EntropyEnergy QualityQuantifying QualityClosed CycleEntropy DefinedMelting IceMixingSecond Law IIIEntropyEntropyEnergy QualityEnergy QualityYou are offered 1000 J of energy. Would you rather You are offered 1000 J of energy. Would you rather have it ashave it asA) mechanical workA) mechanical workB) frictional workB) frictional workC) heat from an object at 1000 KC) heat from an object at 1000 KD) heat from an object at 300 KD) heat from an object at 300 KQuantifying QualityQuantifying QualityA Carnot cycle found a A Carnot cycle found a relationship between the relationship between the temperatures and heat.temperatures and heat.The heat in and out are of The heat in and out are of opposite sign.opposite sign.LLHHHHLLTQTQTQTQ0HHLLHLHLTQTQTTQQClosed CycleClosed CycleAny closed cycle can be Any closed cycle can be approximated by a sum of approximated by a sum of Carnot cycles.Carnot cycles.On a PV diagram this is any On a PV diagram this is any reversible cycle.reversible cycle.•The heat to temperature The heat to temperature ratios can be added.ratios can be added.TQ0Entropy DefinedEntropy DefinedEntropy is defined as the Entropy is defined as the heat flow at an absolute heat flow at an absolute temperature.temperature.The path doesn’t matter, so The path doesn’t matter, so entropy is a macroscopic entropy is a macroscopic state variable.state variable.TQS Melting IceMelting IceThe latent heat of ice is 79.7 The latent heat of ice is 79.7 kcal/kg.kcal/kg.What is the change of What is the change of entropy for a very slowly entropy for a very slowly melting 1.00 kg piece of ice?melting 1.00 kg piece of ice?What is the change in What is the change in entropy for the entropy for the surroundings?surroundings?Find the heat transfer.Find the heat transfer.QQ = = mLmL = 79.9 kcal = 79.9 kcalFind the entropy change.Find the entropy change. SS = = QQ//TT = 0.292 kcal/K = 0.292 kcal/KThe process is reversible.The process is reversible. SSsurrsurr = -0.292 kcal/K = -0.292 kcal/KMixingMixingA sample of 50.0 kg water at A sample of 50.0 kg water at 20.0 20.0 C is mixed with 50.0 kg C is mixed with 50.0 kg water at 24 water at 24 C.C.Estimate the change in total Estimate the change in total entropy.entropy.Find the heat transfer. There are Find the heat transfer. There are equal amounts of heat in each equal amounts of heat in each sample.sample.QQ = = mcmcTT = 100. kcal = 100. kcalFind the entropy change in each Find the entropy change in each sample using the average sample using the average temperature.temperature. SSHH = = QQ//TT = -100. kcal/296K = = -100. kcal/296K = -0.338 kcal/K-0.338 kcal/K SSLL = = QQ//TT = +100. kcal/294K = = +100. kcal/294K = +0.340 kcal/K+0.340 kcal/KThe difference is the net The difference is the net change.change. SS = +0.002 kcal/K = +0.002 kcal/KSecond Law IIISecond Law IIIThe second law of thermodynamics can be described The second law of thermodynamics can be described in terms of entropy:in terms of entropy:The entropy of an isolated system never decreases. The entropy of an isolated system never decreases. It only stays the same for reversible processes. It only stays the same for reversible processes.
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