TITLE.PM5

126 ENGINEERING THERMODYNAMICS

dharm

M-therm/th4-2.pm5

We have, Ql–q–m = (Um – Ul) + Wl–q–m

168 = (Um – Ul) + 64

∴ Um – Ul = 104 kJ. (Ans.)

(i) Ql–n–m = (Um – Ul) + Wl–n–m

= 104 + 21 = 125 kJ. (Ans.)

(ii) Qm–l = (Ul – Um) + Wm–l

= – 104 + (– 42) = – 146 kJ. (Ans.)

The system liberates 146 kJ.

(iii) Wl–n–m = Wl–n + Wn–m = Wl–m = 21 kJ

[Q Wn–m = 0, since volume does not change.]

∴ Ql–n = (Un – Ul) + Wl–n

= (84 – 0) + 21 = 105 kJ. (Ans.)

Now Ql–m–n = 125 kJ = Ql–n + Qn–m

∴ Qn–m = 125 – Ql–n

= 125 – 105 = 20 kJ. (Ans.)

Example 4.11. In a system, executing a non-flow process, the work and heat per degree

change of temperature are given by

dW

dT = 200 W-s/°C^ and^

dQ
dT = 160 J/°C
What will be the change of internal energy of the system when its temperature changes
from
T 1 = 55 °C to T 2 = 95 °C?
Solution. Initial temperature, T 1 = 55°C ; Final temperature, T 2 = 95°C
dW
dT = 200 W-s/°C ;

dQ

dT = 160 J/°C.

Change of internal energy :

Now, dWdT = 200 W-s/°C

∴ WdTT TT

T

T

T

===z 200 200 200

1

2

1

2

55

95

= 200 (95 – 55) = 8000 W-s = 8000 J [Q 1 W-s = 1 J]

Also, dQ

dT

= 160 J/°C

∴ QdTTT

T

T

T

==z 160 160

1

2

1

2

= 160

55

95

T = 160 (95 – 55) = 6400 J

Applying the first law of thermodynamics to the given non-flow system,

Q = ∆ U + W