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Click
here for the real exam for Autumn 2002!!
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From Au99 |
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1. (20 points) A car tire has an initial gauge pressure of exactly 32 lbs/in2 on a cold morning when the initial temperature is exactly 0 0C and the initial internal volume of the tire is 6.00 x104 in3 . What is the final gauge pressure at high noon when the tire has been heated up to exactly 400C and the volume has increased to 6.15x104 in3? Take atmospheric pressure to be 14.70 lbs/in2. Assume no air leaks from the tire. |
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Solution Outline: Use PiVi / Ti = PfVf
/ Tf . Remember to convert all temperatures to Kelvin! You
start the problem by converting Pi to absolute pressure: 32 lbs/in2
+ 14.7 lbs/in2. Find Pf and then
subtract to get the final gauge
pressure: Pf - 14.7 lbs/in2. Note that Patm
= 14.7 lbs/in2. You add this to gauge pressure to get absolute
pressure, and you subtract this from absolute to get gauge
pressure!! Note that Patm = 14.7 lbs/in2 = 1.013 x 105 N/m2 |
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2. (20 points) A 0.100-kg ingot of metal at temperature 200.0 0C is dropped into a beaker containing 0.800 kg of water initially at 20.0 0C. The final equilibrium temperature of the system is 22.4 0C. (Ignore the heat transfer to the beaker.) (a) (17 points) Find the specific heat of the metal. Assume the specific heat of water is (b) (3 points) What is the magnitude of the heat transferred to the water as the ingot cools down ? |
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Solution Outline: (a) This is easy. (b) The magnitude can be written in two ways, both of which will give you the same numerical answer since they are equal. The notation below corresponds to the example in the book about the metal dropped into a beaker : |mxCx(Tix - Tfx)| or |mwCw(Tfw - Tiw)| |
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3. (26 points) An ideal gas at an initial pressure of exactly Pi = 1.0 atm and an initial temperature Ti = 273 K is taken through a process of expansion. The volume increases from an initial value Vi = 5.0 L to a final value of exactly Vf = 10 L. During this process, the pressure P varies as the volume squared according to the following formula: P = 0.5·a·V2 Note: 1 L = 1x10-3 m3 , 1 atm = 1.013x105 N/m2 and R = 8.31 J/mol·K . (a) ( 3 points) Find the constant a. (f) (4 points) What is the change in internal energy (in joules) during the process? Assume an ideal diatomic gas and that the molecules translate, rotate, but do not vibrate. |
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Solution Outline: Parts (a), (b), (c), (d) and (e) are quiz problems. Use the same methods as you did on the quiz, except you are integrating V2 NOT V-2 . (f) delta U = (5/2)nR*( Tf - Ti ). Use the temperatures that you were given ( Ti ) and that you got in part (c). Note 5/2 comes from 5 degrees of freedom: 3 translational and 2 rotational! (g) Q = delta U + W |
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4. (7 points) Extra Credit (From Au 98)
A 1.5 kg iron (fe) horseshoe initially at 1200 0C is dropped into a bucket containing 1.0 kg of water at exactly 250C. The final equilibrium temperature of the system is 100 0C exactly. The specific heat of iron = 448 J/(kg 0C). The specific heat of water = 4186 J/(kg 0C). The latent heat of vaporization for water = 2.26x106 J/kg . (a) (2 points) What is the entropy change for the horseshoe ? (b) (1 point) How many grams of water boil away by the time the horsehoe
reaches 100 0C? (d) (2 points) What is the entropy change for the water as it boils away ? |
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Solution Outline: (a) Delta S = Mfe*Cfe*ln[(100+273)/(1200 +273)] . Use absolute temperature!! I got this by integrating dS = dQ/T= Mfe*CfedT/T from the initial temp. to the final temp. for the horseshoe. (b) MwCw(100 - 25) + m*LF = Mfe*Cfe*(1200 - 100). Solve for m. Note you do not have to use absolute temperature if you are taking the difference of temperatures!! (c) Delta S = Mw*Cw*ln[(100+273)/(25 +273)] . Use absolute temperature. I got this by integrating dS = dQ/T= Mw*CwdT/T from the initial temp. to the final temp. for the water. (d) Integrate this dS = dQ/T to get Delta S = Q/T = m*LF/(100+273) |
