A 1.000 gsample of octane (C8H18) is burned in a bomb calorimeter containing 1200 grams of water. After the reaction, the temperature of the water increases from 25.00 oC to 33.20 oC. The heat capacity of the calorimeter (also known as the “calorimeter constant”) is 837 J/oC. The specific heat of water is 4.184 J/g oC. Calculate the heat of absorbed by the calorimeter.
First, we need to calculate the heat absorbed by the water:
q = mCΔT
where q is the heat absorbed by the water, m is the mass of water, C is the specific heat of water, and ΔT is the change in temperature.
q = (1200 g)(4.184 J/g oC)(33.20 oC - 25.00 oC)
q = 35,220.48 J
Next, we need to calculate the total heat absorbed by the system (water + calorimeter):
Q = q + CcalΔT
where Q is the total heat absorbed by the system, Ccal is the heat capacity of the calorimeter, and ΔT is the change in temperature of the system.
Q = 35,220.48 J + (837 J/oC)(33.20 oC - 25.00 oC)
Q = 36,993.16 J
Finally, we need to calculate the heat released by the combustion of octane:
ΔH = Q/n
where ΔH is the heat released per mole of octane, Q is the total heat absorbed by the system, and n is the number of moles of octane.
n = mass/molar mass = 1.000 g/114.23 g/mol = 0.008748 mol
ΔH = 36,993.16 J/0.008748 mol
ΔH = 4,228,236 J/mol
Therefore, the heat released per mole of octane is 4,228,236 J/mol.
To calculate the heat absorbed by the calorimeter, we need to use the formula:
q = m * c * ΔT
where:
q = heat absorbed by the calorimeter
m = mass of water
c = specific heat of water
ΔT = change in temperature
Given values:
m = 1200 g
c = 4.184 J/g oC
ΔT = 33.20 oC - 25.00 oC = 8.20 oC
Let's calculate the heat absorbed by the calorimeter:
q = 1200 g * 4.184 J/g oC * 8.20 oC
q = 39,347.2 J
Therefore, the heat absorbed by the calorimeter is 39,347.2 J.