Complete the table below on the basis of the conservation of

Complete the table below on the basis of the conservation of energy principle for a closed system.

Solution

Case

Q_in (KJ)

W_out (KJ)

E₁ (KJ)

E₂ (KJ)

M (kg)

e₂- e₁ (KJ/kg)

1

350

511

1021

860

5

32.2

2

350

130

550

770

5

22

Solution :

We will need Energy Conservation equation from first law of thermodynamic

Total Energy input = Total Energy Output

Q_in + E₁ = E₂ + W_{out    ………………………………………………………………………….}(1)

E = m* e

Where Q = Heat Energy

                W = Work Output
                                E = Energy input(1) /Output (2)

                e = Energy input(1)/Output (2) per unit mass.

CASE 1:

                                Q_in + E₁ = E₂ + W_out   

                                350+1021 = 860+ W

                                W = 511 KJ (Work Done)

                                e₂ –e₁   =       (E₁ -E₂)/m

                                                =             (1021-860)/m (taking m = 5kg)

                                                =             (1021-860)/5

                                                =             32.2 KJ/kg

                                               

Case 2:

                                Q_in + E₁ = E₂ + W_out                  

                                                350+550 = E₂ + 130
                                E₂ = 770 KJ

e₂ –e₁   =       (E₁ -E₂)/m

= (770-550)/5

= 44 KJ/kg

Case	Qin (KJ)	Wout (KJ)	E1 (KJ)	E2 (KJ)	M (kg)	e2- e1 (KJ/kg)
1	350	511	1021	860	5	32.2
2	350	130	550	770	5	22