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When heat energy of 1500 J is suppliedto a gas the external work done on the gas is 525 J what is the increase in its internal energy |
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Answer» Solution :Heat energy SUPPLIED `DELTAQ= +1500J` EXTERNAL work done `Deltaw=-525J` By `1^(st)` law of THERMODYNAMICS `DeltaQ = DeltaU +Deltaw` `:. DeltaU =DeltaQ-Deltaw=1500+525=1025J` |
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