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645_Physics ProblemsTechnical Physics

645_Physics ProblemsTechnical Physics - Chapter 22 P22.36...

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Chapter 22 647 P22.36 At a constant temperature of 4.20 K, S Q T L S v == = =⋅ 420 20 5 488 . . . . K kJ kg K kJ kg K P22.37 S dQ T mcdT T mc T T i f T T f i i f = F H G I K J zz ln S ° F H G I K J 250 353 293 46 6 195 g 1.00 cal g C cal K J K bg ln . *P22.38 (a) The process is isobaric because it takes place under constant atmospheric pressure. As described by Newton’s third law, the stewing syrup must exert the same force on the air as the air exerts on it. The heating process is not adiabatic (energy goes in by heat), isothermal ( T goes up), isovolumetic (it likely expands a bit), cyclic (it is different at the end), or isentropic (entropy increases). It could be made as nearly reversible as you wish, by not using a kitchen stove but a heater kept always just incrementally higher in temperature than the syrup. The process would then also be eternal, and impractical for food production.
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