course Phy 202 Question: query intro set problem 14 substance, water, both temperatures and masses known, final temperature known, find spec ht
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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 #$&*: 2 ********************************************* Question: prin phy and gen phy Ch 13.30 air at 20 C is compressed to 1/9 of its original volume. Estimate the temperature of the compressed air assuming the pressure reaches 40 atm. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: P1V1/T1=P2V2/T2 Volume is inversely proportional to pressure. To figure the temperature when pressure is 40 atm, we solve for T2. So, T2 = (P2 / P1) * (V2 / V1) * T1. Be sure to convert C to K. = (40/1)*(1/9)*(273) = 1302.2 K confidence rating #$&*: 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 #$&*: 3 ********************************************* Question: query gen phy ch 13.38 fraction of air released after tire temp increases from 15 to 38 C at 220 kPa gauge YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: First of all, the temperature and pressure changes leave the volume and # of moles constant. Then while going to the next state, pressure goes back to its initial value, and the volume will be constant and the moles decrease. T1 = 288 K and T2 = 311 K T2 / T1 = 311 / 288 = 1.07 220kPa= 311 kPa P2 = 3ll / 288 * 321 kPa = 346 kPa ??????????????????I’m a bit confused on how to make this a percent. N3/n2 = P3/P2= 321 kPa / 346 kPa = 0.927 ?????????????????I understand how to get the decimals, but finding the percent of increase or decrease becomes a bit confusing for me. confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: (Note that the given 220 kPa initial gauge pressure imply 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 obtain 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’m a bit confused on how to make the percents ?????????????????I understand how to get the decimals, but finding the percent of increase or decrease becomes a bit confusing for me.