Question

Tetrahydrofuran (THF) is a common organic solvent with a boiling point of 339 K. Calculate the total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K. The specific heat of liquid THF is 1.70 J/g K, the specific heat of THF vapor is 1.06 J/g K, and the heat of vaporization of THF is 444 J/g.

Answer 1

Explanation:

For the given reaction, the temperature of liquid will rise from 298 K to 339 K. Hence, heat energy required will be calculated as follows.

             $Q_{1} = mC_{1} \Delta T_{1}$

Putting the given values into the above equation as follows.

           $Q_{1} = mC_{1} \Delta T_{1}$

                      = $27.3 g \times 1.70 J/g K \times 41$

                      = 1902.81 J

Now, conversion of liquid to vapor at the boiling point (339 K) is calculated as follows.

           $Q_{2}$ = energy required = $mL_{v}$

    $L_{v}$ = latent heat of vaporization

Therefor, calculate the value of energy required as follows.

             $Q_{2}$ = $mL_{v}$

                         = $27.3 \times 444$

                         = 12121.2 J

Therefore, rise in temperature of vapor from 339 K to 373 K is calculated as follows.

            $Q_{3} = mC_{2} \Delta T_{2}$

Value of $C_{2}$ = 1.06 J/g,    $\Delta T_{2}$ = (373 -339) K = 34 K

Hence, putting the given values into the above formula as follows.

             $Q_{3} = mC_{2} \Delta T_{2}$

                       = $27.3 g \times 1.06 J/g \times 34 K$

                       = 983.892 J

Therefore, net heat required will be calculated as follows.

            Q = $Q_{1} + Q_{2} + Q_{3}$

                = 1902.81 J + 12121.2 J + 983.892 J

                = 15007.902 J

Thus, we can conclude that total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K is 15007.902 J.