1.45 g of a hydrocarbon C20H16 were burned in excess oxygen in a constant-volume calorimeter.?

Question

The heat capacity of the calorimeter was 171.335 J/oC.
The calorimeter was surrounded by 1050. g of water.
As a result the temperature of the water rose from 21.8oC to 25.7 oC.

Calculate DE (kJ/mol) for the combustion of the hydrocarbon based on these readings.

Answer 1

To solve this problem, you must remember that all the energy released from burning the hydrocarbon has to go somewhere: that somewhere is into the calorimeter itself, and the water. Once you determine the amount of energy released into those two places, you can calculate the DE.

Step 1: Working backwards from the temperature of the water, calculate the amount of energy gained by the water:

Heat capacity of water: 4184 J/(kg·K), which means it takes 4184 J of energy to heat 1 kg of water by 1 degree K or C.

Energy (J) = heat capactiy * mass * change in temperature
Energy (J) = 4184 J/(kg·C) * 1.050 kg * 3.9 degrees C
E = 17133.48 J transferred to the water.

Step 2: Calculate the energy used to heat the calorimeter:
In this case, since the water temp rose from 21.8 to 25.7 degrees, the calorimeter also must have started at 21.8 degrees, and ended at 25.7 degrees at the end of the process. Thus:

Energy (J) = 171.335 J/C * 3.9 degrees C
E = 668.2065 J

Step 3: Calculate the total energy released by burning the hydrocarbon:
E = 17133.48 J transferred to the water + 668.2065 J transferred to the calorimeter
E = 17801.6865 J

Step 4: Determine the number of moles you had of C20H16:
C20 = 6 * 20 = 120
H16 = 1 * 16 = 16
1 mol = 136g
1.45g / 136g = 0.01066 mol

Step 5: Using the total energy transferred and the # of moles of C20H16, calculate DE:
DE = kJ/mol
DE = 17801.6865 J/0.01066 mol
DE = 17.8016865 kJ/0.01066 mol
DE = 1669.95 kJ/mol

Step 6: Apply significant numbers.
The lowest # of significant numbers used was 3 (from the temperatures:
DE = 1670 kJ/mol (solution!)