Answer:
(a). The specific volume at the final state is 0.25 m³/kg.
(b). The energy transfer by work is 50.4 kJ.
(c). The energy transfer by heat is -10.4 kJ.
Negative sign shows the direction of heat
Explanation:
Given that,
Weight of carbon monoxide = 4 kg
Volume of tank = 1 m³
Constant rate of 14W for 1 hour
The specific internal energy of the carbon monoxide increases by 10 kJ/kg.
(a). We need to calculate the specific volume at the final state
Using formula of specific volume
![V'=\dfrac{V}{m}](https://tex.z-dn.net/?f=V%27%3D%5Cdfrac%7BV%7D%7Bm%7D)
Put the value into the formula
![V'=\dfrac{1}{4}](https://tex.z-dn.net/?f=V%27%3D%5Cdfrac%7B1%7D%7B4%7D)
![V'=0.25\ m^3/kg](https://tex.z-dn.net/?f=V%27%3D0.25%5C%20m%5E3%2Fkg)
(b). We need to calculate the energy transfer by work
Using formula of work
![W=Pt](https://tex.z-dn.net/?f=W%3DPt)
Where, P = power
t = time
Put the value into the formula
![W=14\times1\times3600](https://tex.z-dn.net/?f=W%3D14%5Ctimes1%5Ctimes3600)
![W=50.4\ kJ](https://tex.z-dn.net/?f=W%3D50.4%5C%20kJ)
(c). The energy transfer by heat transfer, and the direction of the heat transfer
We need to calculate the ![\Delta U](https://tex.z-dn.net/?f=%5CDelta%20U)
Using formula of internal energy
![\Delta U=m\Delta u](https://tex.z-dn.net/?f=%5CDelta%20U%3Dm%5CDelta%20u)
Put the value into the formula
![\Delta U=4\times10](https://tex.z-dn.net/?f=%5CDelta%20U%3D4%5Ctimes10)
![\Delta U=40\ kJ](https://tex.z-dn.net/?f=%5CDelta%20U%3D40%5C%20kJ)
We need to calculate the energy transfer by heat
Using formula of energy transfer
![Q=\Delta U-W](https://tex.z-dn.net/?f=Q%3D%5CDelta%20U-W)
Put the value into the formula
![Q=40-50.4](https://tex.z-dn.net/?f=Q%3D40-50.4)
![Q=-10.4\ kJ](https://tex.z-dn.net/?f=Q%3D-10.4%5C%20kJ)
Negative sign shows the direction of heat that is removed from CO.
Hence, (a). The specific volume at the final state is 0.25 m³/kg.
(b). The energy transfer by work is 50.4 kJ.
(c). The energy transfer by heat is -10.4 kJ.
Negative sign shows the direction of heat.