Calculate the change in the enthalpy of argon, in kj/kg, when it is cooled from 75 to 35°c. if neon had under-gone this same change of temperature, would its enthalpy change have been any different?

Enthalpy almost doubles.

Explanation:

Argon

Cp = Specific heat at constant volume = 0.520 kJ/kgK

T₁ = Initial temperature = 75°C

T₂ = Final temperature = 35°C

Enthalpy

Δh = CpΔT

⇒Δh = Cp(T₂-T₁)

⇒Δh = 0.520×(35-75)

⇒Δh = -20.8 kJ/kg

Neon

Cp = Specific heat at constant volume = 1.03 kJ/kgK

T₁ = Initial temperature = 75°C

T₂ = Final temperature = 35°C

Δh = Cp(T₂-T₁)

⇒Δh = 1.03×(35-75)

⇒Δh = -41.2 kJ/kg

Enthalpy will change because Cp value is differrent.

Enthalpy almost doubles.

answer:

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answer:

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explanation:

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