course PHY 202 fЀыpwassignment #013
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20:07:59 query problem 15 introductory problem sets temperature and volume information find final temperature. When temperature and volume remain constant what ratio remains constant?
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RESPONSE --> Given the ideal gas law, PV=nRT where P is the absolute pressure, v=volume, n=number of moles, R=universal gas constant and T is temp in Kelvin, if temp and volume are constant, then NR/P is constant as well.
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20:09:13 ** PV = n R T so n R / P = T / V; since T and V remain constant T / V and therefore n R / P remain constant; since R is constant it follows that n / P remains constant. **
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RESPONSE --> I had nR/P remains constant, which is true..ALSO since R is constant it follows that n/P remains constant as well.
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20:13:51 why, in terms of the ideal gas law, is T / V constant when only temperature and volume change?
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RESPONSE --> Because in order for T/V to remain constant, they are the only two variables that can change, therefore, V1/T1=V2/T2. If Pressure, n, or R are not constant then the above relationship would not be true.
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20:14:39 ** STUDENT ANSWER AND INSTRUCTOR RESPONSE: They are inverselt 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. **
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RESPONSE --> ok
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20:39:11 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?
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RESPONSE --> I cal = 4.186E3 Joules Therefore, 2500(4.186E3)=1.0E7 1kwh=860 calories, thus 2500/860=2.9kwh At .10 an hour. 2.9kwh(.10)=.29 cents a day.
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20:45:56 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. It's worth noting that you use 85% of this energy just keeping yourself warm, so the total amount of physical work you can produce in a day is worth less than a dime.
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RESPONSE --> I understand....these are just conversions which are pretty straightforward. Remember that a watt is a joule per second and kilowatt is 1000 Joules/second.
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21:24:31 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?
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RESPONSE --> W=change in KE, Where w=work, KE =Kinetic energy. Since final KE is zero, W=1/2mv^2), where m=mass in kg and v=velocity in m/s, So, first must convert km/h to m/s; 100km/hr=27.7 m/s, so W=1/2(1200)(27.7^2)=443, 904 J There are 4.186E3 J in one KCal so 443,904J/4.186E3 Kcal=1.0E2
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21:25:17 **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 ** 100 km / hr is 100 * 1000 meters / (3600 sec) = 28 m/s or so. With units your conversion would be 100km/hr *3600 sec / hr *1/ (1000 m / km) = 360 km^2 sec / ( m hr^2). Correct conversion with conversion factors would be 100 km / hr * (1000 m / km) * (1 hr / (3600 sec) = 28 m/s, approx. Otherwise your solution is correct. 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). **
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RESPONSE --> ok
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21:58:00 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 hossshoe
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RESPONSE --> This answer is based on the conservation of heat, or heat lost=heat gained. Given the formula, m1(C1)(Tfinal-T1)+m2(C2)(Tfinal-T2)+m3(c3)(Tfinal-T3)=0 M1=mass of first object in kg C1=Specific heat of first object Tfinal=Final Temp of all objects in closed system T1=initial temp of first object The variables for object two and three will mean the same as those for object 1 with different values, of course. Solving for specifec heat for the first object gives us: c1=-m2(c2)(Tf-T2)-m3(c3)(Tf-T3)/m1(c1(Tf-T1) c1=-(.3)(450)(25-20)-1.35kg(25-20)/.4(25-20)= 13,790 J/kg, which looks high.....
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22:01:58 ** 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 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. 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. **
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RESPONSE --> I thought that the problem in this query asked us to find the specifec heat of the horseshoe, so I based my answer on that, not on the initial temp of the horseshoe. Also, it seems like you would have to account for the mass of iron pot and its specifec heat in the problem as well because this is a closed system.
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T儣ʕڬ~ቋڂ assignment #013 013. `query 3 Physics II 07-03-2007"