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
