Four kilograms of carbon monoxide (co) is contained in a rigid tank with a volume of 1 m3 . the tank is fitted with a paddle wheel that transfers energy to the co at a constant rate of 14 w for 1 h. during the process, the specific internal energy of the carbon monoxide increases by 10 kj/kg. if no overall changes in kinetic and potential energy occur, determine (a) the specific volume at the final state, in m3 /kg. (b) the energy transfer by work, in kj. (c) the energy transfer by heat transfer, in kj, and the direction of the heat transfer.

a)

b)W= 50.4 KJ

c)Q= -10.4 KJ

It means that heat will be rejected from the system.

Explanation:

m= 4 kg

Work in put = 14 W  14 J/s

1 h = 60 sec

Work in put = 14 x 60 J

W=14 x 60 J =840 J

ΔU= 10 KJ/Kg

ΔU= 10 x 4 = 40 J

a)

We know that specific volume is the ratio of volume and mass.

v= V/m

b)

Energy transfer by  work =W

W=14 x 60 x 60 J =50.4  KJ

W= 50.4 KJ

c)

From first law of thermodynamics

Q = ΔU + W

Here work in added to the system that is why W= -840 J

Q = 40 - 50.4 KJ

Q= 40 - 50.4  KJ

Q= -10.4 KJ

It means that heat will be rejected from the system.

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.