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chemistry aqa a level calorimetry question

A calorimeter was calibrated by burning 2.00 g of methanol (CH3OH) whose enthalpy of combustion is -715 kJ mol-1. The temperature of the calorimeter rose from 19.6°C to 52.4°C. The same calorimeter was used to measure the enthalpy of combustion of propan-2-ol. 1.50 g of propan-2-ol CH3CH(OH)CH3 raised the temperature by from 19.8°C to 56.2°C. Calculate the heat capacity of the calorimeter and then the enthalpy of combustion of propan-2-ol
moles of methanol burned = 2/32 = 0.0625
energy released by methanol = 0.0625 * 715 = 44.7 kJ
temperature change in calorimeter = 32.8ºC
hence heat capacity of calorimeter = 1.36 kJ K-1- why is it this value?

Reply 1

qwert7890

Original post by sierranonis

Are you familiar with the equation Q= mcT?

Q = heat energy
m = mass used
T= temperature

Enthalpy change of combustion means heat energy per mole . Hence you will have to divide by number of moles.

Hence Enthalpy of combustion = (mcT) / (n)
where n = number of moles

You know the Enthalpy value for methanol, the mass burnt, the Temperate, and the number of moles. The only unknown is c which you can now find out.

Does that make more sense now?