Question

If you take Organic Chemistry Laboratory (CHM2211L) you will learn the term reflux. A reaction is considered to be at reflux when it is set at the boiling point of your solvent. a. In order to bring 74.81 mL of chloroform (density = 1.4832 g/cm3 , c = 0.96 J/g*K ) to reflux from room temperature (25 oC) it requires 1.46 kJ of energy. What will be the temperature (in oF) of the chloroform?

Answer 1

Answer : The temperature of the chloroform will be, $101.67^oF$

Explanation :

First we have to calculate the mass of chloroform.

$\text{Mass of chloroform}=\text{Density of chloroform}\times \text{Volume of chloroform}=1.4832g/ml\times 74.81ml=110.958g$

conversion used : $(1cm^3=1ml)$

Now we have to calculate the temperature of the chloroform.

Formula used :

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = amount of heat or energy = 1.46 kJ = 1460 J   (1 kJ = 1000 J)

$c$ = specific heat capacity = $0.96J/g.K$

m = mass of substance = 110.958 g

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $25^oC=273+25=298K$

Now put all the given values in the above formula, we get:

$1460J=110.958g\times 0.96J/g.K\times (T_{final}-298)K$

$T_{final}=311.706K$

Now we have to convert the temperature from Kelvin to Fahrenheit.

The conversion used for the temperature from Kelvin to Fahrenheit is:

$^oC=\frac{5}{9}\times (^oF-32)$

As we know that, $K=^oC+273$ or, $K-273=^oC$

$K-273=\frac{5}{9}\times (^oF-32)$

$K=\frac{5}{9}\times (^oF-32)+273$  ...........(1)

Now put the value of temperature of Kelvin in (1), we get:

$311.706K=\frac{5}{9}\times (^oF-32)+273$

$T_{final}=101.67^oF$

Therefore, the temperature of the chloroform will be, $101.67^oF$