An engine operates at a constant temperature of 110 degrees celsius through a reversible process. The engine work output is 5.30kJ and the heat loss is 6.20 kJ. Determine the change in entropy during the process.
ΔS=QoutTreservior=6200110+273=16.19J/K\Delta S= \frac{Q_{out}}{T_{reservior}}= \frac{6200}{110+273}= 16.19 J/KΔS=TreserviorQout=110+2736200=16.19J/K
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