1.

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

Answer»

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

SOLUTION :CHANGE in internal ENERGY is given by the relation,
`DeltaE=q+w`
`DeltaE=-q+w` (as heat is given out by the system)
`DeltaE=-1+4`
`DeltaE=+3kJ`.


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