Answer to Question #121206 in General Chemistry for Anna

Question

the amount of heat (kJ) transferred when 80g of a new chemical substance (M=53g/mol , freezing point is - 10 degrees * C , boiling point is 78^ C) is heated from 20 degrees * C to 100 degrees * C . The following parameters are reported for this substance: [C l =30.5 J/mol.^ C C s =22.6 J/mol.^ C C g =58.1 J/mol. C, Delta Hvap=13.5 kJ/mol, AHfus = 7.4 J/mol]

Answer 1

According to the values of the boiling and fusion points, the substance heating from 20°C to 100 °C can be divided in the following steps:

1) the heating of liquid from 20°C to 78°C

2) the evaporation of liquid at 78°C

3) the heating of gas from 78°C to 100 °C.

As the molar heat capacities of liquid, gaseous and solid substance are given, let's first calculate the number of the moles of the substance:

$n = \frac{m}{M} = \frac{80\text{ g}}{53\text{ g/mol}} = 1.5094$ mol.

Therefore, the calculation of the heat transferred to the substance can be performed as:

$Q = Q_1 + Q_2 + Q_3$ .

The heat transferred in the first step is:

$Q_1 = c_l·n·(T_2-T_1)$

$Q_1 = 30.5\text{ (J/mol°C)}·1.5094 \text{ (mol)}·(78-20) (\text{°C})$

$Q_1 = 2670$ J.

The heat transferred in the second step:

$Q_2 = ∆H_{vap}·n = 13.5\text{(kJ/mol)}·1.5094 \text{(mol)}$

$Q_2 = 20.377$ kJ, or 20377 J.

The heat transferred in the third step:

$Q_3 = c_g·n·(T_2-T_1)$

$Q_3 = 58.1\text{ (J/mol°C)}·1.5094 \text{ (mol)}·(100-78) (\text{°C})$

$Q_3 = 1929$ J.

Finally, the overall heat transferred to the substance is:

$Q = 2670 + 20377 + 1929 = 24977$ J, or 25 kJ.

Answer: 25 kJ of heat were transferred.