The heat transferred to and the work produced by the steam during this process is 13781.618 kJ/kg
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How to calcultae the heat?</h3>
The Net Change in Enthalpy will be:
= m ( h2 - h1 ) = 11.216 ( 1755.405 - 566.78 ) = 13331.618 kJ/kg
Work Done (Area Under PV curve) = 1/2 x (P1 + P2) x ( V1 - V2)
= 1/2 x ( 75 + 225) x (5 - 2)
W = 450 KJ
From the First Law of Thermodynamics, Q = U + W
So, Heat Transfer = Change in Internal Energy + Work Done
= 13331.618 + 450
Q = 13781.618 kJ/kg
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