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605_Physics ProblemsTechnical Physics

605_Physics ProblemsTechnical Physics - Chapter 21 P21.19...

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Chapter 21 607 P21.19 Consider 800 cm 3 of (flavored) water at 90.0 °C mixing with 200 cm 3 of diatomic ideal gas at 20.0°C: Q Q cold hot = − or m c T T m c T P f i w w w air , air air = − , d i a f T m c T T m c V c V c w P f i w w P w w w a f d i b g a f b g = = ° ° air , air air air , air C C , . . ρ ρ 90 0 20 0 where we have anticipated that the final temperature of the mixture will be close to 90.0°C. The molar specific heat of air is C R P , air = 7 2 So the specific heat per gram is c R M P , air J mol K mol 28.9 g J g C = F H G I K J = F H G I K J = ⋅° 7 2 7 2 8 314 1 00 1 01 . . . b g T w a f e je j b ga f e je j b g = − × ⋅° ° ⋅° 1 20 10 200 1 01 70 0 1 00 800 4 186 3 . . . . . g cm cm J g C C g cm cm J kg C 3 3 3 3 or T w a f ≈ − × ° 5 05 10 3 . C The change of temperature for the water is between C and C 10 10 3 2 ° ° . P21.20 Q nC T nC T P V = + b g b g isobaric isovolumetric In the isobaric process, V doubles so T must double, to 2 T i . In the isovolumetric process, P triples so T changes from 2 T i to 6 T i . Q n R T T n R T T nRT PV i i i i i = F H G I K J + F H G I K J = = 7 2 2 5 2 6 2 13 5 13 5 b g b g . . P21.21 In the isovolumetric process
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