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# Sol-Ch19 - CHAPTER 19 E xercises 1(4186(0.4535(5/9 = 1055 J...

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Exercises 1. (4186)(0.4535)(5/9) = 1055 J 2. Q = mcΔT, so c = 1070 J/kg 3. [(0.090)(385)+(0.5)(4190)](T–15) +(0.08)(450)(T – 180) = 0. Thus T = 17.7 ° C 4. m b c b (T f – 210)+ (m 1 c 1 + m a c a )(T f – 20) = 0. Find c 1 = 1680 J/kg 5. (0.5)(450)(80) + (0.6)(4190)(80) = 1200 t, thus t = 183 s 6. Q = 0.6 × 600 × 2 = 720 kcal = 3.017 × 10 6 J m = Q/L = 1.23 kg. 7. (0.08)(2100)(10) + (0.08)(3.34 × 10 5 ) + (0.08)(4190)(100) + (0.02)(2.26 × 10 6 ) = 107 kJ 8. m p c p (T f – 200) + (m c c c +m w c w )(T f – T i ) =0. with m w = 0.1 kg, m p = 0.2 kg, and m c = 0.07 kg, find T i = 9.5 ° C 9. 5 × 10 8 = (4190)(10)(dm/dt), thus dm/dt = 1.2 × 10 4 kg/s 10. dQ/dt = (6)(10 3 )(0.8) = (dm/dt)(4190)(40) Thus. dm/dt = 28.6 g/s 11. (a) 1/2 mv 2 = 463 kJ; (b) ΔT = 2.78 × 10 5 /(10)(450) = 61.7 ° C 12. E =(0.2)(0.5 mv 2 )(15) = 3J = (6 ×   10 –3 )cΔT, thus ΔT = 1.11 ° C 13. (a) Q = (0.35)mv 2 , ΔT = Q/(130)(0.02) = 330 ° C, so T f = 327 ° C (b) To reach 327 ° C : Q 1 = mc(327 – 30) ° = 772 J. To melt : Q 2 = m 2 L = (857.5 – 772)J = 85.5 J. CHAPTER 19

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Find 3.5 g melts. 14. mgh = mcΔT, so ΔT = gh/c = 0.28 ° C. 15. ΔT = 16mgh/(3.5)(4190) = 0.423 ° C 16. E = 0.8 (m 1 gh) = (m 1 c 1 + m 2 c 2 ) ΔT, find ΔT = 7.84 ×   10 –3 ° C 17. W = PΔV = (101 kPa)(1/920 – 1/1000) = 8.8 J 18. W = PΔV: (a) 600 J; (b) –300 J 19. (a) W = PΔV= 2.4 kJ; (b) W = nRT ln(P 2 /P 1 ) = P 1 V 1 ln(4/5) = –1.34 kJ 20. W = nRT ln(V f /V i ), but V α 1/P, so W = nRTln(1/2.5) = –2R(273)ln(2.5) = –4160 J 21. (a) ΔU = Q – W = 35 – 11 = 24 J, thus U f = 229 J; (b) ΔU = 24 = Q’ – 15, thus Q’ = 39 J 22. (a) W = PΔV = –16.2 J; (b) ΔU = Q – W = 416 J 23. (a) W = (mg + PA)h = (19.6 + 31.4)(0.024) = 1.22 J; (b) ΔU = 5 – 1.22 = 3.88 J 24. (a) For process abc, U c – U a = Q – W = 4500 – W bc = 500 J Thus U c = 1100 J.
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