Archived Thermodynamics problem, calculate energy transfer?

AI Thread Summary
The problem involves calculating energy transfer during the expansion of 5 kg of steam in a piston cylinder assembly. The steam's internal energy decreases from 3000 kJ/kg to 2500 kJ/kg, with a heat transfer of 80 kJ and work input from a paddle wheel of 10 kJ. The equation used is based on the first law of thermodynamics, where the change in energy (ΔE) is related to heat transfer (Q) and work done (W). The correct calculation shows that the work done by the steam on the piston is 2590 kJ. This solution emphasizes the importance of applying energy balance principles accurately.
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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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