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Calculate the heat required to convert 69.0 g of C2Cl3F3 from a liquid at 14.10 ∘C...

Calculate the heat required to convert 69.0 g of C2Cl3F3 from a liquid at 14.10 ∘C to a gas at 79.30 ∘C.

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Answer #1

The normal boiling point of C2Cl3F3 is 47.6 oC

The amount of heat required,q= MCdt + mL. + MC' dt'

Where

M= mass of C2CL3F3= 69.0 g*(1mol/187.5(g/mol)=0.368mol

C= The specific heat capacity of C2Cl3F3 is 170.1 J/mol K for the liquid

C'= The specific heat capacity of C2CL3F3 125.5 J/mol K for the gas

dt= 47.6-14.10= 33.5 oC

L= molar enthalpy of vaporization is 27.49 kJ/mol.= 27.49*10^3 J

dt'= 79.30-47.6= 31.7 oC

Plug the values we get,. q= 13.68*10^3 J= 113.68 kJ

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