course phy202 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: I don’t understand why P is in the denominator when nR was moved to the left side of the equation
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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: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 1C = 4200J 1W= 1J/sec 1kW = 1000J/Sec 1kWh = 1000J/sec * (60sec * 60min) = 3600000J 25C * 4200J = 10,500,000J per day 10,500,000 perday/ 3600000J per hour = 3kwh .10c * 3kwh = 30 cents. Confidence Rating: 3
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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: ********************************************* 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: M = 1200kg V = 100km/hr = 100km*1000m/km / 60*60 V = 28m/s 1C = 4200J KE = .5mv^2 = .5 (1200kg)(28m/s)^2 KE = 470400J C = 470400/4200 = 112 Confidence Rating: 2
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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: The book tells that according to energy conservation · initial KE = final KE + heat or (Q) · 100km/hr *3600*1/1000 = 360 m/s INSTRUCTOR COMMENT: 100km/hr *3600 sec / hr *1/ (1000 m / km) = 360 km^2 sec / ( m hr^2), not 360 m/s. The correct conversion would be 100 km / hr is 100 * 1000 meters / (3600 sec) = 28 m/s or so. STUDENT SOLUTION WITH DIFFERENT ERROR IN UNITS Ke=0.5(1000Kg)(100Km)^2 = 5MJ 1Kcal=4186J 5MJ/4186J==1194Kcal INSTRUCTOR COMMENT: Right idea but 100 km is not a velocity and kg * km^2 does not give you Joules. 100 km / hr = 100,000 m / (3600 sec) = 28 m/s, approx. so 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 = ½(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: I don’t understand how we got kcal from just C wouldn’t the value needed to be divided by 1000 to get kcal?
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Given Solution: ** STUDENT SOLUTION BY EQUATION (using previous version in which the amount of water was 1.6 liters; adjust for the present 1.35 liters): M1*C1*dT1 + M2*C2*dT2 = 0 is the equation that should be used. 0.4kg * 450J/Kg*Celcius *(25celcius -T1) + 0.3kg * 450J/kg*celcius * (25 celcius - 20 celcius) + 1.6kg *4186J/kg*celcius * (25 celcius - 20 celcius) = 0 Solve for T1, T1 = 214.8 Celsius Solution below is 189.8 C. GOOD STUDENT SOLUTION: This problem is very similar to the first video experiment. You can determine in the same way the thermal energy gained by the water. The pot gains a little thermal energy too,and you have to know how many Joules are gained by a kg of the iron per degree per kg, which is the specific heat of iron (give in text). You can therefore find the energy lost by the horseshoe, from which you can find the energy lost per degree per kg. For this problem I think we are assuming that none of the water is boiled off ass the hot horse shoe is dropped into the water. First I will find the initial temperature of the shoe. Since water's density is 1g/ml, 1 milliliter of water will weigh 1 gram. So 1.35 Liters of water will have a mass of 1.35 kg. 1.35kg of water is heated by 5 degrees · The specific heat of water is 4186 J/kg/degrees celsius so 4186 J / kg / C *1.35 kg * 5 C = 28255 J of energy is necessary to heat the water but since it is in equilibrium with the bucket it must be heated too. mass of bucket = 0.30 kg · specific heat of iron = 450 J/kg/degrees · 450 J / kg / C * 0.30 kg * 5 C = 675 J to heat bucket So it takes · 675 J to heat bucket to 25 degrees celsius · 28255 J to heat water to 25 degrees celsius so the horse shoe transferred 675+28255 = 28930 J of energy. Mass of horse shoe = 0.40 kg · horse shoe is also iron · specific heat of iron = 450 J/kg/degree · 28930 J / 0.40kg =72,326 J / kg · 72,326 J/kg / (450 J / kg / C) = 160.7 C, the initial temperature of the horseshoe. STUDENT ERROR: MISSING COMMON KNOWLEDGE: Estimate 1.60L of water = 1KG. Could not find a conversion for this anywhere. INSTRUCTOR RESPONSE: Each of the following should be common knowledge: · 1 liter = 1000 mL or 1000 cm^3. · Density of water is 1 gram/cm^3 so 1 liter contains 1000 g = 1 kg. · Alternatively, density of water is 1000 kg / m^3 and 1 liter = .001 m^3, leading to same conclusion. ** Your Self-Critique: I don’t understand this problem and need to review my notes more. I understand that the heated horseshoe placed in the water would transfer thermal energy not only to the water but also to the pot. I found the dQ of water and the iron pot then solved for T of horseshoe however I was left with 25-T1 which didn’t make sense. I may have missed a step?