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?
Download Thermodynamics Problems With Solutions book pdf free download link or read online here in PDF. Read online Thermodynamics Problems With Solutions book pdf free download link book now. All books are in clear copy here, and all files are secure so don't worry about it.
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
Patch fr sims 3 complete edition. 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