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

Question

Need help solving this.

Answer 1

q = mC∆T

∆H = -3325 kJ/mol

How many moles do we have?

3.65 g butane x 1 mol butane (C4H10) / 58.12 g = 0.0628 moles butane

Total heat needed to burn the butane = 0.0628 mols x -3325 kJ/mol = 209 kJ = q = 209,000 J

m= mass of water = 1000 g (assuming a density of 1 g/ml)

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = ?

q = mC∆T

209,000 J = (1000 g)(4.184 J/gº)(∆T)

∆T = 49.95º

Final temperature = 21.0º + 49.95º = 70.95º = 71.0º (3 sig. figs.)