#$&* course Phy 122 004. `query 4
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Given Solution: ** PV = n R T so n R / P = V / T Since T and V remain constant, V / T remains constant. • Therefore n R / P remain constant. • Since R is constant it follows that n / P remains constant. ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: why, in terms of the ideal gas law, is T / V constant when only temperature and volume change? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: V/T=nR/P if n and P don’t change then that proportion is constant and this so is T/V confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT ANSWER AND INSTRUCTOR RESPONSE: They are inversely proportional. They must change together to maintain that proportion. INSTRUCTOR RESPONSE: You haven't justified your answer in terms of the ideal gas law: PV = n R T so V / T = n R / P. If only T and V change, n and P don't change so n R / P is constant. Therefore V / T is constant, and so therefore is T / V. You need to be able to justify any of the less general relationships (Boyle's Law, Charles' Law, etc.) in terms of the general gas law, using a similar strategy. ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: prin and gen problem 14.3: 2500 Cal per day is how many Joules? At a dime per kilowatt hour, how much would this cost? Your Solution: 1C=4200J so 4200J*2500C=10,500,000 J 1kilowatt hour=3,600,000J 10,500,000/3,600,000= 3kwh*10c=30cents confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: One Cal (with capital C) is about 4200 Joules, so 2500 Cal is about 4200 * 2500 Joules = 10,500,000 Joules. A watt is a Joule per second, a kilowatt is 1000 Joules / second and a kiliowatt-hour is 1000 Joules / second * 3600 seconds = 3,600,000 Joules. 10,500,000 Joules / (3,600,000 Joules / kwh) = 3 kwh, rounded to the nearest whole kwh. This is about 30 cents worth of electricity, and a dime per kilowatt-hour. Relating this to your physiology: • You require daily food energy equivalent to 30 cents’ worth of electricity. • It's worth noting that you use 85% of the energy your metabolism produces just keeping yourself warm. • It follows that the total amount of physical work you can produce in a day is worth less than a dime. Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: prin phy and gen phy problem 14.07 how many Kcal of thermal energy would be generated in the process of stopping a 1200 kg car from a speed of 100 km/hr? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 100km/h=28m/s KE=1/2mv^2 KE=1/2(1200kg)*(28^2)=470,400J/4186= 112 kcal confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: NOTE: The given solution is based on student solutions for a previous edition of the text, in which the mass of the car was 1000 kg and its initial velocity 100 km/hr. The adjustments for the current mass and velocity (1200 kg and 95 km/hr as of the current edition) are easily made (see the student question below for a solution using these quantities). **STUDENT SOLUTION for 1000 kg car at 100 km/hr WITH ERROR IN CONVERSION OF km/hr TO m/s: KE = .5(1000 kg)(28 m/s)^2 = 400,000 J (approx.). or 100 Kcal (approx). ** STUDENT QUESTION: The book and this problem originally states *1200kg* as the mass. The solution uses 1000kg, giving the answer as about 100kcal. The book uses 95km/hr, 1200kg, and gets an answer of 100kcal. This problem shows 100km/hr, with a mass of 1000kg in the solution and still an answer of 100kcal. I just want to make sure I am doing it correctly. Where 26.39m/s should be used, I am using the conversion for this specific problem with 100km/hr = 1000m/1hr = 1hr/3600s = 27.78 ~28m/s. KE = 1/2mv^2 = ˝(1200kg)(28 m/s)^2 = 470,400kg*m/s^2 = 470,400 J 470,400J = 1 cal/4.186J = 1 kcal/1000cal = 112.37 kcal = 112kcal INSTRUCTOR RESPONSE: I apparently missed the change in the latest edition of the text; or perhaps I chose not to edit the student solutions posted here. In any case your solution is good. Your Self-Critique: OK Your Self-Critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: prin phy and gen phy problem 14.07 how many Kcal of thermal energy would be generated in the process of stopping a 1200 kg car from a speed of 100 km/hr? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 100km/h=28m/s KE=1/2mv^2 KE=1/2(1200kg)*(28^2)=470,400J/4186= 112 kcal confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: NOTE: The given solution is based on student solutions for a previous edition of the text, in which the mass of the car was 1000 kg and its initial velocity 100 km/hr. The adjustments for the current mass and velocity (1200 kg and 95 km/hr as of the current edition) are easily made (see the student question below for a solution using these quantities). **STUDENT SOLUTION for 1000 kg car at 100 km/hr WITH ERROR IN CONVERSION OF km/hr TO m/s: KE = .5(1000 kg)(28 m/s)^2 = 400,000 J (approx.). or 100 Kcal (approx). ** STUDENT QUESTION: The book and this problem originally states *1200kg* as the mass. The solution uses 1000kg, giving the answer as about 100kcal. The book uses 95km/hr, 1200kg, and gets an answer of 100kcal. This problem shows 100km/hr, with a mass of 1000kg in the solution and still an answer of 100kcal. I just want to make sure I am doing it correctly. Where 26.39m/s should be used, I am using the conversion for this specific problem with 100km/hr = 1000m/1hr = 1hr/3600s = 27.78 ~28m/s. KE = 1/2mv^2 = ˝(1200kg)(28 m/s)^2 = 470,400kg*m/s^2 = 470,400 J 470,400J = 1 cal/4.186J = 1 kcal/1000cal = 112.37 kcal = 112kcal INSTRUCTOR RESPONSE: I apparently missed the change in the latest edition of the text; or perhaps I chose not to edit the student solutions posted here. In any case your solution is good. Your Self-Critique: OK Your Self-Critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!