Answer:
ΔU= 1069 kJ
Explanation:
Given that
The work done on the system
W= 375 kcal ( 1 kcal = 4.184 kJ)
W= - 1569 kJ
Heat release by system ,Q = 5 x 10² kJ
Q= - 500 kJ
Sign convention :
Work done on the system is taken as negative and work done by the system is taken as positive .
Heat gain by system is taken as positive and heat release by system is taken as negative.
Now by using first law of thermodynamics
Q= W+ ΔU
Q=Heat transfer
W=Work
ΔU =Change in the internal energy
Now by putting the values
-500 = -1569 + ΔU
ΔU= -500+ 1569 kJ
ΔU= 1069 kJ
The change in internal energy (ΔE or ΔU) of the system is 1069 kJ.
Given data:
The amount of work done is, W = - 375 kcal. (Negative sign shows that work is done on the system)
1 kcal = 4.184 kJ
Then,
[tex]W = -375 \times 4.184 = - 1569 \;\rm kJ[/tex]
The Amount of heat released is, [tex]Q = -5.00 \times 10^{2} \;\rm kJ[/tex] (Negative sign shows that heat is released)
Use first law of thermodynamics which says that the amount of heat given to the system is equal to the work done and internal energy change. That is,
Q = W + U
Here, [tex]\Delta[/tex]U is the internal energy change.
Solving as,
[tex]Q = W + \Delta U\\-5 \times 10^{2} = -1569 + \Delta U\\\Delta U = 1569 +500\\\Delta U = 1069 \;\rm kJ[/tex]
Thus, the change in internal energy (ΔE or ΔU) of the system is 1069 kJ.
Learn more about the first law of thermodynamics here:
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