Question

Calculate the magnitude of the heat transfer (kJ) required to cool 65.0 liters of a liquid mixture containing 70.0 wt% acetone and 30.0% 2-methyl-1-pentanol (C6H14O) from 45.0°C to 20.0°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).

Estimate the required heat transfer using Kopp's rule to estimate the heat capacities of both acetone and 2-methyl-1-pentanol. ? kJ

Estimate the required heat transfer using the true heat capacities. ? kJ

Answer 1

Density of 2-methyle -1- pentanol = 826 Kg/m³, density of acetone = 784 Kg/m³ from litrature

Assume total waight is W Kg, so mass of acetone = 0.7 W (70% acetone)

Volume of acetone = Mass / density = 0.7 W/784 m³

Mass of 2 methyle 1 pentanol = 0.3 W (30% 2 methyle 1 pentanol)

Volume of 2 methyle 1 pentanol = Mass/ density = 0.3W/ 826 m³

total volume = volume of acetone + volume of 2 methyle 1 pentanol = 0.7W/784 + 0.3W/826 = 0.065 m³ (65 liter = 0.065 m³ or 1000 liter = 1 m³)

from the above equation W = 51.75 Kg

Heat capacity of 2 methyle 1 pentanol = 248 j/mol oC = 248 Kj/Kmol. oC (j/mol = Kj/ Kmol)

1 Kmole of 2 methyle 1 pentanol = 102 Kg (molecular weight of 2 methyle 1 pentanol=102)

So heat capacity of 2 methyle 1 pentanol = 248/102 = 2.43 Kj/ Kg oC

Heat capacity of acetone = 2.15 Kj/Kg oC from litrature

Kopps rule = Where i is component Xi is mole fraction of component

in oure case total Cp = Acetone + 2 methyle 1 pentanol = 0.7 * 2.15 + 0.30*2.43 = 2.234 Kj/Kg oC

Heat transfer = W = 51.75 Kg , Cp = 2.234 Kj/Kg oC   45 oC - 20 oC = 25 oC

Heat transfer = 51.75*2.234*25 = 2890 KJ

(ii) heat transfer using true heat capacities

Q = heat transfer of acetone + heat transfer of 2 methyle 1 pentanol

Q = W Cp = 0.7 W * 2.15 *25 + 0.3 W * 2.43* 25 ( = 25 oC, W = 51.75)

Q = 2890 Kj    This is the same result.