5 kg of steam contained within a piston-cylinder assembly undergoes an expansion from state 1, where the specific internal energy U(1) is = 2710 [kJ/kg] to state 2, where U(2) = 2660 [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]. Determine the energy transfer by work from the steam to the piston during the process

To determine the energy transfer by work from the steam to the piston during the process, we need to apply the First Law of Thermodynamics:

ΔQ = ΔU + ΔW

Where:
ΔQ is the heat transfer to the steam
ΔU is the change in internal energy of the steam
ΔW is the work done by or on the steam

Given:
ΔQ = 80 kJ (heat transfer to the steam)
ΔU = U(2) - U(1) = 2660 kJ/kg - 2710 kJ/kg = -50 kJ/kg (change in internal energy of the steam, as there is a decrease)
ΔW = 18.5 kJ (work done by the paddle wheel on the steam)

Now, let's determine the energy transfer by work from the steam to the piston:

ΔQ = ΔU + ΔW
80 kJ = -50 kJ/kg * m + 18.5 kJ

Where:
m is the mass of the steam (given as 5 kg)

Rearranging the equation and solving for the energy transfer by work:

80 kJ - 18.5 kJ = -50 kJ/kg * 5 kg

61.5 kJ = -250 kJ/kg

Energy transfer by work from the steam to the piston = -250 kJ/kg

Note: The negative sign indicates that work is done on the steam by the piston.