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.
