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4 July, 00:16

The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air

originally at 21.7°C. Assume that all of the heat produced by the reaction was absorbed by the air

(specific heat = 1.005 J / (g. °C)) in the room.

(c) Determine the final temperature of the air in the room after the combustion.

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  1. 4 July, 00:31
    0
    Combustion reaction is given below,

    C₂H₅OH (l) + 3O₂ (g) ⇒ 2CO₂ (g) + 3H₂O (g)

    Provided that such a combustion has a normal enthalpy,

    ΔH°rxn = - 1270 kJ/mol

    That would be 1 mol reacting to release of ethanol,

    ⇒ - 1270 kJ of heat

    Now,

    0.383 Ethanol mol responds to release or unlock,

    (c) Determine the final temperature of the air in the room after the combustion.

    Given that:

    specific heat c = 1.005 J / (g. °C)

    m = 5.56 * 10⁴ g

    Using the relation:

    q = mcΔT

    - 486.34 = 5.56 * 10⁴ * 1.005 * ΔT

    ΔT = (486.34 * 1000) / 5.56*10⁴ * 1.005

    ΔT = 836.88 °C

    ΔT = T₂ - T₁

    T₂ = ΔT + T₁

    T₂ = 836.88 °C + 21.7°C

    T₂ = 858.58 °C

    Therefore, the final temperature of the air in the room after combustion is 858.58 °C
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