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

In a process, 701J of heatis absorbed by a system and 394 J of work is

1.	In a process, 701J of heatis absorbed by a system and 394 J of work is done by the system. What is the change in internal energy of the process ?
Answer» Solution :`q=+ 701 J ,w = - 394 J , DeltaU =?` By firstlaw of THERMODYNAMICS `Delta U =q +w = + 701 J + ( -394J) = + 307 J` i.e., internal energy of the system increases by 307J