Question

A 2.425 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 24.91 ∘ C to 30.28 ∘ C. The heat capacity (calorimeter constant) of the calorimeter is 28.81 kJ / ∘ C, what is the heat of combustion per gram of the material?

Answer 1

The heat of combustion per gram of the material is -63.8 kJ/ gram

Step-by-step explanation:

Step 1: Data given

Mass of a ew organic material = 2.425 grams

The initial temperature of the calorimeter = 24.91 °C

The final temperature of the calorimeter = 30.28 °C

The heat capacity (calorimeter constant) of the calorimeter is 28.81 kJ /°C

Step 2: Calculate heat

Q = c*ΔT

⇒with c = the heat capacity (calorimeter constant) of the calorimeter is 28.81 kJ /°C

⇒with ΔT = The change of temperature = T2 - T1 = 30.28 - 24.91 = 5.37 °C

Q = 28.81 kJ/ °C * 5.37 °C

Q = 154.7 kJ

Step 3: Calculate the heat of combustion per gram of the material

heat of combustion per gram = -Q / mass (negative since it's exothermic)

heat of combustion per gram = -154.7 kJ / 2.425 grams

heat of combustion per gram = -63.8 kJ/ gram

The heat of combustion per gram of the material is -63.8 kJ/ gram

Answer 2

-63.79 kJ/g

Step-by-step explanation:

According to the law of conservation of energy, the sum of the heat released by the combustion of the new organic material (Qcomb) and the heat absorbed by the bomb calorimeter (Qbc) is zero.

Qcomb + Qbc = 0

Qcomb = -Qbc [1]

We can calculate the heat absorbed by the bomb calorimeter using the following expression.

Qbc = C × ΔT

where,

• C: calorimeter constant

• ΔT: change in the temperature

Qbc = C × ΔT

Qbc = 28.81 kJ/°C × (30.28°C - 24.91°C) = 154.7 kJ

From [1],

Qcomb = -154.7 kJ

The heat of combustion per gram of the material is:

-154.7 kJ / 2.425 g = -63.79 kJ/g