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29 July, 12:00

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?

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  1. 29 July, 14:15
    0
    -63.79 kJ/g

    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
  2. 29 July, 14:56
    0
    The heat of combustion per gram of the material is - 63.8 kJ / gram

    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
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