Assignment 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:

When one substance is placed into a substance of a different temperature, there is thermal energy loss. Energy is transferred from the substance of greater temperature to the other. The equation we use to find the amount of thermal energy loss is:

Energy loss= c*m*dT

Next we use the mass to find out the substance gain per mass unit. We divide the amount of thermal energy loss by the mass of the substance that energy has gone to.

We then use the final and initial temperatures to find out the temperature change. So we now know how much energy is gained by the initial substance as the temperature changes that much.

Finally we divide the amount of energy gained by the substance per mass unit by the temperature change. This will give us the change per unit mass per degree, or specific heat.

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:ok

Your Self-Critique Rating:

3

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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:

Because in the equation PV= nRT all the variables are constant, we can assume that P1V1/T1= P2V2/T2

We want to find T2 so the equation becomes, T2= (P2/P1)*(V2/V1)*T1

I got stuck here because I wasn’t able to figure out what the original pressure was. I think that V2/V1= 9 because I flipped the fraction, and I know that I need to convert the temperature from 20 C to 293.15 K, but I don’t know how to find the original pressure that I need to find the final temperature.

Confidence Assessment: 1

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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:

How did you figure out that the initial pressure was 1 atm? I didn’t see any indication in the problem. Is it supposed to be assumed?

The problem in the text starts with the phrase 'air at atmospheric pressure'.

Your Self-Critique Rating:1

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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:

We first need to change the temperature values from C to K. That gives us and initial temperature of 288.15 K and a final temperature of 311.15 K. We divide T2/T1 to find the factor by which the temperature changed.

dT= 311.15/288.15= 1.08

Since the values are constant, if temperature changes by a certain factor then pressure must change by that same value.

To find the final pressure we multiply the initial pressure by this factor

P2= 220 kPa * 1.08= 237.6 kPa

dP= 237.6-220= 17.6/220= .08

So the air is released at a rate of 8%

Confidence Assessment:2

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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 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 don’t understand the reason why we needed to add 101 kPa to the initial pressure value we have. This really threw me off.

You were given gauge pressure.

The gas laws apply to absolute pressure.

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