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: QQ cold hot =− or mc T T mc T Pf i w w w air , air air −= , di a f T Vc w i ww P ww w a f bg a f = −− = −° ° air , air air air , air CC , .. 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 CR P , air = 7 2 So the specific heat per gram is c R M P , air Jmo lK mol 28.9 g Jg C = F H G I K J =⋅ F H G I K J ° 7 2 7 2 8314 100 101 . . . T w af ej e j e j ×⋅ ° ° ⋅° 120 10 200 700 1 00 800 4 186 3 . gcm cm C Jk g C 33 or T w ≈− × ° 505 10 3 .C The change of temperature for the water is between C and C 10 10 32 °° . P21.20 Qn CT n PV =+ ∆∆ 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 . Qn R TT n R T T n R T P V ii i i i = F H G I K J −+ F H G I K J = 7 2 2 5 2 62 1 3 5 1 3 5 b g P21.21 In the isovolumetric process AB , W = 0 and C T V ==
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