Asked by Anne
The heat capacity of a bomb calorimeter is 87.5 kJ/K (this value is for the total heat capacity including that of the water jacket around the reaction chamber). If 67.2g of CH4 (g), is combusted under such reaction conditions, what will be the increase in temperature of the calorimeter? Delta Ecombustion for CH4 (g) is -885.4 kJ/mol.
Answers
Answered by
drwls
Calculate the heat energy release:
The number of moles of CH4 you are burning is 67.2/16 = 4.20 mol
Multiply that by 885.4 kJ/mol for the heat release. Then divide that by 87.5 kJ/K for the temperature rise.
Or, you could do it all at once
Delta T = (67.2g*885.4kJ/mol)/(16 g/mol*87.5 kJ/K) = ? K
The number of moles of CH4 you are burning is 67.2/16 = 4.20 mol
Multiply that by 885.4 kJ/mol for the heat release. Then divide that by 87.5 kJ/K for the temperature rise.
Or, you could do it all at once
Delta T = (67.2g*885.4kJ/mol)/(16 g/mol*87.5 kJ/K) = ? K
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