External forces performed work on the system of 5 kilojoules and the system received 10 kilojoules

External forces performed work on the system of 5 kilojoules and the system received 10 kilojoules of energy from the environment. How did the internal energy of the body change?

A = 5 kJ = 5000 J.

Q = 10 kJ = 10000 J.

According to 1 law of thermodynamics, the internal energy of a substance U can be changed in two ways: by transferring the amount of heat Q to the substance or by performing mechanical work A on the substance by external forces.

If external forces A> 0 perform work on the system, then the internal energy of the system increases ΔU> 0.

If the system performs work A <0, then the internal energy of the system decreases ΔU <0.

If the system is supplied with the amount of heat Q> 0, then the internal energy increases ΔU> 0.

If the system gives off the amount of heat Q <0, then the internal energy decreases ΔU <0.

ΔU = Q + A.

ΔU = 10000 J + 5000 J = 15000 J.

Answer: the internal energy of the system has increased by ΔU = 15000 J.