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(12 pts) How much heat is involved when 42.0 grams of isopropanol, C3H30, initially at 10.0°C is warmed to 125.0°C? Isopropanol has a melting point of -89°C and a boiling point of 82.6°C. Specific heat of solid C H2O 0.212 J/g.°C Specific heat of liquid C,H,O 2.68 J/g.°C Specific heat of gaseous C3H2O 1.54 J/g.°C 2.68 5.2 kJ/mole AH yaporization 38.5 kJ/mole (42.04) ΔΗρυκόρα. 9 co

Answer 1

Mass of C₂ HQ O = 42gm Molar mass of C₂ H₂O = 60 g/mol so Mole of C Hg 0 = 42 g 60 g/mol 0.7 mole Срі 2.68 Cpg = 10 54 J/goc DHy= 38.5 X 10 mole TE 10°c To= 82.6°C T2= 125°C. so we have to heat the material from I to To the heat 14 sensible heat for vaporization, it is latent heat To Tz - sensible heat T 80, QE mpo (To-T,) + n D Hvt mcpv (T2-T6) more -> Q- 42 946x 2.623 (22:6-10%eb% + 0.7 motex 38.5X13_J + 42 goix 1.547 x (125-82.6)°¢ 0 - [8171.856 + 26950+ 2742.432]J met gehie > Q= 37864•288 J

We are using the moles while calculation of the heat of vaporization because it is given per mole.

So heat involved is 37864.288J