590_Physics ProblemsTechnical Physics

590_Physics ProblemsTechnical Physics - 592 P20.55 Heat and...

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592 Heat and the First Law of Thermodynamics P20.55 (a) Before conduction has time to become important, the energy lost by the rod equals the energy gained by the helium. Therefore, mL mc T v bg ch He Al =∆ or ρρ VL Vc T v He Al so V Vc T L v He Al He = ρ V V He 33 3 He 3 gcm cm c a lg C C g cm J kg cal 4.186 J kg 1 000 g cm liters = ⋅° ° × = 270 625 0210 2958 0125 209 10 100 168 10 168 4 4 .. . . . ... ej e j af e j b g (b) The rate at which energy is supplied to the rod in order to maintain constant temperatures is given by P = F H G I K J =⋅ F H G I K J = kA dT dx 31 0 2 50 295 8 917 . JscmK K 25.0 cm W 2 This power supplied to the helium will produce a “boil-off” rate of L v = × == 917 10 351 0 351 3 4 W g k g Jk g cm s L s 3 3 e j . *P20.56 At the equilibrium temperature T eq the diameters of the sphere and ring are equal: dd T T dd T TT T T T ss i rr i i i +− = + ° ° = ° ° °+ × × = × × ° ×° + × = × °+ = −− αα Al eq Cu eq eq eq eq eq eq eq C cm cm 1 C cm cm 1.70 10 1 C C CC C C e j e j e j 15 5 01 5 01 2 40 10 5 00 5 00 15 0 01 1 202 4 10 1 202 4 10 8 5 10 1 275 10 1 127 5 10 3 524 10 1 202 4 10 319 95
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