course Phys 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: When only temperature and volume change, we are holding pressure and moles constant (R is already a constant). PV=nRT T/V = P/nR Since P and n are held constant and T/V is equal to that constant this ratio is also constant although the temperature or volume can be changing. It is just changing in a constant ratio. 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: 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 1 kcal = 1 Cal 1 cal = 4.186 J 2500 Cal = 2500 kcal = 2500000 cal * 4.186J/1 cal = 1.05 * 10^7 J 1 kWh = 3.6 * 10^6 J (1.05 * 10^7 J)(1 kWh/3.6 * 10^6 J) = 2.91 kWh * 10 cents = 29.1 cents, about 29 cents/day 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 = 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: My answers dont match up as well because I didnt round as much. My answers did match those in the back of the book though. 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 speed of 100 km/hr? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: W = Fd F = ma W = mad Vf^2 = Vo^2 + 2 ad 100 km/hr = 100000m/hr*1hr/60mins *1min/60 seconds = 28 m/s 0 = (28 m/s)^2 +2ad -771 m2/s2 = 2ad -385 m2/s2 = ad (negative because car is slowing) W = 1200 kg * 385 m2/s2 = 463000 J = 463 kJ * 1kcal/4.186 kJ = 110 kcal Confidence Rating: 3
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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 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: iron horseshoe (h): m = 0.4 kg c = 450 kcal/kg C water (w): V = 1.35 = 1350 mL = 1350 cm^3 = 0.00135 m^3 L c = 4186 kcal/kg C (mass of water: p=m/v 1*10^3 kg/m^3 = m/0.00135 m^3 = 1.35 kg) iron pot (p): m = 0.3 kg c = 450 kcal/kg C Tf= 25 C Ti = 20 C (for water and iron pot that started in equilibrium) heat input = heat output Q=mc deltaT mh*ch*(Tf - Ti)h = [mw*cw*(delta T)] + [mp *cp * (delta T)] -(0.4kg)(450 kcal/kgC)(25-Ti)=(1.35kg)(4186 kcal/kgC)(5C) +(0.3kg)(450kcal/kgC)(5C) - 180 kcal/C (25C-Ti) = 28255.5 J + 675 J -180 kcal/C (25 - Ti) = 28930.5 J 25C Ti = -160.7 C Ti = 185 C Confidence Rating: 3
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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: OK Your Self-Critique Rating: OK ********************************************* Question: query univ problem 18.60 (16.48 10th edition) 1.5 L flask, stopcock, ethane C2H6 at 300 K, atm pressure. Warm to 380 K, open system, then close and cool. What is the final pressure of the ethane and how many grams remain? Explain the process you used to solve this problem. ********************************************* Question: univ phy query problem 18.62 (16.48 10th edition) unif cylinder .9 m high with tight piston depressed by pouring Hg on it. How high when Hg spills over? How high is the piston when mercury spills over the edges? ********************************************* Question: query univ phy 18.75 (16.61 10th edition) univ phy problem 16.61 for what mass is rms vel .1 m/s; if ice how many molecules; if ice sphere what is diameter; is it visible? "