What amount of heat (in kJ) is required to convert 10.1 g of an unknown liquid (MM = 83.21 g/mol) at 19.2 °C to a gas at 93.5 °C? (specific heat capacity of liquid = 1.58 J/g･°C; specific heat...

Question

What amount of heat (in kJ) is required to convert 10.1 g of an unknown liquid (MM = 83.21 g/mol) at 19.2 °C to a gas at 93.5 °C? (specific heat capacity of liquid = 1.58 J/g･°C; specific heat capacity of gas = 0.932 J/g･°C; ∆Hvap = 22.5 kJ/mol; normal boiling point, Tb = 57.3°C)

Answer 1

Since the normal boiling point is 57.3º, this problem will require 4 steps.

Step 1. Find heat needed to raise the temperature of the liquid from 19.2º to 57.3º

q = mC∆T

q = heat = ?

m = mass = 10.1 g

C = specific heat of liquid = 1.58 J/gº

∆T = change in temperature = 57.3º - 19.2º = 38.1º

q = (10.1 g)(1.58 J/gº)(38.1º) = 608 J

Step 2. Find the heat needed to convert the liquid to a gas at the boiling point of 57.3º (phase change)

q = m∆Hvap

q = heat = ?

m = mass = 10.1 g

∆Hvap = enthalpy of vaporization = 22.5 kJ/mol x 1 mol / 83.21 g = 0.2704 kJ/g = 270.4 J/g

q = (10.1 g)(270.4 J/g) = 2731 J (NOTE: there is no ∆T since this is just a phase change so no change in T)

Step 3. Find the heat needed to raise the temperature of the gas from 57.3º to 93.5º

q = mC∆T

q = heat = ?

m = mass = 10.1 g

C = specific heat of the gas = 0.932 J/gº

∆T = change in temperature = 93.5º - 57.3º = 36.2º

q = (10.1 g)(0.932 J/gº)(36.2º) = 341 J

Step 4. Add up the heats from each step to get the total amount of heat needed.

608 J + 2731 J + 341 J = 3680 J