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A system absorbs 2 Kcals heat and does 500 J of work, then internal energy of the system will be ……. |
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Answer» 6400 J `THEREFORE` dQ=+2000 cal =2000 x 4.2 J =+8400 J System does work, `therefore DeltaW`=+500 J From first law of thermodynamics, `therefore DeltaW`=+500 J dQ=dU+dW `therefore` dU=dQ-dW =8400-500 `therefore` dU=7900 J It is POSITIVE so internal ENERGY INCREASES . |
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