A work of 0.2 kJ was performed on the body by external forces and the amount of heat of 300 J

A work of 0.2 kJ was performed on the body by external forces and the amount of heat of 300 J was transferred to the body. What is the change in the internal energy of the body?

Given:

A = 0.2 kJ = 200 J (converted to Cu system);

Q = 300 J;

To find:

ΔU -?

Decision:

The first law of thermodynamics says: the change in internal energy is equal to the difference between the transferred amount of heat and the perfect work on external forces;

ΔU = A + Q = 200 + 300 = 500 J;