Chemistry - Thermodynamics - II Question with Solution | TestHub
ChemistryThermodynamics - IICarnot Cycle/Cyclic processMedium2 minQB
ChemistryMediumsingle choice
A Carnot engine operates between 327°C and 117°C. If it absorbs 120 cal heat per cycle from the source, the heat rejected per cycle to the sink is
Options:
Answer:
C
Solution:
Th = 600 K, Tc = 390 K. For a Carnot engine qc/qh = Tc/Th = 390/600 = 0.65. qc = 0.65 × 120 = 78 cal. Matches (c).
Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Carnot Cycle/Cyclic process
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