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Determine the change in the enthalpy of helium, in kJ/kg, as it undergoes a change of state from 100 kPa and 20 oC to 600 kPa and 300 oC
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Determine the change in the enthalpy of helium, in kJ/kg, as it undergoes a change of state from 100 kPa and 20 oC to 600 kPa and 300 oC

User Zamil
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Answer:

The change in enthalpy of helium is 4073.86kJ/kg

Step-by-step explanation:

∆H = Cp(T2 - T1)

Cp = 3.5R = 3.5×8.314 = 29.099kJ/kgmolK ÷ 2 (1kgmol of helium = 2kg of helium) = 14.5495kJ/kgK, T2 = 300°C = 300+273K = 573K, T1 = 20°C = 20+273K = 293K

∆H = 14.5495kJ/kgK(573K - 293K) = 14.5495kJ/kgK × 280K = 4073.86kJ/kg

User Torourke
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