This one I am stuck....High-purity benzoic acid (C6H5COOH; ΔHrxn for combustion = −3227 kJ/mol) is used as a standard for calibrating bomb calorimeters. A 1.221-g sample burns in a calorimeter...

Question

(heat capacity = 1365 J/°C) that contains exactly 1.370 kg of water. What temperature change is observed?

My steps:

-qreaction = - (1.221g HBZ) X 1 mol HBZ over 1221.2g HBZ X -3227 kJ over 1 mol HBZ X 10³ J over 1 kJ = 3.22647 X 10⁴ J

3.22647 X 10⁴ J = 4.184 J over g ⁰C X (1.370 X 10³ g) X ΔT + 1365 J over ⁰C X ΔT

3.22647 X 10⁴ = 5732 X ΔT) + (1365 X ΔT)

ΔT =

Answer 1

Thank you Mr. J.R.S for the suggestion

The change in temperature should be 4.5 degree Celsius.

1.221 grams of Benzoic acid is 0.00991 moles( divide 1.221 grams with the molar mass of Benzoic acid which is 122.12) which gives 0.00991 moles of Benzoic acid.

Based on the enthalpy of combustion of Benzoic acid which is - 3227 KJ/mole.

3227 KJ of heat is released for one mole of Benzoic acid but the moles of the Benzoic acid are only 0.00991, therefore for 0.00991 moles the energy released would be 31.97 KJ or 31970 Joules.

Now we use the equation heat released = (mass of water X specific heat X change in temperature of water )+ (heat capacity of the calorimeter X change in temperature)

31970 Joules = (1365 Joules/degree Celsius) X Change in temperature + (1370 g X 4.184 J/g-c) x Change in temperature

31970 = (1365 + 5732.08) change in temperature

31970 J = (7097.08 J) X Change in temperature

Change in temperature = 31970/7097.08 =4.5 degree Celsius

Thank you Mr. J.R. S

Dividing both sides with 1365 Joules/Celsius will give you 4.5