What amount of heat (in kJ) is required to convert 14.0 g of an unknown liquid (MM = 67.44 g/mol) at 43.5 °C to a gas at 128.2 °C? (specific heat capacity of liquid = 1.18 J/g･°C; specific heat capacity of gas = 0.792 J/g･°C; ∆Hvap = 30.1 kJ/mol; normal boiling point, Tb = 97.4°C)

1.24 kJ is required to convert 14 g of liquid from 43.5°C to 128.2°C

Explanation:

This is a typical calorimetry problem:

We have to assume, no heat is lost to sourrounding.

First of all, we need to go from 43.5°C to 97.4°C, the boiling point.

Q = Ce . m . ΔT

We replace data, 1.18° J/g . 14 g . (97.4°C - 43.5°C)

Heat for the first stage is: 890.4 Joules

Now we have to change the state, and we need the ΔH. As we do not have latent heat, we can proceed like this:

1 mol release 30.1 kJ at vaporization.

We convert the mass to moles → 14 g.  1mol/ 67.44g = 0.207 mol

0.207 mol will release (0.207 . 30.1 kJ) = 6.25 kJ

Now, we are at gaseous phase.

Q = Ce . m . ΔT → 0.792 J/g°C . 14g . (128.2°C - 97.4°C)

Q = 341.5 Joules

To determine the amount of heat, we sum all the obtained values:

890.4 Joules + 6250 Joules + 341.5 Joules = 1238.2 J

We convert to kJ →  1238.2 J . 1kJ / 1000J = 1.24 kJ