Question

(30 pts) Calculate the magnitude of the energy transfer (kJ) required to cool 75.0 liters of a liquid mixture containing 70.0 wt% acetone and 30.0% 2-methyl-1-pentanol (C6H140) from 45.0°C to 25.0°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). HINT: Since you do not know how much of the mixture is in the vessel, you may want to assume a "temporary" weight basis to begin. a. Estimate the required energy transfer (kJ) using Kopp's rule to estimate the heat capacities of both acetone and 2-methyl-1-pentanol. b. Estimate the required energy transfer using the true heat capacities (e.g., Table B.2).

Answer 1

According to Kopp's law, specific heat capacity "C" can be calculated as follows

c=Σ(ci..)

where N is the total number of constituents, C_i is the specific heat of the i^th components and f_i is the mass fraction of the i^th components.

Given the specific gravity of 2 methyl 1 pentanol is 0.826. We also know that the specific gravity of acetone is 0.788. The specific gravity of these materials can be approximated to their densities. The average density of the liquid mixture is 0.807 g/cm³. Using this approximate density we can find out the weight of 75 L of the solution which is equal to 60.525 kg.

We need to find out the mass fraction and specific heat capacities of both the constituents. We have 70% of acetone and 30% of 2methyl1pentanol in the liquid mixture.

Weight of 70 wt.% of acetone = 0.70 * 60.525

= 42.37 kg

Weight of 30 wt.% of 2mtehyl 1pentanol = 0.30 * 60.525

= 18.16 kg

Heat capacity of acetone taken from the literature = 125.5 J mol^-1 K^-1

Specific heat capacity of acetone = 125.5/42.37 J kg^-1 K^-1

= 2.962 J kg^-1 K^-1

Heat capacity of 2mtehyl 1pentanol = 248.0 J mol^-1 K^-1

Specific heat capacity of 2mtehyl 1pentanol = 248.0/18.16 J kg^-1 K^-1

= 13.656 J kg^-1 K^-1

Specific heat capacity of the liquid mixture = (0.70 * 2.962) + (0.30 * 13.656) J kg^-1 K^-1

= (2.073 + 4.097) J kg^-1 K^-1

= 6,17 J kg^-1 K^-1

We know that

C = (ΔΗ/ΔΤ)

from this enthalpy/energy change = C x temperature change

= 6.17 J kg^-1 K^-1 x 20 K

energy change = 123. 4 J/kg

or = 7468.78 J

energy change = 7.468 kJ

b) energy transfer for 2methyl 1 pentanol = C x temperature change

= 248.0 J mol^-1 K^-1 x 20 K

= 4960 J mol^-1

= 4.96 kJ mol^-1