Question

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

Answer 1

$C_4H_{10} + \frac{13}{2}O_2 \to 4CO_2 + 5H_2O$

3.65g of Butane = 3.56/58 = 0.063 mol

Heat evolved = n(∆$H_c$) = 0.063(-3325) = -209.5 kJ

Q = mc∆T = mc($T_2 -T_1$)

V = 1.0 L

m = density × volume = 1kg/L × 1.0 L

= 1kg

Q = 209.5 kJ = 209,500J

$T_1$ = 21.0°C

$T_2$ = T

c = 4200J/kg.°C

209,500 = 1 × 4200 × (T - 21)

209500/4200 = T - 21

49.9 = T - 21

T = 70.9°C