TITLE.PM5

FIRST LAW OF THERMODYNAMICS 129

dharm

M-therm/th4-2.pm5

Let the system flow be in kg/s.

∴ zdQ = 840 m& kJ/s

zdW = 1200 – 6 = 1194 kJ/s
But zdQ = zdW
i.e., 840 m& = 1194
∴ m& =^1194840 = 1.421 kg/s
Q Steam flow round the cycle = 1.421 kg/s. (Ans.)
Example 4.14. A closed system of constant volume experiences a temperature rise of 25°C
when a certain process occurs. The heat transferred in the process is 30 kJ. The specific heat at
constant volume for the pure substance comprising the system is 1.2 kJ/kg°C, and the system
contains 2.5 kg of this substance. Determine :
(i)The change in internal energy ;
(ii)The work done.
Solution. Temperature rise, (T 2 – T 1 ) = 25°C
The heat transferred in the process, Q = 30 kJ
Specific heat at constant volume, cv = 1.2 kJ/kg°C
Mass of the substance, m = 2.5 kg

Now, ∆ U = m cdTv

T

T

1

2

z

= 2.5^12

1

2

. dT
T

T
z = 3.0 × (T^2 – T^1 )
= 3.0 × 25 = 75 kJ
Hence, the change in internal energy is 75 kJ. (Ans.)
According to the first law of thermodynamics,
Q = ∆ U + W
∴ 30 = 75 + W
∴ W = 30 – 75 = – 45 kJ
Hence, the work done = – 45 kJ. (Ans.)
It may be observed that even though the volume is constant the work is not zero. Clearly,
the process is irreversible.
Example 4.15. A system receives 50 kJ of heat while expanding with volume change of
0.14 m^3 against an atmosphere of 1.2 × 105 N/m^2. A mass of 90 kg in the surroundings is also
lifted through a distance of 5.5 metres.
(i)Find the change in energy of the system.
(ii)The system is returned to its initial volume by an adiabatic process which requires
110 kJ of work. Find the change in energy of the system.
(iii)For the combined processes of (i) and (ii) determine the change in energy of the system.