The heat of combustion of ethanol is -29.8 kJ/g. What is the heat of cmbustion expressed in Joules per kilogram?
To convert the heat of combustion from kilojoules per gram (kJ/g) to joules per kilogram (J/kg), you can use the following conversion factor:
1 kJ/g = 1000 J/kg
Therefore, to convert the given heat of combustion from kJ/g to J/kg, you can multiply it by the conversion factor:
-29.8 kJ/g * 1000 J/kg = -29,800 J/kg
The heat of combustion of ethanol, when expressed in joules per kilogram, is -29,800 J/kg. Note that the negative sign represents an exothermic process, meaning heat is released during combustion.
To convert the heat of combustion from kJ/g to J/kg, we need to know the density of ethanol. The density of ethanol is approximately 0.789 g/mL or 789 kg/m³.
First, we convert the heat of combustion from kJ/g to J/g:
-29.8 kJ/g * 1000 = -29,800 J/g
Next, we convert grams to kilograms by dividing by 1000:
-29,800 J/g / 1000 = -29.8 J/kg
Since we want the heat of combustion expressed in Joules per kilogram, we divide the value by the density of ethanol:
-29.8 J/kg / 789 kg/m³ = -0.038 J/kg
Therefore, the heat of combustion of ethanol expressed in Joules per kilogram is approximately -0.038 J/kg. Note that the negative sign indicates the energy release during combustion.