5 kg of steam contained within a piston-cylinder assembly undergoes an expansion from state 1, where the specific internal energy is u1 = 2709.9 kJ/kg, to state 2, where u2 = 2659.6 kJ/kg.

During the process, there is heat transfer to the steam with a magnitude of 80 kJ. Also, a paddle wheel transfers energy to the steam by work in the amount of 18.5 kJ.

There is no significant change in the kinetic or potential energy of the steam.

Determine the energy transfer by work from the steam to the piston during the process, in kJ.

Energy Transfer = 350 kJ

Explanation:

The net work can be determined from an energy balance. That is, with assumption

∆KE + ∆PE + ∆U = Q - W

Where

∆KE = ∆PE = 0 (Since There is no significant change in the kinetic or potential energy of the steam)

The net work is the sum of the work associated with the paddlewheel Wpw

and the work done on the piston Wpiston:

W = Wpw + Wpiston

From the given information, Wpw = - 18.5 kJ,

Collecting results:

Wpw + Wpiston = Q - ∆U

Wpiston = Q - ∆U - Wpw = Q - m (u2 - u1) - Wpw

Where Q=80kJ, m=5kg, u2 = 2659.6 kJ/kg, u1 = 2709.9 kJ/kg

= 80 kJ - 5 kg (2659.6 - 2709.9) kJ/kg - (-18. 5 kJ)

= 350 kJ