Thermodynamics problem, calculate energy transfer?

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SUMMARY

The discussion focuses on calculating energy transfer in a thermodynamic process involving 5 kg of steam in a piston-cylinder assembly. The steam expands from an initial energy state of 3000 kJ/kg to a final state of 2500 kJ/kg, with a heat transfer of 80 kJ and work input from a paddle wheel of 10 kJ. The correct calculation for the work done by the steam on the piston is determined using the first law of thermodynamics, resulting in a total work output of 2590 kJ.

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  • Understanding of the first law of thermodynamics
  • Knowledge of energy transfer concepts in thermodynamics
  • Familiarity with piston-cylinder assembly operations
  • Ability to perform calculations involving heat transfer and work
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  • Study the first law of thermodynamics in detail
  • Learn about energy balance equations in thermodynamic systems
  • Explore steam properties and phase changes in thermodynamics
  • Investigate practical applications of piston-cylinder assemblies in engineering
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Students studying thermodynamics, engineers working with steam systems, and anyone involved in energy transfer calculations in mechanical systems.

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Homework Statement



A mass of 5 kg of steam is contained within a piston cylinder assembly. The steam undergoes an expansion from its initial state with e1 = 3000 kJ/kg to its final state with e2 = 2500 kJ/kg. During the process there is a heat transfer of 80 kJ to the steam. Also, a paddle wheel transfers 10 kJ of work to the steam. Calculate the amount of energy transfer by work from the steam on the piston during the process.

Homework Equations



dE = del Q - del W

The Attempt at a Solution



Would this be correct?
W = del Q - dE
= (80kJ + 10kJ) - (2500 kJ/kg - 3000 kJ/kg)
W = 590 kJ

I don't believe this is correct but it's all I have been able to come up with so far. Any guidance is appreciated
 
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Determining ΔE
\Delta \hat{E} = 2500 \frac{kJ}{kg} - 3000 \frac{kJ}{kg} = -500 \frac{kJ}{kg}
\Delta E = 5 \ kg \left( - 500 \frac{kJ}{kg} \right) = -2500 \ kJ
Now, applying an energy balance on the cylinder (1st Law of Thermodynamics)
\Delta E = Q + W_s - W
Where Ws is the work input by the paddle wheel and W is the work output on the piston. Finally
W = Q + W_s - \Delta E = 80 \ kJ + 10 \ kJ - (-2500 \ kJ) = 2590 \ kJ
 

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