What is the answer?

Question

What amount of heat (in kJ) is required to convert 10.1 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)

Answer 1

First, start by finding the amount of heat it takes to reach the boiling point (from 43.5°C to 97.4°C) which is equal to the specific heat of the liquid * mass * ΔT.

(1) ΔT = 97.4 °C - 43.5 °C = 53.9 °C

(2) H₁ = 1.18 J/(g·°C) * 10.1 g * 53.9 °C = 642.38 J

Looking ahead, the problem asks for the solution in kJ so I'll convert from J to kJ as I go.

(3) H₁ = 642.38 J * 0.001 kJ/J = 0.642 kJ

Now we turn our attention to the heat of vaporization (also known as "enthalpy of vaporization" depending on your textbook. While the specific heat constants are given in terms of mass, this constant is almost universally given in terms of moles. So we have to first determine the number of moles of our unknown substance is present.

(4) n_moles = mass_s / MM_s = 10.1 g / 67.44 g/mol = 0.1498 moles

(5) H_vap = n_moles * ΔH_vap = 0.1498 * 30.1 kJ/mol = 4.509 kJ

Lastly, we calculate the heat absorbed by the substance in its gaseous form. Which follows the same pattern as the liquid phase (equation 2), simply a different constant.

(6) ΔT = 128.2 °C - 97.4 °C = 30.8 °C

(7) H₂ = 0.792 J/(g·°C) * 10.1 g * 30.8 °C = 246.37 J * 0.001 kJ/J = 0.246 kJ

Finally, we simply need to add all 3 heat values:

(8) H_tot = 0.642 kJ + 4.509 kJ + 0.246 kJ = 5.397 kJ

Thus the final answer to transition a 10.1-gram sample of this substance from a 43.7 °C liquid to a 128.2 °C gas is 5.397 kJ of energy.

A few key concepts from this example:

1. The phase change step tends to dominate the energy requirements.
2. Be sure to understand how to readily convert from a mass to the number of moles and vice-versa (this is a calculation you'll be repeating through at least your undergraduate chemistry career [probably further, but I have no direct experience beyond senior physical chemistry].
3. Remember that there isn't a temperature change during the phase transition.