HW 7 solutions (Ch 17,18)

HW 7 solutions (Ch 17,18) - -W (6) =-720J (7) The negative...

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Physics 6C HW#7 solution May 22, 2008 17.13 The bullet will not melt all the ice, so its final tem- perature is 0 C . Then, ( 1 2 mv 2 + mc | Δ T | ) = m w L f m w is the melt water mass m w = 86 . 4 J + 11 . 5 J 333000 J/kg = 0.294 g 17.19 The work is, W if = - Z f i P d V The work done on the gas is the negative of the area under the curve P = αV 2 between V i and V f . W if = - Z f i αV 2 d V = - 1 3 α ( V 3 f - V 3 i ) = - 1 3 ± (5 . 00 atm/m 6 )(1 . 013 × 10 4 Pa/atm ) ² × ± (2 . 00 m 3 ) 3 - (1 . 00 m 3 ) 3 ² = -1.18 MJ 17.20 a) Remember the work is the negative area under the curve. W = - Z f i P d V (1) W = - (6 . 00 × 10 6 Pa )(2 . 00 - 1 . 00) m 3 + (2) - 1 2 (6 . 00 × 10 6 Pa + 4 . 00 × 10 6 Pa )(3 . 00 - 2 . 00) m 3 + (3) - (2 . 00 × 10 6 Pa )(4 . 00 - 3 . 00) m 3 + (4) W i f = -12.0 MJ (5) b) The work along the reverse path will be the same value with opposite sign. W f i = + 12.0 MJ 17.23 From the first law of thermodynamics, Δ E int = Q + W From the given conditions Δ Eint = - 500 J and W = 220 J. Then, Q = Δ E int
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Unformatted text preview: -W (6) =-720J (7) The negative sign indicates that positive energy is transferred from the system by heat. 1 18.3 a) The eciency is, e = W eng | Q h | = | Q h | - | Q c | | Q h | = 1-| Q c | | Q h | From the given conditions, e = 0.250, | Q c | = 8000 J. Then, | Q h | = 10.7 kJ b) W eng = | Q h | - | Q c | = 2667 J and P = W eng t t = W eng P = 2667 J 5000 J/s = 0.533 s 18.7 a) Eciency of Carnot cycle is, e = 1-T c T h = 1-| Q c | | Q h | T c T h = | Q c | | Q h | | Q c | = | Q h | T c T h = 1200 J 323 523 = 741 J b) W eng = | Q h | - | Q c | = 1200 J-741 J = 459 J 2...
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This note was uploaded on 07/02/2008 for the course PHYS 6C taught by Professor Staff during the Spring '01 term at UCLA.

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HW 7 solutions (Ch 17,18) - -W (6) =-720J (7) The negative...

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