course PHY 202 \K|ڍt\¸B۹assignment #012
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21:07:58 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 --> Given the formula: m1(c1)(Tf-T1)+m2(c2)(Tf-T2)=0 Where m1=mass of first substance, c1=specifec heat of first substance, T1=initial temp of substance 1 Tf=final temp of both substances(thermal equilibrium), m2=mass of second substance, c2=specifec heat of second substance, Tf=final temp of both substances, T2=initial temp of second substance. The above equation can be manipulated to find any unknown, including the specifec heat.
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21:08:31 ** 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 --> ok
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21:33:49 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 --> I think I would derive the answer using, PV/nR=T where P=final pressure, V=initial volume, n=number of moles, R=8314J/mol*K(universal gas constant) so: 40(.1111(22,4)/1.0(8314)=.01193. ? Not sure if I did that correctly.
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21:38:31 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 --> Given the formula: P2=P1(T2/T1) Where P1= 220 kPa gauge(initial pressure) T2=Final Temp T1= Initial Temp 220(38-15)=5,060
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21:43:04 ** 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 --> First of all, change Celsius to Kelvin. Also, when working with gauge pressure and ideal gas law, must add atmospheric pressure(101kPA) to get absolute pressure. Finally, convert n of moles of gas to get back to kPa. Must review Problem Solving with Ideal Gas Law again, to refresh my memory on the conversions that are necessary to use it effectively.
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؎mzߘǣz assignment #012 012. `query 2 Physics II 06-28-2007"