Calculate the magnitude of the heat transfer (kJ) required to cool 75.0 liters of a liquid mixture containing 70.0wt% acetone and 30.0% 2-methyl-1-pentanol (C6H140) from 55.0°C to 25.0°CThe specific gravity of 2-methyl-1-pentanol is about 0.826. The true heat capacity of 2-methyl-1-pentanol isabout 248.0 J/(mol °C).Estimate the required heat transfer using Kopp's rule to estimate the heatcapacities of both acetone and 2-methyl-1-pentanol.ikJEstimate the required heat transfer using the true heat capacities.ikJ

Koop's Rule states that the sum of heat capacities of individual components in a compound is equal to the heat capacity of the same compound. We can calculate heat capacity of compounds from the individual compounds present in it. Heat transferred is calculated from product of mass change in temperature and heat capacity. Acetone and methyl pentanol heat capacity are calculated with help of Koop's rule.Using Koop's rule we will calculate the heat capacity of the given compounds. Heat transferred is also calculated. Acetone heat capacity calculation heat capacity of C3H6O=3(heat capacity of C)+6(heat capacity ofH)+1(heat capacity of O)heat capacity of C3H6O=3(11.7)+6(18)+25.1 J/mol°Cheat capacity of C3H6O=168.2 J/mol°Cmolecualr weight of C3H6O=58 g/molheat capacity of C3H6O=168.2 J/mol°C58 g/molheat capacity of C3H6O=2.9 J/g°C 2-methyl -1-pentanol heat capacity calculation heat capacity of C6H14O=6(heat capacity of C)+14(heat capacity ofH)+1(heat capacity of O)heat capacity of C6H14O=6(11.7)+14(18)+25.1 J/mol°Cheat capacity of C6H14O=347.3J/mol°Cmolecualr weight of C6H14O=102 g/molheat capacity of C6H14O=347.3 J/mol°C102 g/molheat capacity of C6H14O=3.405 J/g°C Heat capacity of mixture =(0.7×2.9J/g°C)+(0.3×3.405J/g°C)=3.0515J/g°C Mass determination basis of 100 gm of mixture Volume of acetone=70 g acetonespecific gravity of acetoneVolume of acetone=70 g acetone0.787Volume of acetone=88.9 mLVolume of 2-methyl-1-pentanol=30 g 2-methyl-1-pentanolspecific gravity of 2-methyl-1-pentanolVolume of 2-methyl-1-pentanol=30 g 2-methyl-1-pentanol0.826Volume of 2-methyl-1-pentanol=36.3 mL Volume of mixture =88.9mL+36.3mL=125.2mL=0.1252 L mass required for 75L 0.1252L→100g75L→75L×100g0.1252L=59904.15g heat transfer calculation Heat transferred=mass×heat capcity×∆THeat transferred=59904.15 g×3.0515J/g°C×(55-25)°CHeat transferred=5483925.412 JHeat transferred=5483.925 kJ We have mass of 59904.15 g ∆T of 30°C True specific heat capacity of acetone is 2.15J/g°C True specific heat capacity of 2-methyl-1-pentanol is 2.43 J/g°C Heat capacity of mixture =(0.7×2.15J/g°C)+(0.3×2.43J/g°C)=2.234 J/g°C Heat transferred=mass×heat capcity×∆THeat transferred=59904.15 g×2.234J/g°C×(55-25)°CHeat transferred=4014776.133 JHeat transferred=4014.776 kJ Final answers a) Heat transferred by Koop's rule 5483.925 kJ b) Heat transferred on true heat capacities 4014.776 kJ