Question

A mixture of 20 litres of 1-hexanol (C6H14O) and 15 litres of eugenol (C10H12O2) are mixed and heated from 25°C to 125°C. Calculate the heat input required in (kJ). Assume there is no heat of mixing. The specific gravity of 1-hexanol and eugenol are 1.061 and 1.021, respectively. Hint: Use Kopp’s rule to calculate the specific heat

Answer 1

Given specific gravities of 1-hexanol and eugenol are 1.061 and 1.021 respectively

Temperature difference ΔT= 100K

As specific gravity of liquid = density of the liquid/density of water,

we have, density of 1-hexanol = 1.061*1 Kg/L = 1.061 Kg/L and density of eugenol = 1.021 Kg/L

As density = mass/volume, we have,

Mass of 1-hexanol = 1.061*20L = 21.22 Kg

Moles of 1-hexanol, n1 = 21.22/102.162 = 0.2077 Kmol

Mass of eugenol = 1.021*15L = 15.31 Kg

Moles of eugenol, n2 = 15.31/164.2 = 0.0932 Kmol

At, temp = 25 deg C= 298.15 K,

Kopp's rule

where C_i and f_i denote the specific heat and mass fraction of the i-th constituent

Specific heats of Carbon, Hydrogen, and Oxygen at 298.15 K are 0.71 J/ g. K, 14.304 J/g. K and 0.92 J/ g. K respectively

Therefore, Cp of 1-hexanol (C6H14O) = 6*Cp of Carbon + 14 * Cp of hydrogen + 1* Cp of oxygen

= (6 x 0.71 x 12) + (14 x 14.304 x 1) + (1 x 0.92 x 16)

[Converted J/g. K to J/mol. K by multiplying with gram molecular weight]

Cp1 = 266.096 J/mol. K = 266.096 KJ/Kmol. K

Similarly, Cp of eugenol (C10H12O2) = (10 x 0.71 x 12) + (12 x 14.304 x 1) + (2 x 0.92 x 16)

Cp2 = 286.288 J/mol. K = 286.288 KJ/ Kmol. K

Therefore, heat input is Q= n*cp*ΔT, where n = num of moles

Q = (n1 x Cp1 x ΔT) + (n2 x Cp2 x ΔT)

Q = (0.2077 x 266.096 x 100) + (0.0932 x 286.288 x 100)

Q = 8,195 KJ

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