How much energy is required to convert 1 mole of liquid Ar from 5◦C below its boiling point...?
The specific heat for liquid argon Ar(ℓ) is 25.0 J/ mol C and gaseous Ar(g) is 20.8 J/mol C. The enthalpy of evaporation is 6506 J/mol. How much energy is required to convert 1 mole of liquid Ar from 5◦C below its boiling point to1 mole of gaseous Ar 5◦C above its boiling?
Please help me I'm so lost!
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Verified answer
Use Q = m(mass)*c(specific heat)*deltaT
So
(1*25*5)+(1*6506)+(1*20.8*5)=6735 J