Answer:
[tex]\Delta U = 64218.9 J[/tex]
Explanation:
As we know that power of the heater is given as
P = 100 W
now the energy given by the heater for 11 min of time
[tex]E = P \times t[/tex]
[tex]E = 100 \times 11 \times 60[/tex]
[tex]E = 100\times 11 \times 60[/tex]
[tex]E = 66000 J[/tex]
now from 1st law of thermodynamics we know that
[tex]E = \Delta U + W[/tex]
work done under constant pressure condition we have
[tex]W = P \Delta V[/tex]
[tex]W = (3.527 \times 1.01 \times 10^5)(6 - 1) \times 10^{-3}[/tex]
[tex]W = 1781.13 J[/tex]
now from first equation we have
[tex]66000 = 1781.13 + \Delta U[/tex]
[tex]\Delta U = 64218.9 J[/tex]
The change in the internal energy of the system is 67.787 kJ.
The given parameters;
The heat added to the system by the heater;
Q = Pt = 100 x 660 = 66,000 J = 66 kJ
The work done on the system is calculated as follows;
W = PΔV
W = 3.527(6 - 1)
W = 17.64 L.atm
1 Latm = 0.1013 kJ
17.64 Latm = 1.787 kJ
The change in the internal energy of the system is calculated by applying the first law of thermodynamics as follows;
ΔU = Q + W
ΔU = 66 kJ + 1.787 kJ
ΔU = 67.787 kJ.
Thus, the change in the internal energy of the system is 67.787 kJ.
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