Answer to Question #231035 in General Chemistry for Naz

Question

If 3.65g of butane is burned underneath a cup holding 1.00 L of water at 21.0 oC, what will the final temperature of the water be (Hcomg = -3325kJ/mol)?

Answer 1

The combustion reaction for butane is:

2C₄H₁₀ + 13O₂ = 8CO₂ + 10H₂O + 3325 kJ/mol

We have 3.65/58 = 0.0629 mol of butane, so during combustion this amount of butane will be released 3325*0.0629 = 209.2 kJ of heat.

Q = c*m*Δt, where Q - amount of heat, C - heat capacity for wather (4.19 kJ/kg*K), m - mass of water (we have 1.00 L = 1.00 kg).

Δt = Q/(c*m) = 209.2 / (1*4.19) = 49.94 ^oC

The final temperature of the water be 21.00 + 49.94 = 70.94 ^oC