course phy202 ÄžŽ‹Û†±íÕ´p‰ÅxjÊCÈÖ¯¢›Ýz¼ùassignment #012 012. `query 2 Physics II 02-24-2008
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17:06:59 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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RESPONSE --> c=Q/m*dT Knowing the final and initial T allows us to calculate dT
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17:09:23 ** 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 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 specific heat is not in fact strictly constant and for most substances changes with temperature.) We know that what energy one substance loses, the other gains, and we know the specific heat of water , so we can set the two quantities equal and solve for specific heat of the unknown substance. 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. If we know all the quantities except, say, the specific heat c2 of the second substance we can easily solve for c2. **
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RESPONSE --> I understand. The formula I cited was the same one noted in the answer except solved for c instead of dQ.
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17:20:27 prin phy Ch 13.26. Kelvin temperatures corresponding to 86 C, 78 F, -100 C, 5500 C and -459 F.
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RESPONSE --> 86C = 86 + 273.15 = 359.15K 78F = 5/9*(78-32) +273.15 =298.7K -100C= -100 +273.15 = 173.15K 5500C + 273.15 = 5773.15K 5/9*(-459F -32) +273.15 = 35.93K
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17:22:41 The Kelvin temperature is 273 above 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. 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, or since freezing is at 0 C, just 26 C. The Kelvin temperature is therefore (26 + 273) K = 299 K. -459 F is (459 + 32) F = 491 F below freezing, or (5/9 C / F) * (-491 F) = 273 C below freezing, which is -273 C or (-273 + 273) K = 0 K. This is absolute zero, to the nearest degree.
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RESPONSE --> messed up signs on the last one but I understand the conversions.
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17:41:50 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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RESPONSE --> P1/T1=P2/T2 1atm/293K = 9? or 40?atm/T2 T2=2637K? or 11,720 K?
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18:06:48 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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RESPONSE --> P1/T1 =P2/T2 (220kpa+101kpa)/288K = P2/311K P2= (321kpa*311K) / 288K = 346 kpa %dP=25.6 / 321 = 8% To maintain original pressure, volume of air would have to be reduced by the same 8%
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18:11:56 ** T1 = 288 K, T2 = 311 K so T2 / T1 = 311 / 288 = 1.08, approx. This is approx. an 8% increase in temperature. The pressure will end up at 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. You then have to change the number n of moles of gas to get back to 331 kPa, so n3 / n2 = P3 / P2 = 321 kPa / 346 kPa or approximately .93, which would be about a 7% decrease. Note that the results here are mental estimates, which might not be particularly accurate. Work out the process to see how the accurate numbers work out. **
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RESPONSE --> why do we have to work out number of moles of gas? If the pressure increases by 8%, wouldn't the volume of air have to be decreased by the same percentage to maintain the original pressure? P1*V1/T1 = P2*V2/T2 ??
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