Question

After a system is added 40-J heat, the system does 30-J work. How do you find the change of the system’s internal energy?

Answer 1

`10J`

Explanation:

1st Law of Thermodynamics:
`DeltaU=Q-W`
`DeltaU=` change in internal energy.
`Q=` Heat energy supplied .
`W=` work done by the system.

`DeltaU=40J-30J=10J`

Some physicist and engineers use different signs for `W`.

I believe this is the engineer's definition:
`DeltaU=Q+W` here, `W` is the work done on the system.
The system does work of `30J` hence the work done on the system is `-30J`.