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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?

Given:
- 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, [tex]T_b = 97.4[/tex] °C

Answer :

At 128.2 °C, 10.1 g of the unknown liquid needs 7.582 kJ of heat to become a gas.

The normal melting point is what?

A substance's normal melting point is its melting point at 1 atm of pressure. The melting point of a solid is the same as the freezing point of a pure material.

we need to calculate the amount of energy required to heat the liquid from 43.5 °C to its boiling point, vaporize it at its boiling point, and then heat the resulting gas from its boiling point to 128.2 °C.

Q1 = m × Cp × ∆T1

Substituting the given values, we get:

Q1 = 10.1 g × 1.18 J/g･ °C × (97.4 °C - 43.5 °C) = 633.38 J

Q2 = n × ∆Hvap

n = m / MM

where MM is the molar mass of the liquid (in g/mol).

n = 10.1 g / 67.44 g/mol = 0.1499 mol

Q2 = 0.1499 mol × 30.1 kJ/mol = 4.51299 kJ = 4,512.99 J

Q3 = m × Cp × ∆T2

Q3 = 10.1 g × 0.792 J/g･ °C × (128.2 °C - 97.4 °C) = 2494.956 J = 2.495 kJ

Q = Q1 + Q2 + Q3 = 633.38 J + 4,512.99 J + 2.495 kJ = 7,582.37 J = 7.582 kJ (to three significant figures)

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