Explanation:
We assumes that all the electrical energy produced will be converted into the heat energy.
Electrical energy input = Heat energy absorbed by object = Heat energy required = Q
As, Q = U - W
= U - RT [tex]ln \frac{P_{2}}{P_{1}}[/tex]
Since, it is given that U = 286.1 kJ/kg, [tex]P_{1}[/tex] is 600 kPa, [tex]P_{2}[/tex] is 200 kPa, and T is 400 K.
Therefore, putting these given values into the above formula is as follows.
Q = U - RT [tex]ln \frac{P_{2}}{P_{1}}[/tex]
= [tex]286.1 kJ/kg - 8.314 kJ/kmol K \times 400 K \times ln \frac{200}{600}[/tex]
= [tex]286.1 kJ/kg - \frac{8.314}{28.97} kJ/kg K \times 400 K \times ln \frac{200}{600}[/tex]
= 412.27 kJ/kg
Thus, we can conclude that the electrical energy supplied to air is 412.27 kJ/kg.
