TITLE.PM5

124 ENGINEERING THERMODYNAMICS

dharm

M-therm/th4-2.pm5

Work input to the paddle wheel = 9000 kJ
Heat transferred to the surroundings from the tank = 3000 kJ
As it is a closed system, the first law of thermodynamics can be written as
U 1 – Q + W = U 2 ...(i)
The work enters into the tank in the form of energy only so this should be considered as heat
input.
∴ Q = Q 1 – Q 2
= 3000 – 9000 = – 6000 kJ.
(i) Since volume does not change (being constant volume process)
∴ Work done, W = 0
Putting the value of W = 0 in equation (i), we get
(ii) U 1 – (– 6000) + 0 = U 2
∴ U 2 – U 1 = 6000 kJ
Hence, change in internal energy (increase) = 6000 kJ. (Ans.)
Example 4.9. A stone of 20 kg mass and a tank containing 200 kg water comprise a
system. The stone is 15 m above the water level initially. The stone and water are at the same
temperature initially. If the stone falls into water, then determine ∆U, ∆PE, ∆KE, Q and W, when
(i)the stone is about to enter the water,
(ii)the stone has come to rest in the tank, and
(iii)the heat is transferred to the surroundings in such an amount that the stone and water
come to their initial temperature.
Solution. Refer Fig. 4.17.

Fig. 4.17

Mass of stone = 20 kg

Mass of water in the tank = 200 kg

Height of stone above water level = 15 m

Applying the first law of thermodynamics,

Q = (U 2 – U 1 ) + m

CC 22 12

2

L −

N

M

M

O

Q

P

P + mg (Z^2 – Z^1 ) + W

= ∆U + ∆KE + ∆PE + W ...(1)