Complete question:
If a system absorbs 300.5J of heat and performs 1.50kJ of work on the surroundings, what is the change in internal energy of the system ΔE?
Answer:
Change in internal energy of the system is 1800.5 J
Explanation:
Given:
heat transferred to the system, Q = 300.5J
work done on the surroundings, W = 1.50kJ
Change in internal energy of the system ΔE, can be calculated by applying thermodynamic equation of change in internal energy, work done and heat transferred.
ΔE = Q + W
Where;
ΔE is change in internal energy
Q is heat transfer
W is work done
ΔE = 300.5 J + 1500 J
ΔE = 1800.5 J
Therefore, change in internal energy of the system is 1800.5 J
Answer:
The intern energy of the system is -1199.5 J
Explanation:
Given:
System absorbs 300.5 J of heat
Performs 1.5 kJ = 1500 J of work on the surroundings
Question: What is the ΔE?
When a system absorbs heat, it will have a positive sign, therefore +300.5 J
When the system performs work on the surroundings, its sign will be negative, therefore, -1500 J
According the first law of thermodinamic:
ΔE = q + W = 300.5 - 1500 = -1199.5 J
