{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solution 2

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 =
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}