a chemical engineer placed 1.520 g of a hydorcarbon in the bomb of a calorimeter. The bomb was immersed in 2.550 L of water and the sample was burned. The water temperature rose from 20.00C to 23.55C. If the calorimeter (excluding the water) had a heat capacity of 403 J/K, what was the heat released per gram of hydrocarbon?
I'm unsure of my work but here it is:
Q(bomb) = 403J/K * 3.55K = 1430.65J
Q(H2O) = 2.550L * (1000g/L) * 3.55K * (4.184J/gK) = 37875.66J
Q(hydrocarbon) = -Q(bomb) -Q(H2O)
Q(hydrocarbon) = -(1430.65J) - (37875.66J) = -39306.31J
1gram hydrocarbon = -39306.31J/1.520g = 25859.41J/g = 25900J/g
Does this seem right?
Chemistry - DrBob222, Monday, October 31, 2011 at 7:58pm
That looks ok to me although I didn't check the math.