Ask Question
19 September, 10:08

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?

+4
Answers (1)
  1. 19 September, 11:19
    0
    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.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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 ...” in 📘 Engineering if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers