580_Physics ProblemsTechnical Physics

580_Physics ProblemsTechnical Physics - 582 *P20.14 Heat...

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582 Heat and the First Law of Thermodynamics *P20.14 Vessel one contains oxygen according to PV nRT = : n PV RT c == ×× = 1 75 1 013 10 16 8 10 8314 1194 53 .. . . . Pa m Nm mol K 300 K mol 3 ej . Vessel two contains this much oxygen: n h = = 2 25 1 013 10 22 4 10 8314450 1365 . . . a f mol mol. (a) The gas comes to an equilibrium temperature according to mc T mc T nMcT cf hf ∆∆ af di cold hot K K =− −+ −= 300 450 0 The molar mass M and specific heat divide out: 1 194 358 2 1 365 614 1 0 972 3 380 . TT T ff f −+−= K K 2.559 K (b) The pressure of the whole sample in its final state is P nRT V = 2559 16 8 10 206 10 204 3 5 . . mol 8.314 J 380 K mol K 22.4 m Pa atm 3 . Section 20.3 Latent Heat P20.15 The heat needed is the sum of the following terms: Q needed heat to reach melting point heat to melt heat to reach melting point heat to vaporize heat to reach 110 C
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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