Question

Calculate the energy involved (in kJ) when 27.6 g ethanol (CH3CH2OH) goes from 106.5°C to -162.1°C. (Be sure to include the sign.) Ethanol boils at 78.37°C and freezes at -114°C with ΔHvap = 38.56 kJ/mol and ΔHfus = 4.9 kJ/mol; cliquid= 2.440 J/g°C, cgas = 1.699 J/g°C and csolid = 2.419 J/g°C.

Answer 1

27.6 g ethanol = 27.6 g / 46 g / mole = 0.60 mole

n1 = 0.60 mole.

Energy involved = [ m1 * c_gas * dT1 (from 106.5°C to 78.37°C) + n1 * ΔHvap + m2 * c_liquid * dT2 (from 78.37 to -114°C) + ΔHfus * n1 + m1 * c_solid * dT2 ( -114°C to -162.1°C)

or

Energy involved = [27.6 * 1.699 * (78.37 - 106.5) + 0.60 * (-38.56 * 1000) + 27.6 * 2.440 * (-114 - 78.37) + (4.9 * 1000 * 0.60) + 27.6 * 2.419 * (-162.1 + 114)]

or

Energy involved = [-1319.08 - 23136 - 12955 - 2940 - 3211.4 ]

or

Energy involved = - 43561.4 J = - 43.56 KJ (answer)