Question

30 kg of liquid n-propyl benzoate (C6H5CO2C3H7) and 15 kg of liquid benzene (C6H6) are mixed and heated from 25°C to 75°C. Calculate the required heat input (kJ). Hint: use Kopp’s rule to estimate the heat capacity of liquid C6H5CO2C3H7 (Table B.10); find heat capacity of benzene from Table B.2.

Answer 1

Using the Kopp's rule to determine the heat capacity of the given liquid:

Heat capacity of C₆H₅CO₂C₃H₇ = 10*Cp(C) + 12*Cp(H) + 2*Cp(O) = 10*12 + 12*18 + 2*25 = 386 J/(mole.K)

The above values of specific atomic heat capacities have been taken from literature.

Now, MW of C₆H₅CO₂C₃H₇ = 164 g

Moles of C₆H₅CO₂C₃H₇ taken = Mass/MW = 30,000/164 = 182.9

Moles of benzene taken = Mass/MW = 15,000/78 = 192.30

Heat capacity of benzene = 134.8 J/mol.K

Total heat required for raising the temperature

= Heat required for benzene + Heat required for C₆H₅CO₂C₃H₇

= moles of benzene*Cp*dT + moles of C6H5CO2C3H7*Cp*dT

= 192.30*134.8*(75-25) + 182.9*386*(75.25)

= 4826.078 kJ