A piston cylinder device contains 0.2 kg of air at 500 kPa and 20.5°C. The air is heated electrically for 10 minutes and mechanically does 550 kJ/kg of work by a paddle wheel. If the volume is doubled while maintaining a constant temperature, the amount of heat transferred to the surroundings is -1811.7 kJ/kg to keep the temperature constant during the expansion.

1. Piston cylinder device contains 0.2 kg of air at 500 kPa and 20½C. Now, 4 Amps passes
through the electric heater connected to 110 V for 10 min. The paddle wheel is also turned with
550 kJ/kg of work. If heat is transferred from the system to the surroundings to maintain a
constant temperature while the volume is doubled. Calculate the amount of heat transfer in kJ/kg.
The gas constant of air is 0.287 kJ/kgK.
Solution
Mass, m = 0.2 Kg
P 1 = 500 KPa
T 1 = 20 C = 293 K
I = 4 Amps
Voltage,v = 110 V
time, t = 10 min = 600 sec
Electrical Work, W e = - I x v x t = - 4 X 110 x 600 = - 264000 J = - 264 KJ
(-ve indicates work done on system)
Paddle Work, W p = - m X 550 = - 0.2 X 550 = - 110 KJ
SInce, Volume doubles, V 2 = 2 V 1
Also, Constant temperature T 2 = T 1 = 293 K
Boundary Work, Wb = m R T ln (V 2 /V 1 ) = 0.2 X 0.287 x 293 x ln ( 2V 1 /V 1 )
= 11.66 KJ
Applying conservation of energy

2. Q = m C v (T 2 - T 1 ) + W b + W p + W e
Q = 11.66 - 110 - 264
= - 362.34 KJ
Specific Heat , q = Q/m = - 362.34/0.2 = - 1811.7 KJ/kg....Answer
(-ve indicates heat lost to surroundings)