course Phy 202 Question: query problem 15 introductory problem sets temperature and volume information find final temperature.
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Given Solution: ** PV = n R T so n R / P = T / V Since T and V remain constant, T / V 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? ********************************************* Your Solution: Charles’s law states that Temperature and volume are inversely related meaning that when temperature increases volume decreased. This means that the ratio will remain the same. In relation to the ideal gas law, PV=nRT. V/T=nR/P. In order to keep this ratio, only temperature and volume can change, number of moles and pressure must remain constant. Therefore when temperature changes, volume also has to change to maintain the ratio. T/V is equivalent to V/T. Confidence Rating: 3
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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: I am not sure if I have properly justified my answer using the ideal gas law.
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Given Solution: One Cal (with capital C) is about 4200 Joules, so 2500 Cal is about 4200 * 2500 Joules = 13,000,000 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. 13,000,000 Joules / (3,600,000 Joules / kwh) = 4 kwh, rounded to the nearest whole kwh. This is about 40 cents worth of electricity, and a dime per kilowatt-hour. Relating this to your physiology: • You require daily food energy equivalent to 40 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: We used slightly different conversion factors but the math and work seems similar. I also think the way you converted joules to kilowatt hours is more useful than mine. I didn’t realize that it was 1000 joules/second Your Self-Critique Rating: 3 ********************************************* 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 speec of 100 km/hr? ********************************************* Your Solution: 1200 kgs from 100km/hr. You can solve this using the equation: Initial KE= Final KE + heat KE= ˝ mass (velocity)^2 Therefore KE= ˝ (1200kg) (100km/hr)= 60000kg*km/hr We need to convert 100 km/hr to m/sec so we can use it to calculate joules which is: 100 km / hr = 100,000 m / (3600 sec) = 28 m/s Then we solve for KE= ˝ (1200kg) (28m/s)^2 which gives us KE= 470000 J. We then convert Joules to Kcalories using the conversion factor 1 kcal= 4185.8 joules 470000J *1kcal/4185.8 joules = 112 kcal. Confidence Rating: 2
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Given Solution: **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). ** Your Self-Critique: OK Your Self-Critique Rating: OK ********************************************* Question: query gen phy problem 14.13 .4 kg horseshoe into 1.35 L of water in .3 kg iron pot at 20 C, final temp 25 C, spec ht of horseshoe ? ********************************************* Your Solution: ‘dQ= ‘dt *sp heat *mass ‘dt1*sp heat1 *mass1 + ‘dt2 * sp heat2 * mass2=0 Initial steps, you can convert liters of water to kgs of water because 1ml of water is equal to one gram. Therefore, 1.35L of water is equal to 1.35kgs. Also the water temperature changes by So far we have: Sp heat1 * 0.4kg * ‘dt1=1.35kg* 4186 J/kg/C * 5C So from this we can access the energy received by the H20 which would be about 28250J. The iron pot that the water is in will also absorb some energy from the hot horse shoe. The specific heat of iron (found in the text is) 450 J/kg/C the mass is 0.3kg and the ‘dt is 5C. This means that the pot absorbed 675J of energy. In total the water and the pot absorbed 28925J. I was not sure exactly what we were supposed to solve for seeing as initial temp of the horse show and specific heat were not provided so I assumed the horseshoe was iron and solved for initial temperature. Iron has a specific heat of 450 kg/J/C. We can plug this in as 450kg/J/C * 0.4kg * ‘dt =28925J and solve for ‘dt of the shoe. This gives ‘dt=160.7C. Therefore the change in temperature of the horseshoe was 160.7 meaning the initial temp was 160.7C higher than the final temp of 25C. Adding the two gives an initial temp of 185.7C. Confidence Rating: 2
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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 got a higher initial temperature than the correct student answer given which had less water to heat, so I may have made a mistake in my math. I also did not mention the common knowledge background, I just stated it. Your Self-Critique Rating:3