Answer:
0.833 m³/kg
-14.4 kJ
20 kJ/kg
Explanation:
[tex]\rho[/tex] = Density = 1.2 kg/m³
[tex]V[/tex] = Volume = 0.6 m³
t = Time taken = 1 hour
P = Power = 4 W
Mass is given by
[tex]m=\rho V\\\Rightarrow m=1.2\times 0.6\\\Rightarrow m=0.72\ kg[/tex]
As there is no change in kinetic and potential energy, the specific volume of the tank will be unaffected
[tex]V_{s}=\dfrac{0.6}{0.72}\\\Rightarrow V_s=0.833\ m^3/kg[/tex]
The specific volume at the final state is 0.833 m³/kg
Energy is given by
[tex]E=Pt\\\Rightarrow E=-4\times 1\times 3600\\\Rightarrow E=-14400\ J[/tex]
The energy transfer by work is -14.4 kJ
Change in specific internal energy is given by
[tex]E=-m\Delta u\\\Rightarrow \Delta u=-\dfrac{E}{m}\\\Rightarrow \Delta u=-\dfrac{-14.4}{0.72}\\\Rightarrow \Delta u=20\ kJ/kg[/tex]
The change in specific internal energy is 20 kJ/kg
