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.

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