We are given that the system “releases” heat of 2,500 J, and that it “does work on the surroundings” by 7,655 J.
The highlighted words releases and does work on the surroundings all refers to that it is the system itself which expends energy to do those things. Therefore the action of releasing heat and doing work has both magnitudes of negative value. Therefore:
heat released = - 2, 500 J
work done = - 7, 655 J
Which means that the total internal energy change of the system is:
change in internal energy = heat released + work
change in internal energy = - 2, 500 J + - 7, 655 J
change in internal energy = -10,155 J
Answer:
The change in the internal energy of a system is [tex]-10155\rm joule[/tex]
Explanation:
Given information:
Heat release by the system is ([tex]Q=-2500\rm joule[/tex])
Taken negative sign as heat is release by the system,
Work on the surrounding is ([tex]W=-7655\rm joule[/tex])
Taken negative as work done on the surrounding,
By using First law of Thermodynamics,
[tex]\Delta U=Q+W[/tex],
On substituting,
[tex]\Delta U=-2500\rm joule-7655\rm joule[/tex]
[tex]\Delta U= -10155\rm joule[/tex]
The change in the internal energy of a system is [tex]-10155\rm joule[/tex]
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