just looking at the units, I'd say
20°C * 6.28kJ/°C = 125.8 kJ / 0.250g = 503.2 kJ/g
No idea how many moles that is for your fuel ...
If 0.250 g of fuel increases the temperature of a calorimeter by 20 °C,
and the calorimeter is calibrated at 6.28 kJ/°C, calculate the heat of
combustion of fuel per mole of fuel. Please show full solutions!
1 answer