# The first law of thermodynamics – problems and solutions

1. 3000 J of heat is added to a system and 2500 J of work is done by the system. What is the change in internal energy of the system?

__Known :__

Heat (Q) = +3000 Joule

Work (W) = +2500 Joule

Wanted: the change in internal energy of the system

__Solution :__

**The equation of the first law of thermodynamics**

ΔU = Q-W

The sign conventions :

Q is positive if the heat added to the system

W is positive if work is done by the system

Q is negative if heat leaves the system

W is negative if work is done on the system

The change in internal energy of the system :

ΔU = 3000-2500

ΔU = 500 Joule

Internal energy increases by 500 Joule.

2. 2000 J of heat is added to a system and 2500 J of work is done on the system. What is the change in internal energy of the system?

__Known :__

Heat (Q) = +2000 Joule

Work (W) = -2500 Joule

Wanted: The change in internal energy of the system

__Solution :__

ΔU = Q-W

ΔU = 2000-(-2500)

ΔU = 2000+2500

ΔU = 4500 Joule

Internal energy increases by 4500 Joule.

3. 2000 J of heat leaves the system and 2500 J of work is done on the system. What is the change in internal energy of the system?

__Known :__

Heat (Q) = -2000 Joule

Work (W) = -3000 Joule

Wanted: The change in internal energy of the system

__Solution :__

ΔU = Q-W

ΔU = -2000-(-3000)

ΔU = -2000+3000

ΔU = 1000 Joule

Internal energy increases by 4500 Joule.

Conclusion :

– If heat is added to the system, then the internal energy of the system increases

– If heat leaves the system, then the internal energy of the system decreases

– If the work is done by the system, then the internal energy of the system decreases

– If the work is done on the system, then the internal energy of the system increases