# 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.

[irp]

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.

[irp]

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.

[irp]

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