1.

Change in internal energy, when 4 kJ of work is done on the system and 1 kJ of heat is givenout by the system is____

Answer»

`+1` KJ
`-5` kJ
`+3` kJ
`-3` kJ

Solution :`DELTAU=q+w`
`DeltaU`=-1 kJ+ 4 kJ
`DeltaU` =+3 kJ


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