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8 February, 19:38

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?

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  1. 8 February, 22:03
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    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.
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