Question

Determine the amount of energy (in kJ) removed while changing 46.1g of C2H5OH from a liquid at 53.1C to a solid at -117.3C.

Answer 1

The process of bringing ethanol from 53.1^oC to -1.

Total heat released is the sum of heat released in three17.3^oC will take place in three steps:

(1) Bringing ethanol from 53.1^oC to fusion point (-114.3^oC)

(2) Fusion

(3) Bringing solid ethanol from -114.1^oC to -117.3^oC process

Heat released = (i) + (ii) + (iii)

Heat released = (m X C X ΔT₁) + ΔH_f2 + (m X C X ΔT₃)

m = mass of ethanol = 46.1g = 1 mole

C = specific heat of ethanol = 0.1KJ/mol^oC

ΔT₁ = temperature difference = -114.1^oC - 53.1^oC = -167.2 ^oC

ΔT₂ = temperature difference = -117.3^oC -( -114.1^oC) = -3.2 ^oC

ΔH_f = 4.9KJ/mol

Heat released = (1mole X 0.1KJ/mol  ^oC X -167.2 ^oC) + 4.9KJ/mol +  (1mole X 0.1KJ/mol  ^oC X -3.2)

Heat released = - 16.72 + 4.9KJ/mol - 0.32 = 12.14KJ