course Phys 202 Question: query intro set problem 14 substance, water, both temperatures and masses known, final temperature known, find spec ht
.............................................
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: I used the method of reasoning through the calculation. I could have used the : net change in thermal energy = m1 c1 (delta T) = m2 c2 (delta T) = 0 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. ********************************************* 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: P2 = 40 atm P1V1= P2V2 P1 (9) = (40 atm) (1) P1 = 4.4 atm P1/T1 = P2/T2 4.44atm/293 K = 40atm/ T2 T2 = 2639 K Confidence Assessment: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: I assumed that since the pressure increased 9 times it would have a 9 time fold increase. Not 40 fold (1 atm to 40 atm). I didn’t put the temperature change into my equation. P1V1/T1 = P2V2/T2 T2 = P2V2T1/P1V1 T2 = 40 * 293/ 9 1302 K 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: P1/T1 = P2/T2 P1 = 220 kPa + 101 kPa = 321 kPa 321 kPa/288K = P2 / 311 K P2 = 346.6 kPa 346.6 kPa - 101 kPa = 245.6 kPa 25.6 kPa increase/220 kPa = 11.7 % increase need to let out 11.7% of air Confidence Assessment: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 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 calculated the first state to find the pressure after the temperature change correctly. I divided the wrong values to get the increase. I should have divided the initial pressure by the new pressure: 321 kPa/ 346 kPa = 0.93 This gave my a 7% increase in pressure which would require me to release 7% of the air. Your Self-Critique Rating: 3 ********************************************* Question: query univ phy 17.116 (15.106 10th edition) 1.5 * 10^11 m, 1.5 kW/m^2, sun rad 6.96 * 10^8 m. ********************************************* Question: univ phy 17.115 time to melt 1.2 cm ice by solar radiation 600 w/m^2, 70% absorption, environment at 0 C. "