#$&* course PHY 202 003. query 3 was submitted 10 Jun 2011 around 7:12 PM. 003. query 3If 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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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 ********************************************* Question: prin phy Ch 13.26. Kelvin temperatures corresponding to 86 C, 78 F, -100 C, 5500 C and -459 F. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The Kelvin scale can be determined by adding 273.15 to the Celsius temperature: T(K) = T(C) + 273.15 • 86 C, -100 C and 5500 C are determined as followed: T(K) = 86C + 273.15 = 359.15 K T(K) = -100 C + 273.15 = 159.15 K T(K) = 5500 C + 273.15 = 5773.15 K The freezing point of water is 0 C or 32 F, and a Fahrenheit degree is determined by: T (C) = 5/9 F - 32, Fahrenheit is 5/9 the size of a Celsius degree. • 78 F is determined as followed: T (C) = 5/9 F - 32 = 5/9 (78) - 32 = 26 C above freezing. Since freezing is at 0 C, this means that the temperature is 26 C. The Kelvin temperature is therefore (26 + 273.15) K = 299.15 K. • -459 F is determined as followed: T (C) = 5/9 F - 32 = 5/9 (-459) - 32 = 273 C below freezing. • This is -273 C or (-273 + 273) K = 0 K. • This is absolute zero, to the nearest degree. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ok Your Self-Critique Rating: 3 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Let P1 be the pressure, V1 be the volume, and T1be the temperature of the air in the cylinder before compression. Also, let P2, V2 and T2 be the corresponding quantities after the compression the n we have the equation. 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. Solve for T2. T2 = P2 V2 * P1 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, which is the same as before. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: First we reason this out intuitively: If the air was compressed to 1/9 its original volume and the temperature didn’t change, it would end up with 9 times its original pressure. However the pressure changes from 1 atm to 40 atm, which is a 40-fold increase. The only way the pressure could end up at 40 times the original pressure, as opposed to 9 times the original, would be to heat up. Its absolute temperature would therefore have to rise by a factor of 40 / 9. Its original temperature was 20 C = 293 K, so the final temperature would be 293 K * 40/9, or over 1300 K. Now we reason in terms of the ideal gas law. P V = n R T. In this situation the number of moles n of the gas remains constant. Thus P V / T = n R, which is constant, and thus P1 V1 / T1 = P2 V2 /T2. The final temperature T2 is therefore • T2 = (P2 / P1) * (V2 / V1) * T1. From the given information P2 / P1 = 40 and V2 / V1 = 1/9 so • T2 = 40 * 1/9 * T1. The original temperature is 20 C = 293 K so that T1 = 293 K, and we get • T2 = 40 * 1/9 * 293 K, the same result as before. Your Self-Critique: ok Your Self-Critique Rating: 3 ********************************************* Question: query gen phy ch 13.38 fraction of air released after tire temp increases from 15 to 38 C at 220 kPa gauge YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 1st State: Given: T1 = 288 K, T2 = 311 K T2 / T1 = 311 / 288 = 1.08, which is approximately an 8% increase in temperature. The pressure must therefore rise to: P2 = 3ll / 288 * 321 kPa P2 = 346.7 kPa 2nd State Pressure is then released by releasing some gas, changing the number n of moles of gas in order to get pressure back to 331 kPa. n3 / n2 = P3 / P2 = 321 kPa / 346 kPa = 0.93, which is approximately 7% decrease. So we have to release about 7% of the air. confidence rating #$&*: 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 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. We would in fact expect a temperature change to accompany the release of the air, but this applies only to the air that escapes. The air left in the tire would probably change temperature for one reason or another, but it wouldn't do so as a direct result of releasing the air. STUDENT QUESTION It seems that the air goes from 288 to 311 K, so the ratio should be n2 / n1 = 288 / 311 and the proportional loss should be about (1 - 288 / 311) INSTRUCTOR RESPONSE 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), or about 7.4% of n1. 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), or about 7.2% of n1. The fractions 288/311 and 321 / 346 don't differ by much, not do the percents, but they do differ. Your Self-Critique: ok Your Self-Critique Rating: 3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!