Calculate the heat transfer (kJ) required to cool 65.0 liters of a liquid mixtur

Question

Calculate the heat transfer (kJ) required to cool 65.0 liters of a liquid mixture containing 70 wt% acetone and 30% 2-methyl-1-pentanol (C6H14O) from 50°C to 25°C

The specific gravity of 2-methyl-1-pentanol is about 0.826. The true heat capacity of 2-methyl-1-pentanol is about 248.0 J/(mol °C)

a) Estimate the required heat transfer using Kopp's rule to estimate the heat capacities of both acetone and 2-methyl-1-pentanol. (KJ)  

b)Estimate the required heat transfer using the true heat capacities (KJ)

Explanation / Answer

Basis : 100 gm of mixture

it contains 70gms of acetone and 2 methyl 1- pentanol

specific gravity of acetone is 0.787 and that of 2 methyl pentanol is 0.826

volume of acetone= 70/0.787 =89ml and that of 2 methyl pentanol= 30/ 0.826= 36.3 ml

volume of mixture= 89+ 36.3 = 125.3 ml=125.3/1000 L= 0.1253

0.1253 L corresponds to 100gm

65 L correspond to 65*100/0.1253=51875.5gm

specific heat of acetone = 2.15 j/g.k and that of 2 methyl 1-pentanol= 248 j/102g.deg.c = 2.43

mixture specific heat= 0.7*2.15+0.3*2.43=2.334j/g.K

Heat transferred = mcpdelT= 51875.5 g* 2.334 * (50-25) =2897247 Joules= 2897.247 Kj

b) acetone molecular formula is C3H6O molecular weight = 58

hence as per Koops rule heat capacity=3*12+18^6+ 25 = 169 J/mol.K =169/58 J/g.K =2.91 J/g.K

Molecular formula of 2 methyl 1 pentanol C6H14O , molecular weight= 102

heat capacoty = 6*12+ 14*18+ 25*1 =349 j/mol.K= 349/102.j/g.K = 3.42 j/g,k

mixture specific heat =0.7*2.91+0.3* 3.42 =3.063 j/g.k heat transferred = 51875.5* 3.063*(50-25)= 3972366 Joules

= 3972.366 Kj

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