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.
First I worked out the moles of methanol and propan-2-ol, which I got 0.0625 and 0.025.
Next temperature changes or either and was 32.8 and 36.4K.
I then used the enthalpy of combustion of methanol to get 44687.5=m*c*Delta H
Not sure if this is correct way, any help would be apprecaited.