Answer:
0.833 m³/kg
-14.4 kJ
20 kJ/kg
Step-by-step explanation:
= Density = 1.2 kg/m³
= Volume = 0.6 m³
t = Time taken = 1 hour
P = Power = 4 W
Mass is given by
![m=\rho V\\\Rightarrow m=1.2* 0.6\\\Rightarrow m=0.72\ kg](https://img.qammunity.org/2021/formulas/physics/college/jw8bi2ezy4j9dvz1vfgxekjpd9rxu0tfj8.png)
As there is no change in kinetic and potential energy, the specific volume of the tank will be unaffected
![V_(s)=(0.6)/(0.72)\\\Rightarrow V_s=0.833\ m^3/kg](https://img.qammunity.org/2021/formulas/physics/college/new7t4fm1jp1tnqbfn148xwjncgijtwrp1.png)
The specific volume at the final state is 0.833 m³/kg
Energy is given by
![E=Pt\\\Rightarrow E=-4* 1* 3600\\\Rightarrow E=-14400\ J](https://img.qammunity.org/2021/formulas/physics/college/6srqgxdwak6a1wm2n5qj4i2vghe89hi34l.png)
The energy transfer by work is -14.4 kJ
Change in specific internal energy is given by
![E=-m\Delta u\\\Rightarrow \Delta u=-(E)/(m)\\\Rightarrow \Delta u=-(-14.4)/(0.72)\\\Rightarrow \Delta u=20\ kJ/kg](https://img.qammunity.org/2021/formulas/physics/college/cjn4pku8h89vt3witq0qfp8n44la38m40p.png)
The change in specific internal energy is 20 kJ/kg