What amount of heat(kJ) is rq to convert 14.8 g of a liquid (MM = 83.21 g/mol) at 19.2 °C to a gas at 93.5 °C?

(HC of Lq = 1.58 J/g・°C;HC of G=0.932 J/g・°C; ∆Hvap=22.5kJ/mol; Tb=57.3°C)

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  • 8 months ago

    Heat to raise the temperature of the liquid to its boiling point:

    q = m c (T2-T1)

    q = 14.8 g (1.58 J/gC) (57.3 - 19.2 C) = 891 J

    Heat to vaporize liquid:

    q = moles X ∆Hvap

    q = (14.8 g / 83.21 g/mol) (22.5 kJ/mol) = 4.00 kJ = 4000 J

    Heat to raise temperature of gas to final temp:

    q = m c (T2-T1)

    q = 14.8 g (0.932 J/gC) (93.5 - 57.3 C) = 499 J

    Total heat = 891 + 4000 + 491 = 5.39X10^3 J = 5.39 kJ

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