Query Asst 3

course Phy 202

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:

Specific heat is the amount of energy that is required to convert 1g of substance by 1 degree Celsius.

Change in thermal energy is represented by the equation:

Change in T Energy= ‘dt * sp. Heat * mass

When you have two substances interacting as in the situation above, you can set two of these equations equal to each other. This is because the heat lost by one object will be picked up by the other object if this is an isolated system. This is set up as :

‘dt1 * sp. Heat 1 * m1= -‘dt2 *sp heat2* m2.

We can then use this to solve the problem. Since we know waters specific heat, we can plug that in for sp heat 2. We also know the masses for both substance and the change in heat for both substances, so we can then solve for sp. Heat 1.

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Confidence Assessment: 2

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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: I think I explained the problem pretty well, but the steps are drawn out better in the above solution.

Your explanation is good. My explanations have been through a number of revisions, and if they aren't pretty wee refined by now you should certainly let me know.

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:

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Confidence Assessment:

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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 = 373 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:

Your Self-Critique Rating:

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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:

P1V1/T1=P2V2/T2

T2(P1V1/T1)= P2V2

T2=P2V2T1/P1V1

P2/P1=40

V2/V1= 1/9

So 40atm (1/9) (293K)= 1302 K.

The final volume is 1302K.

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Confidence Assessment:2

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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 a factor 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: Intuitively, we could figure this out by realizing that there was a 40 fold increase in pressure when it should have only been a 9 fold increase. We could then think about how pressure and temperature are related and realized we need to heat the system up by 40/9 in order to get this result.

Your Self-Critique Rating: 3

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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:

PV=nRT

R is constant and volume is constant so P=nT or n=P/T

220kPA/228K=n1 which is 0.96mols.

P / T would be proportional to the number of moles, provided only P and T were varying.

However n = P V / R T. R is of course a universal constant, but to get the number of moles you would need to know V a well as P and T.

In this situation, however, P does not remain constant.

We then need to find the amount mols that would result from increasing the temp to 311K.

220kPA/311K=0.71 mols.

If the temperatures were 228 K and 311 K, then provided volume is constant, the number of moles would be in the proportion .96 / .71. However 220 kPa is the initial gauge pressure, and the pressure cannot be assumed constant.

This shows that in order to keep the kPas at 220, we need to release enough air to reach .71 mols. We can find the amount of air by subtracting .71 moles from .96 moles which gives 0.26 moles. Therefore we need to release .26 moles of air in order to keep the gauge at 220 kpas.

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Confidence Assessment: 2

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Given Solution:

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 third states 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 331 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 botain 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 understand the math that was done above, but was wondering if my solution worked also.

Your Self-Critique Rating: 2

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