The temperature and pressure change between the first and second states, leaving the volume and the number n of moles constant. Between the second and thirdstates pressure returns to its original value while volume remains constant and the number n of moles decreases.

From the first state to the second:

T1 = 288 K, T2 = 311 K so T2 / T1 = 311 / 288 = 1.08, approx.

This is approx. an 8% increase in temperature. The pressure must therefore rise to

P2 = 3ll / 288 * 321 kPa = 346 kPa, approx

(note that we have to use actual rather than gauge pressure so init pressure is 220 kPa + 101 kPa = 321 kPa, approx. )

From the second state to the third, pressure is then released by releasing some gas, changing the number n of moles of gas in order to get pressure back to 321 kPa. Thus

n3 / n2 = P3 / P2 = 321 kPa / 346 kPa or approximately .93, which would be about a 7% decrease. So we have to release about 7% of the air.

Note that these calculations have been done mentally, and they might not be particularly accurate. Work out the process to obtain the accurate numerical results.

Note also that temperature changes from the second to third state were not mentioned in the problem; in reality we would expect a temperature change to accompany the release of the air.

Your Self-Critique:

I arrived at the same end result (about 7% of air being released), but I went about it in entirely a different direction.

?????Should I have gone through the 3 states individually as outlined in the given?????

Your Self-Critique rating #$&* 2

The Kelvin temperature goes from 288 K to 311 K.

If the air is released at constant pressure, then volume and pressure remain constant while temperature and number of moles vary according to

n T = P V / R

so that

n1 T1 = n2 T2, and
n2 = n1 * (T1 / T2) = n1 * (288 / 311)

and the change in amount of gas is n1 - n2 = n1 - 288/311 n1 = n1( 1 - 288 / 311).

If the temperature is first raised to 311 K, then the gas is released, it is the pressure and amount of gas that change. In that case the change in the amount of the gas is n1 - n3 = n1 - (321 kPa / 346 kPa) * n1 = n1 ( 1 - 321 / 346).

The fractions 288/311 and 321 / 346 don't differ by much, but they do differ.

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Question: query univ phy 17.116 (15.106 10th edition) 1.5 * 10^11 m, 1.5 kW/m^2, sun rad 6.96 * 10^8 m.

How did you calculate the total radiation of the Sun and how did you use this result to get the radiation per unit area?

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Your Solution: N/A

confidence rating #$&* N/A

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Given Solution:

Outline of solution strategy:

If we multiply the number of watts per unit of area by the surface area of the Sun we get the number of watts radiated from the Sun.

The energy flows outward in a spherically symmetric manner; at any distance the entire power is distributed over the radius of a sphere concentric with the Sun and of radius equal to the distance.

So if we divide that number of watts by the area of a sphere whose radius is equal to that of the Earth’s orbit, we get the number of watts per unit of area at that distance.

This strategy is followed in the student solution given below:

Good student solution:

Surface area of sphere of radius r is 4 pi r^2; if flux intensity is I then flux = 4 pi r^2 I.

When r = 1.5 * 10^11 m, I = 1500 W / m^2, so the flux is 4 pi r^2 I = 4 pi * (1.5 * 10^11 m)^2 * 1500 W / m^2 = 4.28 * 10^26 watts.

4.28055 x 10 ^ 26 W / (4*`pi * (6.96 x 10 ^ 8 m)^2) = 4.28055 x 10 ^ 26 W / 6.08735 x 10 ^ 18 m^2 = 70318775.82 J/s/m^2 = 7.03 x 10 ^ 7 J/s/m^2

If the sun is radiating as an ideal blackbody, e = 1, then T would be found as follows:

H = `dQ/`dt = 4.28055 x 10 ^ 26 W = (4*`pi * (6.96 x 10 ^ 8 m)^2) * (1) * (5.67051 x 10^-8 W/m^2*K) * T^4

So T^ 4 = 4.28055 x 10 ^ 26 W / 6.087351 x 10 ^ 18 m^2) * 1 * (5.67051 x 10^-8 W/m^2*K)

T^4 = 1.240 * 10 ^ 15 K ^4

T = 5934.10766 K on surface of sun. **

Your Self-Critique: N/A

Your Self-Critique rating #$&* N/A

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Question: univ phy 17.115 time to melt 1.2 cm ice by solar radiation 600 w/m^2, 70% absorption, environment at 0 C.

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Your Solution: N/A

confidence rating #$&* N/A

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Given Solution:

** Thermal energy is not radiating in significant quantities from the ice, so only the incoming radiation needs to be considered, and as stated only 70% of that energy is absorbed by the ice..

• 70% of the incoming 600 watts/m^2 is 420 watts / m^2, or 420 Joules/second for every square meter if ice.

• Melting takes place at 0 C so there is no thermal exchange with the environment. Thus each square meter absorbs 420 Joules of energy per second.

We need to consider the volume of ice corresponding to a square meter. Having found that we can determine the energy required to melt the given thickness:

• A 1.2 cm thickness of ice will have a volume of .012 m^3 for every square meter of surface area; the mass will be close to 1000 kg/m^3, so there are about 12 kg of ice for every m^2 of surface (you can obtain a more accurate result by using the a more accurate density; the density of ice (which floats in water) is actually somewhat less than that of water).

• It takes about 330,000 Joules to melt a kg of ice at 0 C, so to melt 12 kg requires around 4,000,000 J. At 420 Joules/sec this will require roughly 10,000 seconds, or around 3 hours.

All these calculations were done mentally and are therefore approximate. You should check them yourself, using appropriately precise values of the constants, etc. **"

&#Good work. See my notes and let me know if you have questions. &#

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