TITLE.PM5

FIRST LAW OF THERMODYNAMICS 125

dharm

M-therm/th4-2.pm5

Here Q = Heat leaving the boundary.
(i) When the stone is about to enter the water,
Q = 0, W = 0, ∆U = 0
Q – ∆KE = ∆^ PE = mg (Z 2 – Z 1 )
= 20 × 9.81 (0 – 15) = – 2943 J
∴∆ KE = 2943 J
and ∆ PE = – 2943 J. (Ans.)
(ii) When the stone dips into the tank and comes to rest
Q = 0, W = 0, ∆ KE = 0
Substituting these values in eqn. (1), we get
0 = ∆ U + 0 + ∆ PE + 0
∴∆U = – ∆PE = – (– 2943) = 2943 J. (Ans.)
This shows that the internal energy (temperature) of the system increases.
(iii) When the water and stone come to their initial temperature,
W = 0, ∆ KE = 0
Substituting these values in eqn. (1), we get
∴ Q = – ∆ U = – 2943 J. (Ans.)
The negative sign shows that the heat is lost from the system to the surroundings.
+Example 4.10. When a system is taken from state l to state m, in Fig. 4.18, along path
lqm, 168 kJ of heat flows into the system, and the system does 64 kJ of work :
(i)How much will be the heat that flows into the system along path lnm if the work done
is 21 kJ?
(ii)When the system is returned from m to l along the curved path, the work done on the
system is 42 kJ. Does the system absorb or liberate heat, and how much of the heat is absorbed
or liberated?
(iii)If Ul = 0 and Un = 84 kJ, find the heat absorbed in the processes ln and nm.
Solution. Refer Fig. 4.18.

Fig. 4.18

Ql–q–m = 168 kJ

Wl–q–m = 64 kJ