In a process, `701 J` of heat is absorbed by a system and `394 J` of work is done by the system. What is the change in internal energy for the process?

Answer

q=+701 J
w=-394 J, since work is done by the ststem
`DeltaU=307 J`

Explanation

According to the first law of thermodynamics,
`DeltaU=q+W(i)`
Where,
`DeltaU`= change in internal energy for a process
q=heat
W=work
Given,
q=+701 (Since heat is absorbed)
W=-394 J (Since work is done by the system)
Substituting the values in expression, (i), we get
`DeltaU=701 J +(-394 J)`
`DeltaU=307 J`
Hence, the change in internal energy for the given process is 307 J.