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Solution 2 - & = 1 4 is c v = 5 R= 2 Hence the heat...

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Physics 2C Summer Session I Quiz #2 1. From the ideal gas law PV = NkT ! N = PV=kT N O 2 = : 19 133 : 3 ° 10 ° 9 ° 10 ° 3 1 : 38 ° 10 ° 23 ° 293 N O 2 = 6 : 3 ° 10 9 ( C ) : 2. From kinetic theory of ideal gasses we have 1 2 m ° v 2 ± = 3 2 kT ! v 2 = 3 kT=m p h v 2 i = v rms = r 3 N A kT N A m = r 3 RT m mole v rms = r 3 ° 8 : 314 ° 473 : 004 = 1717 m= sec ( B ) 3. The total heat removed from the water to freeze it includes the heat removed to bring the water to the freezing point plus the latent heat of fusion. Hence E = Pt = 1 : 5 kg ° ² 4184 J kgK ° T + 334 kJ kg ³ Pt = 1 : 5 ° ´ 4184 ° 25 + 334 ° 10 3 µ J = 1 : 5 ° 4 : 39 ° 10 5 J t = 6 : 58 ° 10 3 sec = 109 : 7 min ( E ) 4. Since the coe¢ cient of thermal expansion for the relative change in the area is twice that for the linear expansion we have ° A A = 2 ° ° T = 2 ° 3 : 2 ° 10 ° 6 ° 80 = 5 : 12 ° 10 ° 4 ° A = 5 : 12 ° 10 ° 4 ± (3 = 2) 2 = 3 : 6 ° 10 ° 3 cm 2 ( A ) 5. The rate of energy produced by the falling water is ± E = ± mgh = 10 6 ° 9 : 8 ° 50 = 4 : 9 ° 10 8 W: 1
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Since the turbine produces 400 MW of power, the remaining 90 MW is dissipated into the water. Given the speci°c heat of water we °nd ± Q = ± mc ° T = 10 6 ° 4184 ° ° T = 9 ° 10 7 ° T = 90 =
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Unformatted text preview: & = 1 : 4 is c v = 5 R= 2 : Hence the heat required to heat the gas is Q = nc v & T = PV nRT 5 nR 2 & T = 101 : 3 & 5 : 6 273 5 2 40 = 208 J ( A ) 7. The work done along an isotherm is W BA = Z PdV = Z nRT V dV = ( PV ) f Z dV V W BA = 400 ln 5 = 644 J ( D ) 8. The net work is W net = W BA + W AC = 644 ± 4 L & 80 kPa = 324 J ( C ) 9. The pressure volume relation for an adiabatic process is PV & = const: Hence the pressure at point B is P B = P A ( V A =V B ) & = 80 kPa (5) 1 : 4 = 761 kPa ( A ) 10. The net work done along an adiabatic is W = ( PV ) i ± ( PV ) f & ± 1 = 761 J ± 80 & 5 J 1 : 4 ± 1 W = 361 : 4 = 902 : 5 J ( C ) 2...
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