Answer:
a) 1 m^3/Kg  
b) 504 kJ
c) 514 kJ 
Explanation:
<u>Given  </u>
-The mass of C_o2 = 1 kg  
-The volume of the tank V_tank = 1 m^3  
-The added energy E = 14 W  
-The time of adding energy t = 10 s  
-The increase in specific internal energy Δu = +10 kJ/kg  
-The change in kinetic energy ΔKE = 0 and The change in potential energy  
ΔPE =0  
<u>Required  </u>
(a)Specific volume at the final state v_2
(b)The energy transferred by the work W in kJ.  
(c)The energy transferred by the heat transfer W in kJ and the direction of  
the heat transfer.  
Assumption  
-Quasi-equilibrium process.  
<u>Solution</u>  
(a) The volume and the mass doesn't change then, the specific volume is constant. 
  v= V_tank/m ---> 1/1= 1 m^3/Kg  
(b) The added work is defined by.  
W =E * t --->  14 x 10 x 3600 x 10^-3 = 504 kJ  
(c) From the first law of thermodynamics.  
Q - W = m * Δu 
Q = (m * Δu) + W--> (1 x 10) + 504 = 514 kJ 
The heat have (+) sign the n it is added to the system.