66.5k views
5 votes
An air compressor used with a paint sprayer does 64.0 J of work pumping air into a tank. Warmed by the process, the air in the tank gives off 37.0 J of energy to its surroundings. What is the change in the internal energy of the air in the tank as a result of the two processes?

User Osa
by
3.1k points

1 Answer

3 votes

Answer:

27 J

Step-by-step explanation:

We are given that

Work done,w=64 J

Energy,Q=-37 J

We have to find the change in the internal energy of the air in the tank.

By First law of thermodynamics


\Delta U=q+w

Using the formula


\Delta U=-37+64


\Delta U=27J

Hence, the change in the internal energy of the air in the tank=27 J

User SparkyRobinson
by
3.3k points