How Do You Calculate Work Done by Water in a Piston-Cylinder Assembly?

[BillJ3986]

Homework Statement

A piston-cylinder assembly contains water, initially saturated liquid at 150 degrees C. The water is heated at constant temperature.
a) if the rate of heat transfer to the water is 2.28 kW, determine the rate at which work is done by the water on the piston, in kW.

Homework Equations

I am trying to determine what is the best way to solve this question.

The Attempt at a Solution

I used 2 ways and could not finish them both. The first way I used the equation w=INT[P,v,1.0905m3/kg,.3928m3/kg] I looked up the volume values in a table.I also looked up pressure in the table and it is 4.758bar I ended up with -3.3197 however I had trouble with the conversions.
Second attempt I used the formula. Q=dU+w. Since the rate of heat transfer is 2.28 kw. 2.28kW=U2-U1+W. I looked up the energy values in the table. I got 2.28kW=2,559.5kj/kg-631.68kj/kg+w.
Am I even close to getting this answer. I'm hoping that my professor will extend the homework to wednesday because then I can see a tutor.
Please your help will be greatly appreciated.

FYI: This is thermodynamics

 

[LightHero]

and I'm in college. The correct way to solve this problem is to use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the sum of the heat added to the system (Q) and the work done by the system (W). That means that you can calculate the work done by the water on the piston using the equation W = Q - ΔU.For this problem, the rate of heat transfer (Q) is given as 2.28 kW and the change in internal energy (ΔU) can be calculated from the initial and final temperatures of the water. Since the initial temperature is 150°C and the final temperature is also constant at 150°C, the change in internal energy is zero. Therefore, the rate at which work is done by the water on the piston is simply 2.28 kW.