1.00 g octane burned in a bomb calorimeter causes the temperature of the bomb and the 1.20 kg water surrounding it to change from 25.00°C to 33.20°C. Calculate the heat change per mole of reaction.
C(bomb) = 837 J K
–1 and
c(H
2O) = 4.184 J g
–1 K
–1C8H18(l) + 25/2 O2(g) → 8CO2(g) + 9H2O(g)
Calculate q(surroundings). ΔT = 33.20°C - 25.00°C = 8.2°C = 8.2 K
Temperature change for bomb and the calorimeter.
q(surroundings) = q(H2O) + q(bomb)
q(surroundings) = c(H2O) × m(H2O) × ΔT + C(bomb) × ΔT
q(surroundings) = 41170.6 J + 6863.4 J = 48034 J
q(system) = -q(surroundings)
Assume that heat transfer is complete.
q(system) = -q(reaction)
q(reaction) = -48034 J